Higher Physics
Our Dynamic Universe
Notes
Teachers Booklet
Previous knowledge
This section builds on the knowledge from the following key areas from
Dynamics and Space Booklet 1 - Dynamics
• Velocity and Displacement – vectors and scalars
• Velocity-time graphs
• Acceleration
Learning Outcomes
At the end of this section you should be able to:
o Carry out calculations using the following kinematic relationships
• s = vt
• v = u + at
• s = ut + ½ at2
• v2 = u2 + 2as
• s = ½ (u + v) t
o Draw displacement-time graphs
o Draw velocity-time graphs
o Draw acceleration-time graphs given a velocity-time graph
o Draw the velocity time graph for objects thrown vertically upwards
o Draw the velocity time graph for bouncing objects
• All graphs are restricted to constant acceleration in one
dimension, inclusive of change of direction
o State that the gradient of a displacement-time graph gives
velocity
o State that the gradient of a velocity-time graph gives acceleration
o State that the area under a velocity-time graph is equal to the
displacement
Equations of Motion
The equations of motion are
1. v = u + at
2. s = ut + ½ at2
3. v2 = u2 + 2as
where s = displacement (m)
u = initial velocity (ms-1)
v = final velocity (ms-1)
a = acceleration (ms-2)
t = time (s)
If you have any three pieces of information from the list ‘suvat’ you
are able to find the other two using the appropriate equation.
Deriving the Equations (for information)
1. Acceleration is defined as the rate of change of velocity
so a = v-u
t
ð v – u = at
ð v = u + at
velocity
2.
v
Displacement = area under
Velocity-time graph
s = Area 1 + Area 2
= ut + ½ (v-u) t
2
u
1
0
3.
But v-u = at (from equation 1)
t (time)
v = u + at
v2 = (u + at)2
ð s = ut + ½ at x t
ð s = ut + ½ at2
square both sides of the equation
v2 = u2 + 2uat + a2t2
But s = ut + ½ at2
v2 = u2 + 2a (ut + ½ at2)
v2 = u2 + 2as
Motion – Equations and Graphs
3
Equations of Motion
Example 1
A car accelerates from rest. Its velocity after 8 seconds is 26ms-1.
Calculate: (a) the acceleration
(b) how far it travels in 15 seconds.
s=?
u = 0 ms-1
v = 26 ms-1
a=?
t = 8s
(a)
a = v-u = 26 – 0
t
8
= 3.25ms-2
(b) Using the value for a calculated in part (a)
s = ut + ½ at2
ð
s = (0 x 15) + (½x 3.25 x 152)
ð
s = 365.6m
Example 2
SQA 2008 Q21 adapted
A car is driven at a constant speed of 30ms-1 until it reaches the start
of the braking zone at P. The brakes are then applied.
Calculate the length of the braking zone if the car slows down at a
constant rate of 9 ms-2 and the speed at Q is recorded as 12 ms-1.
s=?
u = 30 ms-1
v = 12 ms-1
a = -9ms-2
t=?
v2 = u2 + 2as
122 = 302 + 2 x (-9) x s
144 = 900 + (-18s)
18s = 900 – 144
s = 756 = 42m
18
Motion – Equations and Graphs
4
ODU Problem booklet 1
P7 Q1 - 10
Graphs
Displacement, velocity and acceleration are all interrelated.
The gradient of a displacement-time graph gives the velocity.
The gradient of a velocity-time graph gives the acceleration.
The area under a velocity-time graph gives the displacement.
When drawing graphs
o Use a ruler
o Label the axes
o Put zero at the origin
Displacement
Constant
Velocity
s
0
t
Acceleration
Velocity
v
0
Decreasing
velocity
s
s
0
t
0
v
t
a
0
Increasing
velocity
v
0
t
a
t
0
t
a
t
0
Motion – Equations and Graphs
t
5
0
t
Comparison between v-t and a-t graphs
for an object moving in a straight line
A
Velocity (ms-1)
C
D
1 2 3 4 5 6 7 8 9 10 9 10 11 E
Time (s)
A
B
E
F
4 B
F
7 6 5 4 3 2 1 0 -­‐1 0 -­‐2 -­‐3 -­‐4 -­‐5 -­‐6 -­‐7 Acceleration (ms-­‐2) 3 2 C
1 0 -­‐1 0 1 2 3 4 5 6 7 8 11 D
-­‐2 -­‐3 -­‐4 Time (s)
From A to B
a = v-u = 6 – 0
t
2
= 3 ms-2
From B to C
a = v-u = 6 – 6
t
2
= 0 ms-2
From C to E
a = v-u = -6 – 6
t
4
= -3 ms-2
From E to F
a = v-u = 0- (-6)
t
2
= 3 ms-2
Motion – Equations and Graphs
6
ODU Problem booklet 1
P9 Q1 - 7
Object thrown upwards
B
v
A
Displacement
upwards
B
0
t
Displacement
downwards
A
C
C
Gravity acts downwards at a constant value of 9.8 ms-2.
