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Physics revision 2015
11
Motion in one and two dimensions
•
apply Newton’s three laws of motion in situations where two or more coplanar
forces act along a straight line and in two dimensions;
2005
2006
2007
2008
4
1, 2
1, 6, 7
1, 2
2009
2010
2011
2012
2013
2014
3
1, 2, 3,
7, 8
3,
4a, b,
5a, b,
c, d.
1a, b
2a, b
1a, b
The following relationships hold for straight line motion.
x -x
vav = 2 1
Δt
a=
v -u
Δv
=
Δt
t
v = u  at
v2 = u2  2ax
x = ut 
x=
1
2
at2
(u + v)
t
2
When given a graph in the exam, look at three things on the graph before even reading the question:
type of graph (x - t, v - t,)
the units on the axis
the limit on each axis.
Graph type
Found from
x-t
v-t
'x' at any 't'
't' at any 'x'
Instantaneous velocity at any point.
vav between any two points
'v' at any 't'
't' at any 'v'
Instantaneous 'a'
Average 'a'
Change in
position
Direct reading
Gradient
Area under graph
a-t
'a' at any 't'
't' at any 'a'
Change in velocity
To convert from km/h to m/s you need to divide by 3.6, this should be on your cheat sheet.
colinhop@bigpond.net.au
Physics revision 2015
12
A car of mass 2000 kg is driven along a straight flat road by an engine that provides a constant driving
force. It accelerates from rest to a speed of 30 m s-1 in 10 seconds. Assume that friction and air
resistance forces remain constant at a total of 500 N.
Example 1: 1984 Question 1
(1 mark)
What is the acceleration of the car?
Example 2: 1984 Question 2
(1 mark)
What is the magnitude of the total driving force acting on the car?
Example 3: 1984 Question 3
(1 mark)
How far does the car travel in the 10 seconds?
Example 4: 1984 Question 4
(1 mark)
How much work has the car engine done in the 10 seconds?
Example 5: 1984 Question 5
(1 mark)
What is the power developed by the engine when the car is travelling at 20 m s-1?
colinhop@bigpond.net.au
Physics revision 2015
13
Newton's Laws of motion
1st Law
Every object remains at rest, or in uniform motion in a straight line (the velocity may be a non-zero
constant), unless acted on by some external force.
A cyclist pedals up a 150 slope at a constant speed as shown. The total mass of rider and bicycle is
100 kg.
Example 6: 2001 Question 1
(2 marks)
What is the magnitude of the net force on the bicycle and rider when travelling up the slope?
Example 7: 2001 Question 3
(3 marks)
Sometime later the rider turns around and rolls at constant speed down the slope.
Calculate the magnitude of the total frictional forces that are acting on the rider and bicycle while
travelling down the slope.
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Physics revision 2015
14
Newton’s 2nd Law
The rate of change of momentum is equal to the resultant force causing the change.
mv - mu
v -u
F=
since a =
 F.t = mv – mu
t
t
Impulse (change in momentum) i.e. I = p2 - p1. is always the area under F-t graph.
When the force is constant, it also equals Ft.
Area under "F - t" graph = Impulse =  momentum = F  t
TOTAL MOMENTUM BEFORE IMPACT EQUALS THE TOTAL MOMENTUM AFTER IMPACT.
 P(total) is constant before, during and after the collision.
Example 8: 2002 Question 7
(2 marks)
A car travelling at 11.0 m s-1 collides head-on with a concrete pillar. The car comes to rest in a
time of 0.10 s. The car comes to rest against the pillar. The mass of the car and occupants is
1.30 tonne. Determine the average force on the car during the impact with the pillar.
Example 9: 2002 Question 8
(3 marks)
Explain how the crumple zone of the car can minimise the extent of injuries experienced by
the occupants of the car. (Assume that the occupants are wearing seatbelts.)
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Physics revision 2015
15
Newton’s 3rd Law
Action and reaction are equal and opposite. The force of A on B is equal and opposite to the force of B
on A.
This is summarised as FA on B = -FB on A
(the minus sign is to give the direction of the force vector).
For the forces to be an action reaction pair, the subjects of the two statements need to be
interchanged.
This concept is often tested using a book on a table. The action reaction pair of forces are: the weight
force of the Earth on the book, and the force of the book on the Earth. Many students get this
incorrect, because they use the normal reaction as the other part of the pair.
Two masses, 10 kg and 20 kg, are placed in contact on a rough surface as shown. A person exerts a
force of 45 N on the 10 kg mass.
The magnitude of the frictional force acting on the 10 kg mass is 10 N and the magnitude of the
frictional force acting on the 20 kg mass is 20 N.
Example 10: 1984 Question 28
(1 mark)
What is the acceleration of the system of two masses?
Example 11: 1984 Question 29
(1 mark)
What is the force exerted by the 20 kg mass on the 10 kg mass while they are in motion?
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Physics revision 2015
16
A bushwalker is stranded while walking. Search and rescue officers drop an emergency
package from a helicopter to the bushwalker. They release the package when the helicopter
is a height (h) above the ground, and directly above the bushwalker. The helicopter is moving
with a velocity of 10 m s-1 at an angle of 30º to the horizontal, as shown below. The package
lands on the ground 3.0 s after its release. Ignore air resistance in your calculations.
Example 12: 2004 Question 1
(3 marks)
What is the value of h?
Example 13: 2004 Question 3
(2 marks)
Which of the graphs below best represents the speed of the package as a function of time?
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Physics revision 2015
17
A model rocket of mass 0.20 kg is launched by means of a spring, as shown below. The spring is
initially compressed by 20 cm, and the rocket leaves the spring as it reaches its natural length.
The force-compression characteristic of the spring is shown below.
When the rocket reaches its maximum height, the parachute opens and the system begins to fall.
In the following questions you should still ignore the effects of air resistance on the rocket, but of
course it is critical to the force on the parachute. This retarding force due to the parachute is
shown below as R, and its variation as a function of time after the parachute opened is also shown
below
Example 14: 2005 Question 4
(3 marks)
What is the acceleration of the rocket at a time 5 s after the parachute opens?
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Physics revision 2015
18
Connected Bodies
Problems involving the motion of two bodies connected by strings are solved on the following
assumptions; the string is assumed light and inextensible so its weight can be neglected and there is
no change in length as the tension varies. In these cases, tension is the condition of a body subjected
to equal but opposite forces which attempt to increase its length, and tension forces are pulls exerted
by a string on the bodies to which it is attached.
To solve these problems you need to consider the vertical
direction first.
m1g – T = m1a
The direction of this acceleration must be downwards.
This leads to:
T = m1g – m1a
The tension in the string is the same in both directions, therefore
T = m2a.
Since both bodies are connected by an inextensible string, both
bodies must have the same acceleration.
The vertical forces acting on m2, (not shown) cancel each other out, and do not impact on its motion.
m1
Combing these two equations gives a =
g
m1 + m2
T=
m1m2
m1 + m2
g
A fireproof door of mass 40 kg closes across an opening 1.0 m wide, in an emergency. Heat from the
fire causes the door to be released and it’s then moved by a falling mass of 10 kg as shown in the
diagram. Ignore friction.
Example 15: (1987 Question 13)
(1 mark)
What is the acceleration of the door when it is released?
colinhop@bigpond.net.au
Physics revision 2015
Example 16: (1987 Question 14)
(1 mark)
Calculate the time (in seconds) it takes the door to fully close after it is released.
A tugboat is towing a ship with a tow rope as shown below.
The tugboat exerts a constant force of 9.0 × 104 N on the tow rope.
The water resistance on the ship as a function of speed is shown below.
Example 17: 2008 Question 1
(2 marks)
What is the acceleration of the ship when the tugboat and ship are travelling at 2.0 m s–1?
You must show your working.
19
colinhop@bigpond.net.au
Physics revision 2015
Example 18: 2008 Question 2
20
(2 marks)
After a time, the tugboat and ship are travelling at a constant speed.
What is this constant speed?
Two physics students are conducting an experiment in which a block, m1, of mass 0.40 kg is being
pulled by a string across a frictionless surface. The string is attached over a frictionless pulley to
another mass, m2 of 0.10 kg. The second mass, m2, is free to fall vertically. This is shown below.
The block is released from rest.
Example 19: 2010 Question 3
What is the acceleration of the block m1?
(2 marks)
colinhop@bigpond.net.au
Physics revision 2015
21
A tractor, including the driver, has a mass of 500 kg and is towing a trailer of mass 2000 kg as
shown above. The tractor and the trailer are accelerating at 0.50 m s-2. Ignore any retarding
friction forces. Ignore the mass of the towing rope. The tractor and the trailer start from rest.
Example 20: 2011 Question 2
(2 marks)
What is the tension in the rope connecting the tractor and trailer?
Three children’s toy blocks A (0.050 kg), B (0.10 kg) and C (0.20 kg), are sitting on a
table as shown in the diagram below.
Example 21: 2011 Question 7
(2 marks)
What is the force by block C on Block B? Explain your answer in terms of Newton’s laws.
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Physics revision 2015
Example 22: 2011 Question 8
22
(2 marks)
The table is now removed and the blocks free fall.
What is now the force by block B on Block C? Ignore air resistance. Explain your answer.
Two metal spheres hang from the ceiling as shown below. Cable A runs between the ceiling and
the upper sphere of mass 2.0 kg. Cable B runs between the 2.0 kg sphere and the 1.0 kg sphere.
Assume that the cables have no mass.
Example 23: 2012 Question 4b
(2 marks)
Newton’s third law is sometimes stated as ‘To every action there is an equal and opposite
reaction’. If the weight (the gravitational force by Earth) of the 2.0 kg sphere is taken as the ‘action’
force, identify the corresponding ‘reaction’ force and give its direction.
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Physics revision 2015
23
A truck is dragging two logs along level ground in a straight line. The mass of each log is 600 kg
and each log experiences a constant retarding friction force of 400 N with the ground.
The connections between the truck and the logs are made with ropes that have a breaking force of
2400 N. T1 and T2 are the tensions in the ropes as shown below. The truck and the logs are moving
towards the left.
Example 24: 2012 Question 5d
(3 marks)
The ropes have a breaking force of 2400 N.
Rope 1 connects the truck to the front log and rope 2 connects the two logs.
The truck, still on level ground, increases its acceleration until one of the ropes is about to break.
Identify which rope is about to break, and calculate the magnitude of the acceleration of the truck
and the logs at this instant.
Rope --------------
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Physics revision 2015
24
Students set up an inclined plane surface, as shown below. It is angled at 10° to the horizontal.
They release the block of wood, of mass 0.50 kg at a distance of 3.5 m from the stopper. When
the block is sliding down the incline, there is a constant frictional force between it and the surface.
They find that it takes 6.0 s to reach the stopper at the bottom.
Example 25: 2013 Question 1b (3 marks)
Calculate the magnitude of the frictional force of the plane surface acting on the block.
colinhop@bigpond.net.au
Physics revision 2015
25
Students set up an experiment that consists of two masses, m1, of 2.0 kg, and m2, of 6.0 kg,
connected by a string, as shown below. The mass of the string can be ignored. The surface is
frictionless. The pulley is frictionless.
At the start of the experiment, the bottom of mass m1 is 1.2 m above the floor and both masses
are stationary.
Example 26: 2013 Question 2b (2 marks)
Calculate the tension in the string as m1 is falling.
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Physics revision 2015
26
A small locomotive is used in a railway yard to arrange rail trucks on trains. The locomotive has a
mass of 40 tonnes (40 000 kg).
In one situation, the locomotive is pulling two trucks, each of mass 10 tonnes, as shown below.