At A – Ball has velocity, which decreases as it goes up.
At B – Ball has zero velocity (at top of flight)
At C – Ball increases in velocity as it falls (opposite direction – so
negative). Velocity has the same magnitude at A but is in the
opposite direction.
Total displacement = zero
speed
Speed-time
A
Graph
for comparison
B
0
t
Distance
up
Distance
down
Motion – Equations and Graphs
7
C
Velocity-time graph for bouncing ball
Velocity ms-1
A
D
B
E
Time (s)
F
C
AB
B
BC
CD
DE
E
EF
– ball leaves hand and decelerates upwards
- top of flight
– ball accelerates downwards
– ball touches ground with a large force causing it to change
direction in a very short time period.
– ball decelerates upwards
– top of flight
- ball falls again
Each time this repeats some energy is lost as heat (and sound), so the ball
does not rise to the same height as before.
The gradient of the graph is negative and constant – which shows that
acceleration due to gravity is constant at -9.8 ms-2.
The area under the graph represents the displacement during each part
of the motion.
Displacement
Velocity –time
graph
Area under
the graph
Displacement- Read
time graphs
straight
from line of
best fit
Motion – Equations and Graphs
Velocity
Read straight
from line of
best fit
Δs
Δy
v = = =m
Δt
Δx
Acceleration
Δv v − u
Δy
a = =
= =m
Δt
t
Δx
a = gradient of graph
v = gradient of
graph
8
ODU Problem booklet 1
P15 Q8 - 10
Previous knowledge
This section builds on the knowledge from the following key area from
Dynamics and Space Booklet
• Projectiles
Learning Outcomes
At the end of this section you should be able to
o Resolve initial horizontal and vertical velocity from velocity acting
at an angle to the horlzontal
o State that horizontal motion for projectiles is constant
o State that vertical motion for projectiles is affected by the force
due to gravity
o Compare projectiles to objects in freefall
Gravitation
9
1.
No horizontal motion
Direction of
motion
Forces
Velocity
Acceleration
Vertical
Weight
Changes with
time
-9.8 ms-2
No horizontal motion
Initial vertical motion = 0 ms-1
Final vertical motion => v = u + at = at (since u = 0)
Acceleration = -9.8 ms-2
(Remember – if upwards is positive, acceleration due
to gravity is negative because it is a vector quantity)
Example 3
A ball is dropped off a bridge. It takes 4 seconds to land in the
water. How high is the bridge?
s = ut + ½ at2
ð
s = (0 x 4) + (½x 9.8 x 42)
s=?
ð
s = 0 + 78.4
u = 0 ms-1
ð
s = 78.4m
v=?
a = 9.8 ms-2
t = 4s
Example 4
A ball is dropped 1.6 m onto a force
sensor. Show that the final velocity is
5.6m.
SQA 2006 Q22 adapted
s = 1.6m
u = 0 ms-1
v=?
a = 9.8 ms-2
t=?
Gravitation
v2 = u2 + 2as
v2 = 0 + 2 x 9.8 x 1.6
v = √31.36
v = 5.6 ms-1 as required.
10
1.
No horizontal motion
B
A
A
At A - the object has maximum upwards velocity - uv
At B – the object has reached its maximum height and
vertical velocity is zero.
At C – the objects velocity has the same magnitude as
at A but is in the opposite direction.
C
Example 5
A ball is thrown into the air with a velocity of 7ms-1. How long does it
take to get to the maximum height? What is that height?
s=?
u = 7 ms-1
v = 0 ms-1
a = -9.8ms-2
t=?
Time
t = v-u = 0 – 7
a
-9.8
= 0.71 s
Height
v2 = u2 + 2as
02 = 72 + (2 x -9.8 x s)
0 = 49 -19.6s
19.6s = 49
s = 49 = 2.5 m
19.6
Example 6
A person on the end of a bungee cord is fired upwards to a height of
29.9m.. They take 2.47s to reach their maximum height.
What was their initial velocity?
s = 29.9m
u = ? ms-1
v = 0 ms-1
a = -9.8ms-2
t = 2.47s
Gravitation
u = v – at
= 0 – (- 9.8 x 2.47)
= 24.2 ms-1
11
2. An object thrown horizontally
height
range
Look at horizontal and vertical velocity separately.
Horizontal velocity is constant. The higher the horizontal velocity the
further the object travels as it falls.
Initial vertical velocity = zero - use v= u + at to calculate final velocity.