Example 27: 2014 Question 1b
(2 marks)
Calculate the tension in the coupling between the locomotive and Truck 1 as they accelerate.
colinhop@bigpond.net.au
•
•
Physics revision 2015
27
analyse the uniform circular motion of an object moving in a horizontal plane
(Fnet = mv2/R) such as a vehicle moving around a circular road; a vehicle moving
around a banked track; an object on the end of a string;
apply Newton’s second law to circular motion in a vertical plane; consider forces at
the highest and lowest positions only;
2005
5, 6, 7
2006
4, 5
2007
2008
3, 4
2009
3, 4
12
2010
1, 2, 5,
6, 9
2011
4, 5, 6
2012
7a, b, c
2013
4a, b
5a, b
2014
4c
Uniform circular motion results when a resultant force of constant
magnitude acts normal to the motion of the body.
When a body travels in a circle, the force is always towards the centre,
while the velocity is tangential. The force is always at right angles to
the velocity. The force is often called a tension (particularly when
using a string)
A mass moving with a uniform speed in a circular path of radius ‘r’ and
with a period ‘T’ has an average speed given by
distance travelled
2πr
=
time taken
T
 F = ma
=
mv 2
r
=m
a=
=
4π 2r
T2
v2
r
4π 2r
T2
In circular motion the movement is always at right angles to the force, since work done = force ×
distance moved in the direction of the force. The work done in circular motion is zero.
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Physics revision 2015
28
A car is travelling in a horizontal path around a circular curve. The car's speed is constant and it is
travelling from left to right in the diagrams below.
Example 28: 1992 Trial
(2 marks)
Which of the following diagrams best shows the horizontal forces (shown as solid lines) and their
resultant force (shown as a dashed line)? Give reasons for your choice.
The safe speed for a train taking a curve on level ground is determined by the force that the
rails can take before they move sideways relative to the ground. From time to time trains
derail because they take curves at speeds greater than
that recommended for safe travel.
The figure below shows a train at position P taking a
curve on horizontal ground, at a constant speed, in the
direction shown by the arrow.
Example 29: 2005 Question 7
(2 marks)
At point Q the driver applies the brakes to slow down
the train on the curve. Which of the arrows (A - D)
indicates the direction of the net force exerted on the
wheels by the rails?
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Physics revision 2015
29
Banked surfaces
Normal
Surfaces (eg. roads, velodromes) are banked to
reduce the need for frictional forces to assist in circular
motion. The forces acting are shown below.
Friction

For the object to travel in circular motion the net force
acting needs to be radially inwards.
If we want the frictional force to be zero, then the
diagrams becomes:
The normal force is resolved into perpendicular and
horizontal components.
Here Ncos = mg (because the object is not accelerating
in the vertical direction) and the unbalanced force is Nsin
mv 2
 Nsin =
dividing this equation by Ncos = mg
R
gives
Nsin mv 2
1


Ncos
R
mg
tan =
mg
Nsin
Normal
Ncos

mg
v2
this formula is very convenient and should be on your own formula sheet.
Rg
In designing a bicycle track at a racing track, the designer wants to bank the track on a
particular corner so that the bicycles will go around the corner with no sideways frictional
force required between the tyres and the track at 10 m s-1.
Figures 2a and 2b below show the banked track and the bicycle.
Example 30: 2010 Question 5
(2marks)
On Figure 2b draw two arrows to show the two forces acting on the bicycle and rider (treated
as a single object).
colinhop@bigpond.net.au
Physics revision 2015
30
A demonstration at a show involved a bike being ridden around a circular banked track.
The horizontal path that the bike takes is a circle of radius 20 m, and the bike travels at a
constant speed of 15 m s-1. The bike and the rider have a total mass of 300 kg. Ignore
retarding friction.
Example 31: 2011 Question 5
(2 marks)
The diagram above shows the only two forces on the bike and the rider. On this diagram
draw an arrow to show the direction of the net force on the bike and rider. Explain how the
two forces shown cause this net effect.
Vertical circles.
You feel your ‘weight’ as the normal reaction of the surface on you, because you can only feel things
that act on you. So if the normal reaction increases you ‘feel’ heavier, and if the normal reaction
decreases you ‘feel’ lighter.
Consider an object travelling around a vertical loop
mg
At the top
 F = N + mg =
mv
r
N
2
F
N
At the bottom
 F = N - mg =
mv 2
r
F
mg
colinhop@bigpond.net.au
Physics revision 2015
31
Person on a swing
At the bottom of the swing, the forces on the person are the reaction from the seat and the weight
force.
mv 2
R - mg =
. In this case R > mg, so the person ‘feels’ heavier.
r
Person in a car going over a hump.
N
v
mg
mg – R =
mv 2
r
R = mg -
mv 2
r
The person ‘feels’ lighter.
An object is rotated in a vertical circle of radius r at constant
speed of v. The centre of the circle is at height 2r above a
horizontal surface. The acceleration due to gravity is ‘g'.
Example 32: 1977 Question 15
(1 mark)
If the object is released at point P, how far will it land from
point X?
Example 33: 1977 Question 16
(1 mark)
If instead, the object were released from point Q, how far would it land from point X?
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Physics revision 2015
32
A fairground amusement machine consists of cages turned at a constant speed in a vertical circle. The
cage just clears the ground at the bottom of its path. The cage pivots during the motion, so that an
observer in the cage always remains in a vertical standing position.
A
Cage
Mass of observer
=m
Radius of rotation of cage
=r
Speed of rotation
=v
Acceleration due to gravity
=g
B
Use either A or B to answer the following question. If neither position is appropriate, write N.
Example 34: 1978 Question 18
(1 mark)
At which position(s) will the observer exert a force on the cage floor greater than his/her normal
weight? (One or more answers)
Example 35: 1978 Question 19
(1 mark)
By how much will the force on the cage floor by the observer differ between position A and B?
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Physics revision 2015
33
Kim has attached a ball of mass 0.20 kg to a piece of string of length 1.8 m and is making it move
in a horizontal circle on a frictionless surface. Kim gradually increases the speed, v, of the ball.
The situation is shown from above in the diagram below. The string has a breaking force of 4.0 N.
Example 36: 2012 Question 7c
(3 marks)
On another occasion, Kim swings the same ball of mass of 0.20 kg in a vertical circle at constant
speed. She uses a much stronger string, which does not break.
She notices that the tension in the string is greater at the bottom of the circle than it is at the top of
the circle.
Explain why the tension at the bottom is greater than the tension at the top. You may include a
diagram as part of your explanation.
colinhop@bigpond.net.au
Physics revision 2015
34
A mass of 2.0 kg is being swung by a light rod in a vertical circle of radius 1.0 m at a constant
speed of 7.0 m s-1, as shown below.
Example 37: 2013 Question 5b
(3 marks)
Calculate the magnitude of the tension in the light rod at point S. Show the steps of your
working.
colinhop@bigpond.net.au
•
Physics revision 2015
35
investigate and analyse the motion of projectiles near Earth’s surface including a
qualitative description of the effects of air resistance;
2005
11, 12
2006
6, 7, 8
2007
14,15,
16, 17
2008
5, 6, 7
2009
10, 11
2010
10, 11
2011
12, 13
2012
6a, b
2013
8a, b
2014
3a, b
Projectile motion is motion under a constant unbalanced force. A projectile is a body that has been
thrown or projected. Air resistance is to be considered as negligible in the quantitative questions
involving calculations, but may need to be included in qualitative questions.
For both the vertical and inclined projectiles:
 the only force acting is the weight, ie. the bodies are in free fall, acceleration is always 10 m s-2,
including at the top
 instantaneous velocity is tangential to the path
 the total energy (KE & PE) is constant, between any two points KE = -PE
 paths are symmetrical for time, if air resistance can be ignored
Inclined or oblique projections
 the only force acting is vertically down, so the acceleration and change in velocity are vertical
 horizontally there is no component of force, so constant horizontal velocity
 maximum range is when angle of projection is 45O
Horizontal projection
For projectiles thrown horizontally and dropped from rest, the vertical motions are the same.
Horizontal:
velocity always = vhorizontal
acceleration = 0
displacement = x = vhorizontal × t
Vertical:
Velocity changing v = u - gt
acceleration
= -g
displacement
y = ut - ½ gt2
To find the 'total' velocity, add vvertical and vhorizontal using vectors.
Symmetrical flights
If there is no air resistance, and the projectile starts and ends at the same height, then the range is
given by:
v 2 sin2θ
R=
R is the range, v is the initial speed and  the angle of projection.
g
Total Energy (TE)
If air resistance is negligible, then the total energy is constant. TE = KE + PE. At ground level PE = 0,
so TE = KE. As the projectile rises it gains PE, so it must lose KE. At the top of its flight, the PE is
maximum and the KE is minimum. (KE is not zero, because the projectile still has some KE due to its
horizontal motion).
At any point on the way up or the way down, the TE is constant. If you know the horizontal component
of the velocity, then you can use this to find the maximum height, by using the TE at ground level and
working out what he PE must be at the top when vvertical = 0, but vhorizontal = constant.
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Physics revision 2015
36
Projectile problem methodology
1. Draw a fully labelled diagram
2. Treat the motion as two separate motions, horizontal and vertical. List the data under vertical and
horizontal
3. For vertical motion use, v2 =u2 + 2ax, and v = u + at. (use a = -g)
4. To go from one direction to another, the common link is the time of flight, t.
5. Remember that it will take the same time to go up as to come down (if air resistance can be
ignored).
6. Label the direction (+ or -) for all variables except time.
7. Sometimes the question is phrased 'with a strong tail wind.' This means that the tail wind cancels
out any effects of air resistance.
A cannon mounted in a trench with its
muzzle at ground level fires a shell. X is a
specific point on the trajectory. Air
resistance is negligible.
mass of shell = m
kinetic energy at Point X = K
total energy at X = T
momentum at X = p
potential energy at X (relative to ground level as arbitrary zero) = U
Example 38: 1971 Question 4
Write an expression for K in ter
ms
Example 39: 1971 Question 5
(1 mark)
of p and m only.
(1 mark)
The initial kinetic energy of the shell is
A. T
B.
T+U
C.
T-U
D. T + K
E.
T–K
F.
not determinable from the
information given.
Example 40: 1971 Question 6
(1 mark)
The velocity of the shell at its maximum height is
A. Zero
B.
2T
m
C.
p
m
2K
m
E.
T–K
F.
dependent on the angle of
D.
elevation of the cannon.
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Physics revision 2015
37
In the movie, Car Escape, Taylor and Jones drove their sports car across a horizontal car park in
building 1 and landed it in the car park of building 2, landing one floor lower.
Building 2 is 20 metres from building 1, as shown.
The floor where the car lands in building 2 is 4.0 m below the floor from which it started in building 1.
In the questions below, treat the car as a point particle and assume air resistance is negligible.
Example 41: 2002 Question 5
(3 marks)
Calculate the minimum speed at which the car should leave building 1 in order to land in the car
park of building 2.
Example 42: 2002 Question 6
(2 marks)
In order to be sure of landing in the car park of building 2, Taylor and Jones in fact left building 1
at a speed of 25 m s-1. Calculate the magnitude of the velocity of the car just prior to landing in
the car park of building 2.
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Physics revision 2015
38
A rocket of mass 0.50 kg is set on the ground, pointing vertically up, as shown below. When
ignited, the gunpowder burns for a period of 1.5 s, and provides a constant force of 22 N. The
mass of the gunpowder is very small compared to the mass of the rocket, and can be ignored.
The effects of air resistance can also be ignored.
Example 43: 2006 Question 6
(2 marks)
What is the magnitude of the resultant force on the rocket?
Example 44: 2006 Question 7
(2 marks)
After 1.5 s, what is the height of the rocket above the ground?
A second identical rocket, that again provides a constant force of 22 N for 1.5 s, is now launched
horizontally from the top of a 50 m tall building. Assume that in its subsequent motion the rocket
always points horizontally.
Example 45: 2006 Question 8
(4 marks)
After 1.5 s, what is the speed of the rocket, and at what angle is the rocket moving relative to
the ground?
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Physics revision 2015
39
Daniel and John are playing paintball. Daniel fires a ‘paintball’ at an angle of 25° to the
horizontal and a speed of 40.0 m s-1. The paintball hits John, who is 127 m away.
The height at which the ball hits John and the height from which the ball was fired are the
same. The situation is shown below.
The acceleration due to gravity should be taken as 10 m s-2, and air resistance should be
ignored.
Later in the game, Daniel is twice as far away from John (254 m). John fires an identical
paintball from the same height above the ground as before.
The ball hits Daniel at the same height as before.
In both cases the paintball reaches the same maximum height (h) above the ground. (Note:
this is not as shown in the diagram).
Example 46: 2007 Question 17
(2 marks)
Which one or more of the following is the same in both cases?
A. flight time
B. initial speed
C. acceleration
D. angle of firing
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Physics revision 2015
40
A batsman hits a cricket ball (from ground level) at a speed of 30.0 m s–1 and at an angle of
36.9° to the horizontal.
An advertising board is now placed on the boundary of the cricket ground at a distance 72.0
m from the batsman as shown below. Air resistance can be ignored.
Example 47: 2008 Question 7
(3 marks)
Assuming the ball is hit exactly the same way, at what height above the ground will the ball
strike the advertising board? You must show your working.
colinhop@bigpond.net.au
Physics revision 2015
41
Max hits a ball from the edge of a cliff. The ball has an initial speed of 60 m s–1 at an angle of
30° to the horizontal as shown. Ignore the effects of air resistance.
Example 48: 2009 Question 11
(3 marks)
The ball takes 9.0 s from the time Max hits it until it lands in the water. What is the height, h,
of the cliff?
colinhop@bigpond.net.au
Physics revision 2015
42
A stone is thrown from the top of a 15 m high cliff above the sea at an angle of 30° to the
horizontal. It has an initial speed of 20 m s–1. The situation is shown below. Ignore air
resistance.
Example 49: 2013 Question 8b
(3 marks)
Calculate the magnitude and direction of the velocity of the stone immediately before it reaches
the sea. Give the direction as the magnitude of the angle between the velocity and the
horizontal.
colinhop@bigpond.net.au


Physics revision 2015
43
apply laws of energy and momentum conservation in isolated systems;
analyse impulse (momentum transfer) in an isolated system, for collisions between
objects moving in a straight line (FΔt = mΔv);
2005
8, 9 10
2006
10, 11
2007
3, 4, 5
2008
8, 9,
10, 11
2009
1
2010
12
2011
14, 15
2012
2
2013
3a, b,
c
2014
1c
Momentum is conserved in all collisions. If a collision is elastic, kinetic energy is also conserved. If KE
is lost, the collision is inelastic. The energy that has gone 'missing' is usually converted into heat,
sound or energy of deformation.
Consider a body of mass 'm' changing its velocity from 'u' to 'v' in time 't' under the action of a constant
force 'F'.
(v - u)
F = ma
since a =
t
mv - mu
F=
t
mΔv
F=
t
Ft = mv
Since a net force gives a change in momentum over a change in time, it can be written that a change
in momentum is given by:
p = Ft
The product of a constant force and the time for which it acts is called the IMPULSE (I) of the force.
I = p = F Δt . The unit is Newton second, (but this is identical to kg m s-1).
 I = p2 - p1.
The impulse of a force can be measured by the change in momentum. Impulse and momentum are
vector quantities. Both impulse and momentum have the same units.
Note: NO collision can result in an INCREASE in the TOTAL KINETIC ENERGY of a system.
Momentum transfer involving the Earth
1. Body rises under gravity - slows down and loses momentum to the earth.
2. Body falling under gravity - speeds up giving the earth equal and opposite momentum change.
3. Falling body hits the ground - its p is transferred to the earth.
4. Body slowed due to friction - gives the earth and equal and opposite p.
5. Body accelerated due to friction - gives the earth an equal and opposite p.
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Physics revision 2015
44
A moving railway truck (X) of mass 10 tonnes, moving at 6.0 m s-1, collides with a stationary railway
truck (Y) of mass 5.0 tonnes. After the collision they are joined together and move off as one. This
situation is shown below.
Example 50: 2002 Question 9
(2 marks)
Calculate the final speed of the joined railway trucks after the collision.
Example 51: 2002 Question 10
(2 marks)
Calculate the magnitude of the total impulse that truck Y exerts on truck X during the collision.
Example 52: 2002 Question 11
(3 marks)
Explain why this is an example of an inelastic collision. Calculate numerical values to justify your
answer.
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Physics revision 2015
45
A small truck of mass 3.0 tonne collides with a stationary car of mass 1.0 tonne. They are
locked together as they move off.
The speed immediately after the collision was known to be 7.0 m s-1 from the jammed
reading on the car speedometer.
Robin, one of the police investigating the crash, uses ‘conservation of momentum’ to
estimate the speed of the truck before the collision.
Example 53: 2005 Question 8 (3 marks)
What value did Robin obtain?
Example 54: 2005 Question 9 (2 marks)
The calculated value is questioned by the other investigator, Chris, who believes that
‘conservation of momentum’ only applies in elastic collisions.
Explain why Chris’s comment is wrong.
colinhop@bigpond.net.au
Physics revision 2015
46
The figure below shows a space shuttle docking with the International Space Station.
Imagine that you are an astronaut floating in space at rest relative to the International
Space Station. You watch the space shuttle, of mass 6000 kg, dock. You observe the
shuttle approaching the space station with a speed of 5.00 m s-1.
After docking, the space station's speed has increased by 0.098 m s-1.
Example 55: 2006 Question 10 (3 marks)
Show that the mass of the space station is 3 x 105 kg.
Example 56: 2006 Question 11
(3 marks)
After first making contact, it takes 20 s for the shuttle to come to rest with the space station.
Calculate the average force exerted on the shuttle by the space station.
colinhop@bigpond.net.au
Physics revision 2015
47
Meredith and Hilary are studying collisions by sliding blocks together on a frictionless table.
Meredith slides a block of mass 2 kg with a speed of 3 m s-1 that collides with a block of mass
1 kg, which was at rest. After the collision the 1 kg block has a speed of 4 m s-1.
The situations before and after are shown below.
Example 57: 2007 Question 3
(2 marks)
Show that, after the impact, the velocity of the 2 kg block is 1 m s-1.
Example 58: 2007 Question 5
(2 marks)
What average force does the 2 kg block exert on the 1 kg block during the contact time of
0.01 s?
colinhop@bigpond.net.au
Physics revision 2015
48
A locomotive, of mass 20 × 103 kg, moving at 8.0 m s–1 east, collides with and couples to
three trucks, each of mass 20 × 103 kg, initially stationary, as shown.
Example 59: 2008 Question 8 (2 marks)
What is the speed of the coupled locomotive and trucks after the collision?
You must show your working.
Example 60: 2008 Question 9 (3 marks)
What is the impulse given to the locomotive by the trucks in the collision (magnitude and
direction)? You must show your working.
colinhop@bigpond.net.au
Physics revision 2015
49
A 1.2 kg block moves to the right along the frictionless surface and collides elastically with a
stationary block of mass 2.4 kg as shown below.
After the elastic collision, the 1.2 kg block moves to the left as shown.
Example 61: 2012 Question 2 (3 marks, 30%)
After the collision, the momentum of the 2.4 kg block is greater than the momentum that
the 1.2 kg block had before the collision.
Explain why the greater momentum of the 2.4 kg block is consistent with the principle of
conservation of momentum.
colinhop@bigpond.net.au
•
Physics revision 2015
50
apply the concept of work done by a constant force
– work done = constant force × distance moved in direction of net force
– work done = area under force - distance graph;
2005
2006
3
2007
8
2008
2009
2010
2011
2012
2013
2014
A fundamental principle of nature is that energy cannot be created or destroyed, only transformed or
transferred to another body.
A body that has energy may transfer some, or all, of its energy to another body. The total amount of
energy remains constant (conserved), even if it has been transformed to another type. The amount of
energy transformed is called work. The body losing energy does work, the body gaining energy has
work done on it. Work = Force  displacement, it is a scalar quantity, with the units of Joule.
The area under a force-displacement graph shows the work done. If the force is constant then the
area under the graph is given by W = F  d, where F is the force, d is the distance over which the
force acts. This assumes that the force and the displacement are in the same direction. If they aren't
then the work is the product of the resolved part of the force (in the direction of motion)  the
displacement.
If the force isn't constant, then the work done is equal to the area under the F-d graph.
W
Fd
Power is the rate of doing work. P =
=
= Fv. Power (scalar) with units of joule sec-1 or Watt.
Δt
Δt
A seaplane of mass 2200 kg
takes off from a smooth lake as
shown below. It starts from rest,
and is driven by a constant force
generated by the propeller. After
travelling a distance of 500 m,
the seaplane is travelling at a
constant speed, and then it lifts off after travelling a further 100 m.
The total force opposing the motion of the seaplane is not constant. The figure below shows the total
force opposing the motion of the seaplane as
a function of the distance travelled.
Example 62: 1999 Question 5
(4 marks)
Estimate the work done by the seaplane against
the opposing forces in travelling for a distance of
500 m from the start.
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Physics revision 2015
51
A cyclist is towing a small trailer along a level bike track show. The cyclist and bike have a
mass of 90 kg, and the trailer has a mass of 40 kg. There are opposing constant forces of
190 N on the rider and bike and 70 N on the trailer. These opposing forces do not depend on
the speed of the bike.
The bike and trailer are initially travelling at a constant speed of 6.0 m s-1. The cyclist stops
pedalling, and the bike and trailer slow down
Example 63: 2006 Question 3
(3 marks)
How far will the bike and trailer travel before they come to rest?
colinhop@bigpond.net.au
•
Physics revision 2015
52
analyse transformations of energy between: kinetic energy; strain potential energy;
gravitational potential energy; and energy dissipated to the environment
considered as a combination of heat, sound and deformation of material
–kinetic energy, i.e. ½mv2; elastic and inelastic collisions in terms of conservation
of kinetic energy
–strain potential energy, i.e. area under force-distance graph including ideal
springs obeying Hooke’s Law, ½k(Δx)2
–gravitational potential energy, i.e mgΔh or from area under force-distance graph
and area under field-distance graph multiplied by mass;
2005
1, 2, 3
2006
2007
9, 10,
11
2008
12, 13,
14
2009
2, 5, 6
2010
4, 7,
13, 14,
15, 16,
17
2011
16, 17,
18,19,
20
2012
1a, b,
c, d
2013
6a, b, c
2014
1d,
2a, b,
c, d
Kinetic Energy (KE), the energy of a body due to its motion.
KE = ½mv2
WD = KE = 21 mv2 - 21 mu2
Gravitational Potential Energy. When changes in height 'h' are small compared to the radius of the
Earth, the potential energy Ug of a body near the earth's surface is given by
Ug = mgh.
Elastic potential energy (strain energy) is the energy stored in any material that has been stretched
or compressed from its normal shape. Springs and elastic bands are good examples. For a spring not
stretched beyond its elastic limit, the force, F, applied is proportional to the extension, Δx, produced.
F = kΔx
where k is called the spring constant.
(the gradient of the F - Δx graph)
This is known as Hooke's Law. The spring constant (force constant), k, gives a measure of the
stiffness of the material.
The energy stored in the spring is = 21 base  height
=
1
2
Δx.kΔx
=
1
2
k(Δx)2 (the strain energy).
The energy stored in the spring is also the work done, which is the area under the graph.
colinhop@bigpond.net.au
Physics revision 2015
53
In a storeroom a small box of mass 30.0 kg is loaded onto a slide from the second floor,
and slides from rest to the ground floor below, as shown below. The slide has a linear
length of 6.0 m, and is designed to provide a constant friction force of 50 N on the
box. The box reaches the end of the slide with a speed of 8.0 m s-1.
Example 64: 2004 Pilot Question 14
(4 marks)
What is the height, h, between the floors?
Example 65: 2004 Pilot Question 15
(3 marks)
The box then slides along the frictionless floor, and is momentarily stopped by a spring of
stiffness 30 000 N m-1. How far has the spring compressed when the box has come to rest?
colinhop@bigpond.net.au
Physics revision 2015
54
In a laboratory class at school, Lee is given a spring with a stiffness of 20 N m-1 and
unstretched length of 0.40 m. He hangs it vertically, and attaches a mass to it, so that the new
length of the spring is 0.60 m.
Example 66: 2007 Question 9
(3 marks)
Assuming the spring has no mass, what was the value of the mass he attached?
Example 67: 2007 Question 10 (3 marks)
Lee pulls the mass down a further distance of 0.10 m.
By how much has the potential energy stored in the spring changed?
He now releases the mass, so that the mass-spring system oscillates. Ignore air resistance.
Example 68: 2007 Question 11 (2 marks)
Which one of the curves (A – D) below could best represent the variation of the total energy
of the oscillating mass-spring system as a function of position?
colinhop@bigpond.net.au
Physics revision 2015
55
A novelty toy consists of a metal ball of mass 0.20 kg hanging from a spring of spring
constant k = 10 N m–1.
The spring is attached to the ceiling of a room as shown. Ignore the mass of the spring.
Without the ball attached, the spring has an unstretched length of 40 cm. When the ball is
attached, but not oscillating, the spring stretches to 60 cm.
Example 69: 2008 Question 12
(2 marks)
How much energy is stored in the spring when the ball is hanging stationary on it?
You must show your working.
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Physics revision 2015
56
The ball is now pulled down a further 5 cm and released so that it oscillates vertically over
a range of approximately 10 cm. Gravitational potential energy is measured from the level
at which the ball is released. Ignore air resistance.
Use Graphs A – E in answering the next two questions.
Example 70: 2008 Question 13
(2 marks)
Which of the graphs best represents the shape of the graph of kinetic energy of the
system as a function of height?
Example 71: 2008 Question 14
(2 marks)
Which of the graphs best represents the gravitational potential energy of the system as a
function of height?
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Physics revision 2015
57
A block of mass 0.20 kg is held at point A against a spring which has been compressed by
10 cm as shown below. The block is released, and is pushed by the spring across a
smooth surface. When the block leaves contact with the spring at point B the block has a
speed of 5.0 m s–1.
Example 72: 2009 Question 5
(2 marks)
What is the spring constant, k, of the spring?
The first figure shows an ideal spring with a 2.0 kg mass attached. The spring-mass system
is held so that the spring is not extended. The mass is gently lowered and the spring
stretches until, the spring-mass system is at rest. The spring has extended by 0.40 m.
Example 73: 2010 Question 14
(2 marks)
What is the difference in the magnitude of the total energy of the spring-mass system
between the two figures? Show your working.
colinhop@bigpond.net.au
Physics revision 2015
58
Physics students are conducting a collision experiment using two trolleys, m1 of mass 0.40 kg
and m2 of mass 0.20 kg.
• Trolley m1 has a light spring attached to it. When uncompressed, this spring has a length
0.20 m.
• Trolley m1 is initially moving to the right. Trolley m2 is stationary.
• The trolleys collide, compressing the spring to a length of 0.10 m.
• The trolleys then move apart again, and the spring reverts to its original length (0.20 m), and
both trolleys move off to the right.
• The collision is elastic.
• The trolleys do not experience any frictional forces.
Example 74: 2010 Question 15
(2 marks)
Which graph best shows how the total kinetic energy of the system varies with time
before, during and after the collision? Explain your answer.
Example 75: 2010 Question 17
(2 marks)
If the collision had been inelastic, which graph would best show how the magnitude of the
total momentum of the system varies with time before, during and after the collision?
Explain your answer.
colinhop@bigpond.net.au
Physics revision 2015
59
Physics students are conducting an experiment on a spring that is suspended from the
ceiling. Ignore the mass of the spring.
Without the mass attached, the spring has an unstretched length of 40 cm. A mass of 1.0
kg is then attached. When the 1.0 kg mass is attached, with spring and mass stationary,
the spring has a length of 70 cm.
Example 76: 2011 Question 16
(2 marks)
What is the spring constant, k, of the spring?
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Physics revision 2015
60
The spring is now pulled down a further 10 cm from 70 cm to 80 cm and released so that
it oscillates. Gravitational potential energy is measured from the point at which the spring
is released (80 cm on the diagram above).
Use the following graphs (A – H) to answer the following questions.
Example 77: 2011 Question 17
(2 marks)
Which of the following graphs (A – H) best shows the variation of the kinetic energy of the
system of spring and mass plotted against the length of the stretched spring?
Example 78: 2011 Question 19
(2 marks)
Which of the following graphs (A – H) best shows the variation of the gravitational
potential energy of the system of spring and mass (measured from the lowest point as
zero energy) plotted against the length of the stretched spring?
colinhop@bigpond.net.au
Physics revision 2015
Example 79: 2011 Question 20
61
(3 marks)
Which of the following graphs (A – H) best shows the variation of the spring (strain)
potential energy plotted against the length of the stretched spring? Give reasons for
choosing this answer for the spring (stain) potential energy.
A spring is resting against a wall. The spring is compressed by a distance of 8.0 cm from its
uncompressed length. Jemima holds a block of mass 1.2 kg stationary against the compressed
spring as shown below.
Jemima releases the block and it slides to the right on a frictionless surface. It leaves the spring
with a kinetic energy of 5.4 J and slides at constant speed as shown below.
Example 80: 2012 Question 1c
(2 marks)
Calculate the spring constant, k, of the spring. Assume that the spring obeys Hooke’s law.
colinhop@bigpond.net.au
Physics revision 2015
62
Students hang a mass of 1.0 kg from a spring that obeys Hooke’s law with k = 10 N m–1.
The spring has an unstretched length of 2.0 m. The mass then hangs stationary at a distance
of 1.0 m below the unstretched position (X) of the spring, at Y, as shown at position 6b in
Figure 6. The mass is then pulled a further 1.0 m below this position and released so that it
oscillates, as shown in position 6c.
The zero of gravitational potential energy is taken to be the bottom point (Z).
The spring potential energy and gravitational potential energy are plotted on a graph, as
shown in Figure 7.
Example 81: 2013 Question 6a
(1 mark)
Calculate the total energy of the system when the mass is at its lowest point (Z).
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Physics revision 2015
Example 82: 2013 Question 6b
63
(2 marks)
From the data in the graph, calculate the speed of the mass at its midpoint (Y).
Without making any other changes, the students now pull the mass down to point P, 0.50 m
below Y. They release the mass and it oscillates about Y, as shown in Figure 8.
The students now take the zero of gravitational potential energy to be at P and the zero of spring
potential energy to be at Q. They expect the total energy at P to be equal to the total energy at Q.
They prepare the following table.
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Physics revision 2015
64
However, their calculation of the total energy (GPE + SPE + KE) at Q (10 J) is different from
their calculation of the total energy at P (5.0 J).
Example 83: 2013 Question 6c
(3 marks)
Explain the mistake that the students have made.
Jo and Sam are conducting an experiment using a mass attached to a spring. The spring has
an unstretched length of 40 cm. The situation is shown below.
They begin their experiment by measuring the spring constant of the spring by progressively
adding 50 g masses to it, as shown in Figure 3b. They measure the resultant length of the
spring with the mass stationary and record the following data.
Example 84: 2014 Question 2a
(2 marks)
Show that the spring constant is equal to 5.0 N m–1.
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Physics revision 2015
65
Jo and Sam now attach four 50 g masses to the spring and release it from its unstretched
position, which is a length of 40 cm. They allow the masses to oscillate freely, as shown in
Figure 3c.
Example 85: 2014 Question 2b
(3 marks)
Find the extension of the spring at the lowest point of its oscillation (when it is momentarily
stationary). Ignore frictional losses. Show your reasoning.
Jo and Sam measure the position of the four masses as they oscillate freely up and down, as
described previously. From this data, they plot graphs of the gravitational potential energy and
spring potential energy. Their results are shown below
Jo says their calculation must be wrong because the graphs should add to a constant amount,
the total energy of the system. However, Sam says that the graphs are correct.
Example 86: 2014 Question 2c
(2 marks)
Explain why Jo is incorrect. Your explanation should include the reason that the spring
potential energy and the gravitational potential energy do not add to a constant amount at
each point.
colinhop@bigpond.net.au
Physics revision 2015
Example 87: 2014 Question 2d
(4 marks)
Calculate the maximum speed of the masses during the oscillation. Show your working.
66
colinhop@bigpond.net.au
•
•
Physics revision 2015
67
GM1
GM1M2
and F =
;
2
r
r2
model satellite motion (artificial, moon, planet) as uniform circular orbital motion
v2
4π2r
a=
=
;
r
T2
apply gravitational field and gravitational force concepts, g =
2005
14, 15,
16
2006
15, 16
2007
12, 13
2008
15, 16,
17
2009
13, 14
2010
18, 19
20
2011
21, 22,
23
2012
8a
2013
7a, b
2014
5a, b
Newton determined that gravity is a force of attraction between two bodies that varied as the inverse
of their separation squared and in direct proportion to their masses. It is a very weak force and is only
noticeable when at least one of the objects is extremely massive.
GMm
The force acts equally on both masses.
F=
r2
The moon is the Earth's only natural satellite but there are thousands of artificial satellites orbiting. For
a satellite in a stable circular orbit, the only force acting on a satellite is the gravitational attraction
between it and the central body. This force acting on it is always perpendicular to its motion. Therefore
the energy of the satellite is unchanged as it orbits. The kinetic energy and gravitational potential
energy both stay the same. The force of gravity holds the satellites in their orbits and causes them to
have acceleration towards the central mass.
Since satellites are in a continual state of free-fall, their acceleration will equal the gravitational field
strength at that point.
2πr
v2
Using v =
r
T
GMm
Using F =
and F = mg
R2
a=
a=

4π 2R
T2
GM
=g
R2
The acceleration of the satellite is independent of the mass of the satellite.
The other sort of question that they ask here is to calculate the speed of the satellite. The two main
problems that students have are:
1) Determining the radius/distance (it is always the distance between the two centres of mass) and
2) Managing the powers of ten on their calculator.
To calculate the value of 'g' at the Earth's surface:
G = 6.67  10-11 N m2 kg-2,
M= 6.0  1024 kg,
r = 6.4  106 m.
 g = 9.8 N kg-1.
At the surface of the Earth, the gravitational field strength, g, is 9.8 N kg-1, it becomes weaker further
from the Earth. 400 km above the Earth g is 8.7 N kg-1. Any object that is falling freely through a
gravitational field will fall with acceleration equal to the gravitational field strength at the point.
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Physics revision 2015
68
The Mars Odyssey spacecraft was launched from Earth on 7 April 2001 and arrived at Mars on 23
October 2001. The figure below is a graph of the gravitational force acting on the 700 kg Mars
Odyssey spacecraft plotted against height above Earth’s surface.
Example 88: 2002 Question 1
(3 marks)
Estimate the minimum launch energy needed for Mars Odyssey to escape Earth’s gravitational
attraction.
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Physics revision 2015
69
The International Space Station (ISS) is in orbit around the
Earth at an altitude 380 km.
Radius of Earth = 6.4 × 106m
Mass of Earth = 6.0 × 1024 kg
Total mass of ISS = 5 × 105 kg
Universal gravitational constant G = 6.7 × 10-11 N m2 kg-2
Example 89: 2004 Pilot Sample Question 12
(1 mark)
What is the period of the ISS in seconds?
Example 90: 2004 Pilot Sample Question 13
(3 marks)
Which row of the table (A – E) best describes the acceleration and speed of the ISS, and the net force
acting on it while in orbit around Earth?
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Physics revision 2015
70
Newton was the first person to quantify the gravitational force between two masses M and
GMm
m, with their centres-of-mass separated by a distance R as F =
where G is the
R2
universal gravitational constant, and has a value of 6.67 × 10-11 N m2 kg-2.
GM
For a mass m on the surface of Earth (mass M) this becomes F = gm, where g = 2 .
R
Example 91: 2005 Question 14
(2 marks)
Which one of the expressions (A – D) does not describe the term g?
A. g is the gravitational field at the surface of Earth.
B. g is the force that a mass ‘m’ feels at the surface of Earth.
C. g is the force experienced by a mass of 1 kg at the surface of Earth.
D. g is the acceleration of a free body at the surface of Earth.
The dwarf planet Pluto was discovered in 1930, and was thought to be the outermost
member of our solar system. It can be considered to orbit the Sun in a circle of radius 6
billion kilometres (6.0 × 1012 m).
In 2003 a new dwarf planet, Eris, was discovered. It has approximately the same mass as
Pluto, but the average radius of its orbit around the Sun is 10.5 billion kilometres (10.5 ×
1012 m).
Example 92: 2007 Question 12
(2 marks)
Which of the choices (A – D) below gives the best estimate of the ratio
gravitational attraction of the Sun on Eris
?
gravitational attraction of the Sun on Pluto
A. 0.33
B.
0.57
C.
1.75
D.
3.06
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Physics revision 2015
71
The period of Pluto around the Sun is 248 Earth-years.
Example 93: 2007 Question 13
(2 marks)
How many Earth-years does Eris take to orbit the Sun?
The figure shows the orbit of a comet around the Sun.
Example 94: 2008 Question 15
(2 marks)
Describe how the speed and total energy of the comet vary as it moves around its orbit
from X to Y.
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Physics revision 2015
72
The Jason 2 satellite reached its operational circular orbit of radius 1.33 × 107 m on 4 July
2008 and then began mapping the Earth’s oceans.
mass of the Earth = 5.98 × 1024 kg
mass of Jason 2 = 525 kg
G = 6.67 × 10–11 N m2 kg–2
Example 95: 2009 Question 13
(2 marks)
On the figure below, draw one or more labelled arrows to show the direction of any force(s)
acting on Jason 2 as it orbits Earth. You can ignore the effect of any astronomical bodies
other than the Earth.
Assume that somewhere in space there is a small spherical planet with a radius of 30 km. By
some chance a person living on this planet visits Earth. He finds that he weighs the same on
Earth as he did on his home planet, even though Earth is so much larger.
Earth has a radius of 6.37 x 106 m and a mass of 5.98 x 1024 kg. The universal gravitational
constant, G, is 6.67 x 10-11 N m2 kg-2.
The acceleration due to gravity (g), or the gravitational field at the surface of Earth, is
approximately 10 N kg-1.
Example 96: 2011 Question 21
(1 mark)
What is the value of the gravitational field on the surface of the visitor’s planet?
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Physics revision 2015
Example 97: 2011 Question 22
73
(2 marks)
What is the mass of the visitor’s planet? Explain your answer by showing clear working out.
Example 98: 2011 Question 23
(2 marks)
The visitor’s home planet is in orbit around its own small star at a radius of orbit of
1.0 × 109 m. The star has a mass of 5.7 × 1025 kg. What would be the period of the orbit of
the visitor’s planet? Show your working.
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74
A satellite is in a geostationary circular orbit over Earth’s equator. It remains vertically above
the same point X on the equator, as shown below.
Example 99: 2013 Question 7a
(1 mark)
Calculate the period of the orbit of the satellite.
Example 100: 2013 Question 7b
(2 marks)
Calculate the radius of the orbit of the satellite from the centre of Earth.
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75
•
apply the concepts of weight (W = mg), apparent weight (reaction force, N),
weightlessness (W = 0) and apparent weightlessness (N = 0);
2005
2006
2007
2008
2009
2010
2011
2012
2013
2014
7, 8, 9
8, 9
9, 10,
11
8b
7c
4a, b
Mass and weight are different quantities. Mass is defined as the amount of matter in a body, it is
measured in kilograms. Since F = ma, a larger force is needed to give a larger mass the same
acceleration.
Weight is not the same as mass. Weight is the force of gravity on an object. The acceleration due to
gravity is 'g', so the ‘weight’ of a mass 'm' is given by W = mg (Newtons). The amount of matter in an
object does not change, hence the mass of an object is always the same. However, 'g' varies from
place to place on the earth's surface depending on the distance from the centre of the earth, so weight
does change.
Apparent Weight
When you are accelerating vertically, the normal reaction force on you will not equal 'mg', this will alter
your perception of your weight; you will feel either lighter or heavier. If you are travelling at a constant
velocity you will feel your usual weight. The apparent weight of a person is equal in magnitude to the
normal force 'N' that the supporting surface exerts.
N
W = mg
ΣF = ma
In this case, the person is accelerating down, F = mg – N
 N = mg - ma
The Normal reaction force is less than mg, so the person feels lighter.
If ma = mg, then the person feels 'weightless'. This happens when the person is in 'free fall' under the
influence of gravity.
Apparent Weightlessness
True weightlessness is only possible where the strength of the gravitational field is zero, this is only
approximated deep in space, a long way from all celestial bodies. Any person or object will experience
apparent weightlessness when they have an acceleration equal to that of gravity; ie. when they are in
free fall. The person or object will experience zero normal force at this time.
This explains why orbiting astronauts are apparently weightless. Both they and their spacecraft, due to
their circular orbit accelerate towards the centre of the Earth. This acceleration is equal to the
gravitational field strength at that altitude. The astronauts are in free fall as they orbit, and the normal
reaction force acting on them is zero.
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76
A space shuttle orbits the Earth at a distance of 6800 km
from its centre. The gravitational field strength at the
orbit is 8.7 N kg-I.
Example 101: 1994 Question 3
(1 mark)
Astronauts carry out experiments inside the shuttle in
unfamiliar conditions they feel weightless. Which one of
the statements shown below (A - E) best explains why
the astronauts feel weightless?
A.
The space shuttle and its contents have the same acceleration towards the Earth: they are in
free fall.
B.
The effect of gravity outside the Earth's atmosphere is negligible.
C.
The gravitational field at the shuttle's orbit is negligible: it decreases rapidly with distance from
the Earth according to an inverse square law.
D.
According to observers on the Earth's surface the gravitational attractive force on the astronauts
in orbit is cancelled by another force acting on them and pointing away from the Earth.
E.
The net force on the astronauts must be zero since their velocity remains constant.
Example 102: 1997 Question 3
(4 marks)
What is the apparent weight of an astronaut of mass 60 kg travelling in a space shuttle under the
following conditions?
a. When the shuttle is stationary on the launching ramp (on the Earth).
b. When the shuttle is accelerating vertically with acceleration 3.0 m s–2, just after take-off.
c. In a stable circular orbit, in a region of space where the gravitational field strength is 8.4 N kg–1.
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77
A softball has a mass m and rests on the
floor of a lift. The weight of the softball,
mg, is taken as a downwards force, and
the reaction force, N, as an upwards force
exerted on the softball by the floor of the
lift. If the softball, and the floor on which it
rests, have a common acceleration
downwards of a, Newton’s second law
gives
mg – N = ma
Example 103: 1998 Question 3 (2
marks)
If both the softball and the floor are
stationary, draw the weight force and
reaction force on the figure below. Label
the two forces mg and N respectively and
mark the points through which the forces
act, using a cross (x) for mg, and a large
dot (•) for N.
Example 104: 1998 Question 4
(2 marks)
If the softball has a mass of 0.20 kg and the softball and floor are stationary, calculate the apparent
weight of the softball. (g = 10 N kg–1)
Example 105: 1998 Question 5
(2 marks)
The softball and floor now move downwards together with an acceleration of 5.0 m s–2.
Calculate the apparent weight of the softball in this situation. (g = 10 N kg–1)
Example 106: 1998 Question 6
(2 marks)
Under what conditions will the apparent weight of the softball be zero?
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78
A ride in an amusement park allows a person to free fall without any form of attachment. A
person on this ride is carried up on a platform to the top. At the top, a trapdoor in the
platform opens and the person free falls.
Approximately 100 m below the release point, a net catches the person. A diagram of the
ride is shown above.
Helen, who has a mass of 60 kg, decides to take the ride and takes the platform to the top.
The platform travels vertically upward at a constant speed of 5.0 m s–1. As the platform
approaches the top, it slows to a stop at a uniform rate of 2.0 m s–2
Example 107: 2009 Question 8
(1 mark)
What is Helen’s apparent weight as the platform slows to a stop?
Helen next drops through the trapdoor and free falls. Ignore air resistance.
During her fall, Helen experiences ‘apparent weightlessness’.
Example 108: 2009 Question 9
(1 mark)
Explain what is meant by apparent weightlessness. You should make mention of
gravitational weight force and normal or reaction force.
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79
A ride at a fun fair involves a car travelling around a vertical circle of radius 15 m. The curved
arrow in the diagram below indicates the direction of travel of the car.
Example 109: 2011 Question 9
(2 marks)
What is the minimum speed that the car can have at position A if it is not to leave the rails and
start to fall?
Example 110: 2011 Question 10
(2 marks)
Melanie, who has a mass of 60 kg, is a passenger in the car. What will be Melanie’s
apparent weight at position A when the car is travelling at the minimum speed? Explain your
answer.
Example 111: 2011 Question 11
(2 marks)
As the car passes point B at the bottom of the circle, it has a speed of approximately
5.5 m s-1. What is Melanie’s apparent weight at this point? Show your working.
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80
A satellite is in a geostationary circular orbit over Earth’s equator. It remains vertically above
the same point X on the equator, as shown below.
Example 112: 2013 Question 7c
(3 marks)
An astronaut of mass 65 kg is on board the satellite. She is interviewed on television and says
that she is weightless.
Explain how the terms ‘weight’, ‘weightlessness’ and ‘apparent weightlessness’ either apply or
do not apply to the astronaut.
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81
Mary and Bob are riding in a car on a roller-coaster. The roller-coaster is designed so that the
riders will feel ‘weightless’ at the top of the ride (point A).
It can be assumed that the track is circular in shape at and near the points A and B. The radius
of these circular paths at A and B is 20 m. This is shown in Figure 7. Ignore air resistance and
frictional forces.
Example 113: 2014 Question 4a
(2 marks)
Calculate the minimum speed at which Mary and Bob must be moving at point A for them to
feel weightless.
Example 114: 2014 Question 4b
(3 marks)
Explain why Mary and Bob feel weightless at point A at this speed.
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Physics revision 2015
Motion in one and two dimensions
solutions
Example 1:
(1984 Question 1)
Use v = u + at
 30 = 0 + a × 10
 a = 3 m s-2 (ANS)
Example 2:
(1984 Question 2)
Use ΣF = ma
 FDriving – FResistance = ma
 FDriving – 500 = 2000 × 3
 FDriving = 6000 + 500
 FDriving = 6500 N (ANS)
Example 3:
(1984 Question 3)
Use x = ut + ½at2
 x = 0 + ½ × 3 × 102
 x = 150 m (ANS)
Example 4:
(1984 Question 4)
Use WD = F × d
 WD = 6500 × 150
 WD = 975000
 WD = 9.8 × 105 J (ANS)
Example 5:
(1984 Question 5)
Use Power = F × v
 P = 6500 × 20
 P = 130000
 P = 1.3 × 105 W (ANS)
Example 6: (2001 Question 1, 46%)
This question is designed to be a nice starter,
so don’t fall into the trap of making it too
complicated. As you were reading the
question, you should have highlighted the
words “constant speed”. This means that
acceleration = 0,
 net force = 0 (ANS)
Example 7: (2001 Question 3, 58%)
When the rider is going down the hill at a
constant speed, then the sum of the forces is
again zero. So, the frictional forces opposing
the motion (acting up the slope) must be equal
to the weight component acting down the slope
82
 Frictional forces = mg sin
= 100 × 10 × sin150
= 258.8
= 259 N (ANS)
Example 8:
(2002 Question 7, 54%)
a = Δv = 11
t
0.1
= 110 m s-2
F = ma =1.30 × 103 × 110
= 1.43 × 105 N
[actually 1.4 × 105 N due to sig. figs]
143 kN (ANS)
Example 9:
(2002 Question 8, 47%)
Crumple Zone enables passenger cell to
travel a greater distance while stopping, in a
longer time.
The change in momentum [pf – po] is the
same regardless of the time taken.
Longer time allows lower average force [same
impulse – force × time]
Example 10: (1984 Question 28)
F = Σma
The net force = 45 – Ffriction
= 45 – 30
= 15
 15 = (10 + 20)a
15
a=
30
= 0.5 m s-2 (ANS)
Example 11: (1984 Question 29)
The force exerted by the 20 kg mass on the l0
kg mass is equal and opposite to the force
exerted by the 10 kg mass on the 20 kg mass.
Since the 20 kg mass is accelerating at
0.5 m s-2, the net force on it must be
ΣF = ma.
 ΣF = 20 × 0.5 = 10N.
The frictional force of 20 N still needs to be
overcome, so F10 on 20 = 10 + 20 = 30 N
 The force exerted by the 20 kg mass on the
10 kg mass = 30 N (ANS)
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Physics revision 2015
83
Example 12: (2004 Question 1, 57%)
Example 18: (2008 Question 2, 60%)
Use h = ut + 21 at2
Where u = 10sin30, a = -10 m s-2, t = 3.0 s
h = 10 × 0.5 × 3 – ( 21 × 10 × 32)
Constant speed  ΣF = 0
 Fresistance = 9 × 104 N
Using the graph
 4 m s-1 (ANS)
 h = 15 – 45
 h = -30 m.
This answer is consistent as we have assumed
the upwards is positive.
 The height was 30 m (ANS)
Example 13: (2004 Question 3, 40%)
The minimum speed (not zero because of the
constant horizontal component) occurs at the
top of the flight of the parcel. It will then
continue to speed up.
 B (ANS)
Example 14: (2005 Question 4, 57%)
At t = 5 secs, the retarding force is 1.8 N (from
graph)
F = ma
ma = mg – R
0.2a = 0.2 × 10 – 1.8
0.2a = 2 – 1.8
0.2a = 0.2
a = 1 m s-2 (ANS)
Example 15: (1987 Question 13, 52%)
The weight of the 10 kg mass is 100 N.
 This weight force needs to accelerate both
the mass and the door (at the same rate)
 100 = Σm × a
 100 = 50 × a
 a = 2 m s-2 (ANS)
Example 19: (2010 Question 3, 35%)
Consider mass 2
m2g – T = m2a
(1)
Consider mass 1
T = m1a
(2)
Combining gives
m2g – m1a = m2a
 0.1 × 10 = (m1 + m2)a
= 0.5 a
 a = 2 m s-2 (ANS)
Example 20: (2011 Question 2, 55%)
The only force that we need to consider that is
acting on the trailer is the tension.
Therefore T = ma
 T = 2000 × 0.5
 T = 1000 N (ANS)
Example 21: (2011 Question 7, 55%)
Use Newton’s third law.
 FC on B = -FB on C
The three forces acting on C are:
Its weight, mg
The normal reaction
The weight of A and B.
 FB on C = (0.050 + 0.10) × 10
 FB on C = 1.5 N
 FC on B = 1.5 N (upwards) (ANS)
Example 16: (1987 Question 14, 79%)
Use x = ut + ½at2
where u = 0
 1.0 = ½ × 2 × t2
 t = 1.0 s
(ANS)
Example 17: (2008 Question 1, 50%)
Fship = Ftug - Fresistance
= 9 × 104 – 2 × 104 (from the graph)
= 7 × 104
 7 × 104 = 100 × 104 × a
 a = 0.07 m s-2 (ANS)
Example 22: (2011 Question 8, 50%)
The blocks will be in free fall, so their
acceleration will be g.
 FB on C = 0, because they are both in free fall,
and accelerating at the same rate.
 FB on C = 0
(ANS)
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Physics revision 2015
Example 23: (2012 Question 4b, 15%)
Newton’s third law can be written in the form;
FA on B = -FB on A
Then the weight of sphere is
FEarth on Sphere
In terms of Newtons action and reaction pairs,
the ‘reaction’ will be
FSphere on Earth UP (ANS)
84
 0.5 x 0.1944 = 0.5 x 10 x sin100 – Friction
 0.0972 = 0.8682 – Friction
 Friction = 0.773 N (ANS)
Example 26: (2013 Question 2b, 30%)
T
Example 24: (2012 Question 5d, 53%)
Rope 1 needs to provide the tension to
accelerate both logs, so the tension in it will be
greater than that of Rope 2.
 Rope 1
(ANS)
Use ΣF = ma
 T1 – T2 – 400 = 600a
and
T2 – 400 = 600a
 T2 = 600a + 400
 T1 – (600a + 400) – 400 = 600a
 T1 = 600a + (600a + 400) + 400
Use T1 = 2400 N at its breaking point.
 2400 = 1200a + 800
 1600 = 1200 x a
 a = 1.33 m s-2 (ANS)
Example 25: (201b Question 1b, 40%)
m1
g
Use the equations m1a = m1g – T
and T = m2a
 2a = 20 – 6a
 8a = 20
 a = 2.5 ms-2
Substitute into T = m2a
 T = 6 x 2.5
= 15 N (ANS)
Example 27: (2014 Question 1b, 50%)
The tension in the coupling is the only force
accelerating the two trucks.
 Use T = ∑ma
= 2 × 10 × 103 × 0.2
= 4 × 103 N (ANS)
N
Example 28: (1992 Trial Question)
Friction
mg
mgcos100
mgsin100
From the diagram we get
ΣF = ma
 0.5a = mgsin100 – Friction
Need to find ‘a’.
Use x = ut + ½at2 to get
3.5 = ½ x a x 62
 a = 0.1944
Since the car is moving in circular motion,
the resultant force must be acting radially
inwards. The air resistance is going to
oppose the motion; hence, it has to be to
the left. To obtain a resultant force, you
need to add the forces that are acting
together. This means that the ‘Total force
of road on tyres’ plus the ‘air resistance’
need to combine to give a ‘resultant
force’ that acts radially inwards.
E (ANS)
Example 29: (2005 Question 7, 33%)
There are now two forces acting, the radially
inwards force from the rails, and the braking
force from the brakes.
The braking force is opposing the motion
therefore acting backwards.
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Physics revision 2015
Braking force
85
They must add together to give the resultant
force F, which is radially inwards.
F
mg
Force of rails on
wheels
N
You could resolve the Normal vector into two
components, one perpendicular and the other
horizontal.
Net force
Nsin
 B (ANS)
mg
There are only two forces acting, the Weight,
(drawn from the centre of mass) and the
Normal (drawn perpendicular to the surface)
Example 31: (2011 Question 5, 50%)
N
N
ΣF
From this we get:
Nsin = mg and
mv 2
= Ncos
r
Example 32: (1977 Question 15)
If the object is released at the point P, it will go
vertically up. Therefore it will land directly
under the point P. This means that it will land
a distance r from the point X.
 “r” (ANS)
Example 33: (1977 Question 16)
If the object is released at the point Q, it will
be a projectile with an initial horizontal velocity
of v. It will be launched from a height of 3r.
Use x = ut + 21 gt2 to find how long it will take
to land.
 3r = 0 + 21 gt2
6r = gt2
6r
t2 =
g
6r
g
To calculate the horizontal distance
travelled in time ‘t’, use d = vt.
6r
d=v
(ANS)
g
t=
mg
The only two forces acting are the weight force,
acting down and the normal reaction from the
surface, acting perpendicular to the surface.

Ncos
Example 30: (2010 Question 5, 45%)
NN
N
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86
Example 34: (1978 Question 18)
Example 36: (2012 Question 7c, 33%)
The force on the cage floor is equal and
opposite to the force of the cage floor on
the observer (N)
At the bottom  F = N - mg
mv 2
=
r
mv 2
 N = mg +
r
Therefore the observer will feel heavier at
the bottom.
 B (ANS)
Consider the motion to be vertical, at constant
speed.
mg
T
Example 35: (1978 Question 19)
The force on the cage floor is equal and
opposite to the force of the cage floor on
the observer (N)
At the top  F = N + mg
mv 2
=
r
mv 2
 N = -(mg ).
r
The negative sign indicates that the normal
reaction is down.
The magnitude of the Normal reaction is
mv 2
mg r
At the bottom  F = N - mg
mv 2
=
r
mv 2
 N = mg +
r
The difference between the top and the
2mv 2
r
bottom is
 A (ANS)
At the top of the path, there are two forces
acting on the sphere, its weight, mg, and the
tension from the string T.
 ΣF = ma
mv 2
Becomes
T + mg =
r
2
mv
T=
- mg.
r
At the bottom of the circle, the situation looks
like.
T
mg
 ΣF = ma
Becomes
T=
T - mg =
mv 2
r
mv 2
+ mg.
r
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Physics revision 2015
mv 2
is constant, the tension at the
r
bottom is 2mg larger than the force at the top.
 the tension in the string is greater at the
bottom of the circular path.
Since
Example 37: (2013 Question 5b, 47%)
At the point S there are two forces acting on
the mass. The weight is acting down, and the
tension force is up.
Since the net force is up,
ΣF = ma
 ma = T – mg
v2
r
72
=
1
= 49 m/s2
 2.0 x 49 = T – 2.0 x 10
 T = 118 N (ANS)
a=
Example 38: (1971 Question 4)
This requires you to express the KE in terms of
the momentum and the mass.
p2
Use the relationship KE =
2m
2
p
K=
(ANS)
2m
Example 39: (1971 Question 5)
The total energy, T, is constant.
Initially U = 0, therefore
T = KE + PE
 T = KE + 0
 KE = T
 A (ANS)
87
Example 41: (2002 Question 5, 47%)
The car must travel the 20 m horizontally while
it ‘falls’ the 4 m vertically.
Time to fall [use x = ut + 21 gt2]
=> t = 0.89 s
v=
x
20
=
0.89
t
= 22.36 m s-1 (ANS)
Example 42: (2002 Question 6, 54%)
Have to use vectors to add the horizontal
and vertical velocities OR use energy.
Vertical velocity is found using
v = u + gt
 v = 0 + 10 × 0.89
 v = 8.9 m s-1
Use Pythagoras.
25 m s-1
8.9 m s-1
252 + 8.92 = 625 + 79.2
= 704.2
 v = 26.5 m s-1
Energy conservation gives
KE + PE (initially) = KE (at end)
 21 × m × 252 + m × 10 × 4
=
1
2
× m × v2
 v2 = 705
 v = 26.5 m s-1 (ANS)
Example 43: (2006 Question 6, 57%)
22 N
F = 22 – mg
= 22 – 0.5 × 10
= 17 N (ANS)
Example 40: (1971 Question 6)
To find the velocity at the highest point, you
need to also know the PE at that point. This is
not given.
At the top, the vertical component of the
velocity will be zero. The horizontal component
will depend on the initial angle. This is not
given.
 F (ANS)
W = mg
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Physics revision 2015
Example 44: (2006 Question 7, 57%)
The rocket has a constant acceleration given
F
17
by;
a=
=
= 34 m s-2
0.50
m
 x = ut + 21 at2
x=0+
1
2
 34  (1.5)2
 x = 38.25
 x = 38 m (ANS)
Example 45: (2006 Question 8, 33%)
To answer this question you need to separate
the vertical motion form the horizontal motion.
Horizontal
Since air resistance can be ignored, the only
horizontal acceleration is due to the constant
force of 22 N.
 vH = u + at
 vH = 0 + 44  1.5
 vH = 66 m s-1
Vertical
v = u + gt
= 0 + 10  1.5
= 15 m s-1
Use vectors to add these and to find the
direction.
60
θ
speed
15
 speed = 60 +15
= 67.7 m s-1 (ANS)
The angle relative to the ground is θ
15
 tan θ =
60
 θ = 12.8º (ANS)
(Remember to have calculator in degree mode)
2
2
88
Example 47 (2008 Question 7, 57%)
Consider this motion in both the horizontal and
the vertical directions.
Horizontal
Find the time it takes to travel the 72 m
horizontally.
vhorizontal = 30cos(36.9)
= 24 m s-1
d
Using v =
t
72
t=
= 3 secs
24
Vertical
Use x = ut + 21 gt 2
= 18 × 3 –
1
2
× 10 × 32
= 54 – 45
= 9 m (ANS)
Example 48: (2009 Question 11, 52%)
First step: assign a direction to be positive,
take ‘up’ to be positive.
Second step: fill in the data table, (only working
through to the top of the flight).
y = ??
uy = 60 sin(30) = 30 m s-1
vy = ??
ay = -10 m s-2
ty = 9 s
Third step: find an equation that does not
involve the final velocity.
y = ut + 21 gt2
y = 30 × 9 +
1
2
× 10 × 92
 y = -135 m
The height of the cliff is 135 m.
 h = 135 m (ANS)
Example 46: (2007 Question 17, 53%)
Example 49: (2013 Question 8b, 40%)
Since the second paintball reaches the same
height, it must take the same time as before to
reach this height and then all back to the
starting height.
The acceleration due to gravity is constant.
 A, C (ANS)
The horizontal component of the velocity will
remain constant at
vH = 20cos300
= 17.32 m/s.
The vertical component will be
v = u + gt
on the way down, u = 0, g = 10, t = 2
 v = 20 m/s.
Use Pythagoras to find the magnitude of the
velocity.
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Physics revision 2015
 202 + 17.322 = v2
 v2 = 700
 v = 26.5 m/s
To find the angle use
89
28000
3000
utruck = 9.3 m s -1
 v = 9.3 m s-1 (ANS)
utruck =
Example 54: (2005 Question 9, 65%)
20 m/s
26.5 m/s
θ
17.32 m/s
20
Use tanθ =
17.32
0
 θ = 49.1
 v= 26.5 ms-1 at an angle of 49.10
 26.5 m/s
49.10 (ANS)
Example 50: (2002 Question 9, 62%)
Momentum is conserved
10 000 × 6 + 5 000 × 0 = 15 000 × vf
 60 000 = 15 000 × fv
60000
vf =
15000
vf = 4 m s-1 (ANS)
Example 51: (2002 Question 10, 62%)
Impulse = change in momentum.
po = 60 000 Ns
pf = 40 000 Ns
 p = 20 000 Ns
 Total impulse = 2.0 × 104 N s (ANS)
Example 52: (2002 Question 11, 50%)
Ek before collision
= ½ × 10000 x 62 = 180 000 J
Ek after collision
= ½ × 15000 x 42 = 120 000 J
 60 000 J of mechanical energy has been lost
so interaction is not elastic.
Example 53: (2005 Question 8, 65%)
You must convert all masses into kilograms.
Find the initial and final momenta.
mcarucar + mtruckutruck = (mcar + mtruck)vf
1000 × 0 + 3000  utruck = (1000 + 3000) × 7
3000utruck = 4000 × 7
Momentum is conserved in all collisions. In
elastic collisions mechanical energy is also
conserved. In inelastic collisions mechanical
energy is lost as heat and sound and energy
used in deformation.
Example 55: (2006 Question 10, 37%)
From conservation of momentum
pi = pf
6000 × 5 + M × 0 = (6000 + M) × 0.098
30000 = 588 + 0.098 × M
29412 = 0.098 M
 M = 300122
 M = 3.0 × 105 kg (ANS)
Example 56: (2006 Question 11, 37%)
F t = mv
 F × 20 = 6000 × (5 – 0.098)
6000  4.902
F=
20
 F = 1.47 × 103 N (ANS)
Example 57: (2007 Question 3, 62%)
Momentum is conserved in all collisions. So the
initial momentum must equal the final
momentum.
Initial momentum is 2 × 3 + 1 × 0 = 6 Ns
Final momentum is 2 × x + 1 × 4 = 2x + 4
 2x + 4 = 6
x=1
 2kg block is moving at 1 m s-1 (ANS)
Example 58: (2007 Question 5, 62%)
The impulse (change in momentum) is given by
Ft = mv
1× 4
F =
0.01
= 400 N (ANS)
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90
Example 59: (2008 Question 8, 85%)
Momentum is conserved in this collision
 pi = pf
 20 × 103 × 8 = 80 × 103 × v
 v = 2 m s-1 (ANS)
Example 60: (2008 Question 9, 63%)
The impulse given to the locomotive by the
trucks is equal and opposite to the impulse
given to the trucks by the locomotive.
Impulse on trucks is change in momentum of
trucks
= mΔv
= 60 × 103 × 2
= 1.2 × 105 kg m s-1 West (ANS)
Example 61: (2012 Question 2, 30%)
Initial momentum is given by
mv = 1.2 × v to the right
The block rebounds, so its new momentum is
to the left.
As the change in momentum of the 1.2 kg
block is given by final momentum – initial
momentum, when the change in direction is
taken into consideration, this is effectively the
scalar sum of the momenta, but to the left.
This will be greater than the initial momentum
of the 1.2 kg mass.
From conservation of momentum, this change
in momentum of the 1.2 kg mass must be
equal to the change in momentum of the 2.4 kg
mass.
Example 62: (1999 Question 5, 55%)
Work done = area under force v displacement
graph.
The examiners actually want to make these
exams to be straight forward, so these curves
often have a very simple solution to them. The
graph was specifically drawn to be
symmetrical, to simplify the area calculation. If
you drew a straight line from the origin to the
coordinates (500, 10 000), you should see that
the ‘curve’ is symmetrical.
This means that the shape is actually a
triangle, so the area = ½ × (10 000 + 0) × 500
= 2.5 × 106
This can also be done by counting squares.
Each square is 2 000 × 100 = 200 000J
There are 13 squares
 WD = 13 × 200 000 = 2 600 000J
2.5 × 106J (ANS)
A range of answers was accepted for this
question.
Example 63: (2006 Question 3, 60%)
The work done by the force is the change in KE
 KE = KEfinal - KEinitial
= - 21 (90 + 40) × 62
= - 2340 J
Use the modulus
 F  d = 2340
= (190 + 70) × d
2340
d=
260
 d = 9 m (ANS)
Example 64: (2004 Pilot Question 14, 23%)
Initial energy = mgh.
We need to find ‘h’.
Work done to overcome friction
f × d = 50 × 6 = 300 J
KE gained = 21 mv2
= 21 × 30 × 82
= 960 J
So total work done by gravity
= 300 + 960 = 1260 J
 mgh = 1260
1260
h =
30 ×10
= 4.2 m (ANS)
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91
Example 65: (2004 Pilot Question 15, 53%)
Example 71: (2008 Question 14, 50%)
The KE of the of the box is given by
K E = 21 mv2
The gravitational potential energy is measured
from the point of release.
 at the bottom, PE = 0.
 A (ANS)
= 21 × 30 × 82
= 960 J
Elastic potential of the spring is given by:
E = 21 k(Δx)2
 21 × 30 000 × (Δx)2 = 960
(as no energy is lost sliding across the floor)
 15 000 × (Δx)2 = 960
960
 x2 =
15 000
 Δx = 0.25 m (ANS)
Example 66: (2007 Question 9, 43%)
The extension (change in length) is
0.6 – 0.4 = 0.2 m
F = kΔx
 mg = kΔx
On substitution, m × 10 = 20 × 0.2
 m = 0.4 kg (ANS)
Example 72: (2009 Question 5, 55%)
The kinetic energy gained by the mass is the
elastic energy that was stored in the spring.
The spring was compressed 10 cm.
From this we can find the following.
1
1
 2 mv2 = 2 k(Δx)2
1
1
 2 × 0.20 × 5.02 = 2 × k × (0.10)2
 2.5 = 0.005 × k
 k = 500 Nm-1 (ANS)
Example 73: (2010 Question 14, 35%)
The total energy is the sum of the
PEElastic + mgh.
1
PEelastic = 2 k(Δx)2
1
Example 67: (2007 Question 10, 43%)
WD = area under the graph.
It is also given by
1 k( x )2 - 1 k( x )2
f
i
2
2
=
1 ×20×0.32
2
-
1 ×20×0.22
2
= 0.5 J (ANS)
Example 68: (2007 Question 11, 47%)
Since we can ignore air resistance (stated in
the question) the system does not lose energy.
 the total energy is constant.
 D (ANS)
Example 69: (2008 Question 12, 55%)
1
The energy stored in the spring is 2 k(Δx)2
1
= 2 × 10 × (0.2)2
= 0.2 J (ANS)
(use metres)
Example 70: (2008 Question 13, 40%)
The ball will be stationary (momentarily) at the
top and the bottom of the oscillation. Therefore
the KE will be zero at these points. The KE will
be a maximum at the midpoint.
 D (ANS)
= 2 × k × 0.42
Also F = k × x
20 = k × 0.4
 k = 50
1
PEelastic = 2 × 50 × 0.42
= 4 Joule (gain)
PEgravitational = mgh
= 2 × 10 × 0.4
= 8 Joule loss.
 Change in energy
= 4 J (ANS)
Example 74: (2010 Question 15, 55%)
The total energy will remain constant
throughout the interaction.
TE = KE + PEelastic
PEelastic increases when the spring is
compressed (in the middle of the interaction)
Therefore if PEelastic increases, KE must
decrease.
PEelastic returns to zero after the interaction,
therefore the KE will return to its initial value.
A (ANS)
Example 75: (2010 Question 17, 55%)
Momentum is ALWAYS conserved.
B (ANS)
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Physics revision 2015
Example 76: (2011 Question 16, 55%)
The force acting is mg = 1.0 × 10
= 10 N
The extension is 70 – 40 = 30 cm (0.3 m)
F = k × x
 10 = k × 0.3
 k = 33.3 Nm-1 (ANS)
Example 77: (2011 Question 17, 42%)
The kinetic energy will start from zero, go to a
maximum and then return to zero.
 D (ANS)
Example 78: (2011 Question 19, 46%)
The gravitational energy will start at a
minimum, when L = 80 cm, and increase as L
decreases to 60 cm.
 B (ANS)
Example 79: (2011 Question 20, 27%)
The strain energy is minimum (but not zero) at
60 cm. It will increase until 80 cm.
92
 v = 3.16 m/s
(ANS)
Example 83: (2013 Question 6c, 17%)
They have assumed that the SPE at Q = 0.
This is not correct because the spring has
already been extended from its original length
of 2.0 m.
SPEQ = ½k(∆x)2, where ∆x = 0.5.
 SPEQ = ½ x 10 x 0.52
= 1.25 J
 TEQ = 10 + 1.25 = 11.25J
SPEP = ½k(∆x)2, where ∆x = 1.5.
 SPEP = ½ x 10 x 1.52
= 11.25 J
 TEP = 0 + 1.25
= 11.25 J
Example 84: (2014 Question 2a, 60%)
Use the equation F = k × ∆x
F = mg, so
0.05 × 10 = k × 0.1
 0.5 / 0.1 = k
 k = 5 N m-1
1
The increase is not linear as, E = 2 k(Δx)2
 F (ANS)
Example 80: (2012 Question 1c, 55%)
Energy stored in the spring is given by
E = ½k(∆x)2
 5.4 = ½ x k x 0.082
(Make sure that you use ∆x in metres)
2  5.4
k=
0.082
 k = 1688 Nm-1
 k = 1.7 x 103 Nm-1 (ANS)
Example 81: (2013 Question 6a, 80%)
From the graph, at Z, the GPE is zero, so the
total energy is 20 J
 20 J
(ANS)
Example 82: (2013 Question 6b, 45%)
This is a conservation of energy question.
From the graph, at the point Y, the SPE = 5 J,
the GPE = 10 J. Since the TE = 20 J, and
remains constant, the KE must be 5 J.
 ½mv2 = 5 J
 ½ x 1 x v2 = 5
 v2 =10
0.1 × 10 = k × 0.2
 0.1 / 0.2 = k
 k = 5 N m-1
0.15 × 10 = k × 0.3
 0.15 / 0.3 = k
 k = 5 N m-1 (ANS)
Example 85: (2014 Question 2b, 20%)
If the four 50 g masses were allowed to hang
under their weight, the spring would be
extended to 80 cm.
If they are released from 40 cm, then the spring
will extend 40 cm past the 80 cm point.
 the extension will be 80 cm.
 0.8 m
(ANS)
(This is on the assumption that the spring has
not exceeded its elastic limit.)
Example 86: (2014 Question 2c, 35%)
The total energy is the sum of three forms of
energy, spring potential energy, gravitational
potential energy and Kinetic energy.
Since Jo is not including the kinetic energy, she
is wrong, and the varying kinetic energy needs
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Physics revision 2015
to be added to the other two to get a constant
total energy at any point.
93
R = 3.8 ×105 m + 6.4 × 106 m = 6.78 × 106 m.
42R3
This gives T =
GM
Example 87: (2014 Question 2d, 13%)
The maximum speed will occur in the middle of
the oscillation.
This is when ∆x = 0.4. This is when a = 0,
because ∑F = 0.
At the lowest point, when it is momentarily
stationary, the KE = 0. We also take this point
to have GPE = 0.
Total Energy(lowest point) = KE + GPE + SPE
= 0 + 0 + ½ × 5 × 0.82
= 2.5 × 0.64
= 1.6
At ∆x = 0.4
TE = KE + GPE + SPE
1.6 = KE + 0.2 × 10 × 0.4 + ½ × 5 × 0.42
 KE = 1.6 – (0.8 – 0.4)
 KE = 0.4
 ½ × 0.2 × v2 = 0.4
 0.1 × v2 = 0.4
 v2 = 4
 v = 2 m/s (ANS)
4π2 (6.78 ×106 )3
T =
6.67 ×10-11 × 6 ×1024
 T = 5.5 × 103 s.
 5.5 × 103 s (ANS)
Example 90: (2004 Pilot Sample Q 13, 27%)
This satellite is moving at constant speed as
the altitude (given) implies a circular orbit. The
only force acting is from the satellite towards
the centre of the earth, and as R is constant
this force is constant. This constant force
implies a constant acceleration.
D (ANS)
Example 91: (2005 Question 14, 40%)
mg is the force that a mass m feels at the
surface of Earth.
 g is the acceleration that a mass m feels at
the surface of the Earth. Hence statement B is
incorrect.
 B (ANS)
Example 88: (2002 Question 1, 53%)
Area under graph is something like 11 to 13
squares
Each square has a value of
1000 × 3 × 106 = 3 × 109 J
=> at least 3.3 × 1010 J (ANS)
Example 92: (2007 Question 12, 47%)
Using gEris = GM and gPluto = GM
2
rEris
gives
Allowing for a variation in the number of
squares counted, a range of values
3.3 to 4.4 × 1010 J, was accepted.
gEris
gPluto
GM
2
r2
= rEris = Pluto
2
GM
rEris
2
rPluto
On substitution
Example 89: (2004 Pilot Sample Q 12, 80%))
The gravitational force, F, acting on the satellite
(the only one) is directed from the satellite
towards the centre of the earth.
GM
This is F = mg, where g = 2 , M is the mass
R
of the earth, R the distance from the satellite to
the centre of the earth, and G is the
gravitational constant. We also know (given)
2
that g = 4 2R , where T is the period of the
T
satellite orbit.
The value of
2
rPluto
2
(6.0×1012 )2
rPluto
=
2
(10.5×1012 )2
rEris
2
= 6.0 2
10.5
= 0.33
 A (ANS)
Example 93: (2007 Question 13, 40%)
Using Kepler’s Law (which is not on the course,
but is very useful)
The ratio of
R3
T2
is the same or all bodies
orbiting the Sun.
12 3
12 3
)
 (6.0 ×102 ) = (10.5 ×10
2
248
T
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Physics revision 2015
Example 97: (2011 Question 22, 50%)
3
3
 6.0 2 = 10.5
2
248
94
T
Use F = mg
GMm
=
r2
(simplifying the calculation)
3
2
 T2 = 10.5 × 3248
6.0
 gVisitor =
 T = 329623
 T = 574 Earth-years (ANS)
2
also gEarth =
Example 94: (2008 Question 15, 50%)
The comet will basically have a constant
energy as it orbits the Sun. (Even though it is
losing a minute amount as it burns).
The total energy is the sum of the PE and KE.
The comet will gain PE as it moves away from
the sun, therefore it must lose KE. This means
that it will travel faster at X than Y.
As the comet moves from X to Y, the speed
will decrease and the energy will remain
constant (ANS)
GME
rE2
MV
ME
= 2
2
rV
rE
 MV =
=
MErV2
rE2
5.9 ×1024 ×(3.4×10 4 )2
(6.37×106 )2
 MV = 1.33 × 1020 kg (ANS)
Example 98: (2011 Question 23, 50%)
Example 95: (2009 Question 13, 40%)
There is only one force acting on Jason 2 as it
orbits, that is its weight force and this acts
towards the centre of the Earth.
direction of movement
Earth

GMV
rV2
mg
Jason 2
Students were required to draw and label
arrow(s) to represent any force(s) acting on the
satellite orbiting the Earth.
The required answer was one arrow from the
satellite and pointing towards the Earth, with a
label weight or gravitational force or mg or Fg.
It was not acceptable to label it Fnet.
Example 96: (2011 Question 21, 74%)
W = mg, if the visitor weighs the same then g
must equal 10 on both the planet and the
Earth.
 10 N kg-1 (ANS)
GMm
mv 2
=
r
r2
GM

= v2
r
2πr
Use v =
T
GM
4π 2r 2

=
T2
r
2 3
4π r
 T2 =
GM
4π 2 ×(1×10 9 )3
 T2 =
6.67×10 -11 ×5.7×10 25
2
 T = 1.04 × 1013
 T = 3.22 × 106 secs (ANS)
Use
Example 99: (2013 Question 7a, 41%)
Period = 24 hours
= 24 x 60 x 60
= 8.64 x 104 s (ANS)
Example 100: (2013 Question 7b, 40%)
This is solved quickest using Kepler’s Law,
which isn’t on the course.
r3
GM
Use 2 =
which is Kepler’s Law.
4 2
t
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Physics revision 2015
6.7  1011  6.0  1024 (8.64  104 )2
42
 r3 = 7.60 x 1022
 r = 4.24 x 107 m (ANS)
r3 =
Example 101: (1994 Question 3)
The reading on the spring balance is zero
because both the mass and the spring balance
are in free fall towards Triton. They both have
the same acceleration. Therefore, the tension
in the spring balance is zero.
 A (ANS)\
95
Example 106: (1998 Question 6, 65%)
Apparent weight will be zero when the lift is in
free fall. This is because the apparent weight is
the value of the normal contact force.
This is given by mg – N = ma.
In this case mg = ma, so N must be zero.
Example 107: (2009 Question 8, 50%)
The acceleration is 2.0 m s-2 down.
ma
Example 102: (1997 Q 3, 90%, 40%, 40%)
a.
Normal weight:
60 × 10 = 600 N
b.
Heavier, actually
60 × 13 = 780 N
c.
Zero: apparent weight is always zero in
stable orbit conditions, because the astronaut
and the satellite both have the same
acceleration towards the centre of the earth.
N
mg
Example 103: (1998 Question 3, 75%)
The weight force is shown acting at the centre
of mass, while the reaction force acts at the
point of contact with the ground. The weight
force is down, the reaction force is up. The two
forces have the same magnitude.
X
.
Example 104: (1998 Question 4, 45%)
Apparent weight is measured by the value of
the normal contact force.
ie 0.2 × 10 = 2 N (ANS)
Example 105: (1998 Question 5, 45%)
Apparent ‘g’ is now 5 less than normal.
Apparent weight is thus:
0.2 × 5 = 1.0 N (ANS)
The net force ΣF = ma. For this to be in the
downwards direction, mg needs to be greater
than N.
 mg – N = ma
 N = mg – ma
 N = 60 × 10 – 60 × 2
 N = 600 – 120
 N = 480N (ANS)
Example 108: (2009 Question 9, 50%)
Helen’s apparent weight is the normal reaction
force that acts on her. Helen’s weight force
remains that same throughout the fall, since
the gravitational field strength is constant
through that distance. When Helen drops
through the trapdoor there is no normal
reaction acting on her so her apparent weight
is zero.
Example 109: (2011 Question 9, 50%)
When the car is just about to leave the rails,
the normal reaction is zero.
At A, F = N + mg
mv 2
= N + mg, but N = 0
r
mv 2

= mg
r

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
Physics revision 2015
Example 114: (2014 Question 4b, 40%)
v2
=g
r
 v = gr
 v = 10×15
 v = 12. 2 m s-1 (ANS)
Example 110: (2011 Question 10, 50%)
Since the normal is zero (from previous
question), Melanie will be apparently
weightless. This is because she is in free fall
accelerating down at 10 m s-2.
0 N (ANS)
Example 111: (2011 Question 11, 50%)
At the bottom,
N – mg =
96
mv 2
r
 N = 600 +
60 × 5.52
15
 N = 721 N (ANS)
Example 112: (2013 Question 7c, 40%)
Weight is the force of attraction between her
and the Earth. It is given by W = mg, where g is
the acceleration due to gravity at that height
above the Earth’s surface. Therefore she has
weight.
Weightlessness exists only where the
gravitational field strength is zero. This is not
true in this situation.
Apparent weightlessness occurs when force
due to gravity is the only force acting, the
object is in free fall or the normal reaction force
is zero. She is in orbit around the Earth so she
will experience apparent weightlessness.
Example 113: (2014 Question 4a, 70%)
For Mary and Bob to feel weightless, the
Normal reaction must be zero.
 mg = mv2/r
 g= v2/r
 10 = v2/20
 v2 = 200
 v = 14.14
 v = 14 m/s (ANS)
At this speed, Mary and Bob are in free fall.
The normal reaction is zero, so the only force
acting on them is the force due to gravity.
This is the condition for apparent
weightlessness.