Calculate displacement using area under a graph OR one of the equations
of motion with ‘s’ in it.
Direction of
motion
Horizontal
Vertical
Forces
Velocity
None
Weight
Constant
Changing with
time
Acceleration
None
-9.8 ms-2
Example 7
A ball rolls off a table with a velocity of 2.3 ms-1. What is the height of
the table if it lands 1.2 m away from the table?
Horizontal
sh = 1.2 m
uh = 2.3 ms-1
vh = 2.3 ms-1
ah = 0 ms-2
th = ?
t = sh
uh
= 1.2
2.3
= 0.52s
Gravitation
Vertical
sv = ?
uv = 0 ms-1
vv = ?
av = 9.8 ms-2
tv = 0.52 s
s = ut + ½ at2
s = (0 x 0.52) + (½x 9.8 x 0.522)
s = 1.3 m
ð
ð
12
2. An object thrown horizontally
Example 8
A stuntman on a motorcycle jumps a river, which is 5.1 m wide. He lands
on the edge of the far bank, which is 2.0 m lower than the bank from
which he takes off. Calculate his minimum take off speed to ensure a
safe landing.
Vertical
(SQA
2004 Q2 adapted)
s = ut + ½ at2
sv = 2.0m
ð
2 = (0 x t) + (½x 9.8 x t2)
uv = 0 ms-1
ð
2 = 4.9t2
vv = ? ms-1
ð
t = √(2/4.9)
av = 9.8 ms-2
ð
t = 0.64s
tv = ?
Horizontal
sh = 2.0m
uh = ? ms-1
vv = uv ms-1
ah = 0 ms-2
th = 0.64s
Gravitation
uh = sh
t
= 5.1
0.64
= 7.96875
= 8.0 ms-1
13
Graphs for an object thrown horizontally
The information about the horizontal and vertical velocity for an object
can be combined to calculate the resultant velocity.
Vh (ms-1)
Vv (ms-1)
10
40
0
4
0
4
time (s)
time (s)
From the two graphs we can tell that at 4s the horizontal velocity is
10ms-1 and the vertical velocity is 40ms-1.
To find the resultant velocity either
A
draw a vector diagram. Or B – use Pythagoras and sohcahtoa
10ms-1
Length of X = √10! + 40!
= 41.2 ms-1
Tan θ = opp = 40
adj
10
40ms-1
X
=> θ = 76°
θ
Gravitation
14
ODU Problem booklet 1
P17 Q1 - 3
3. An object thrown upwards at an angle
θ
Resolve the velocity into its horizontal and vertical components
(this is possible because any vector can be split up into two other vectors
at right angles to one another)
Vh = cos𝜃 = adj = Vh
hyp v
Vv =
=> Vh = v cos𝜃
sin𝜃 = opp = Vv
hyp v
=> Vv = v sin𝜃
height
uv
vv
range
•
The path of a projectile is symmetrical about the highest
point in the horizontal plane.
•
Initial vertical velocity = -final vertical velocity
Uv = -Vv
•
The time of flight = twice the time to the highest point
•
Vvertical at highest point = zero
Direction of
motion
Horizontal
Vertical
Gravitation
Forces
Air resistance is
negligible so no
forces
Air resistance is
negligible so only
the force of
gravity
Velocity
Acceleration
Constant
Zero
Changing at a
constant rate
Constant or uniform
acceleration of
9.8 ms-2
15
3. An object thrown upwards at an angle
Example 9
A gun is fired with a velocity of 16 ms-1 at an angle of 40° above the
horizontal.
Find
(a)
the horizontal and vertical components of the balls velocity
(b)
The maximum height reached by the ball
(c)
The horizontal distance the ball travels.
a)
Uh = 16 cos 40 = 12.3 ms-1
16 ms-1
uv
40°
Uv = 16 sin 40 = 10.3 ms-1
uh
b) Vertical
s=?
uv = 10.3 ms-1
vv = 0 ms-1
a = -9.8 ms-2
t=?
c)
v2 = u2 + 2as
02 = 10.32 + (2 x -9.8 x s)
19.6 s = 10.32
s = 5.4 m
Vertical
vv = uv + at
0 = 10.3 + (-9.8 x t)
t = -10.3 = 1.05 s
-9.8
Gravitation
Total time in air = 2 x 1.05 = 2.1s
Horizontal
s = vt = 12.3 x 2.1 = 25.8m
16
3. An object thrown upwards at an angle
Example 10
A basketball player throws a ball with initial velocity of 6.5ms-1 at an
angle of 50° to the horizontal. The ball is 2.3 m above the ground
when released.
The ball travels a horizontal distance of 2.9m to reach the top of the
basket. The effects of air resistance can be ignored. (SQA H 2009
adapted)
(a)
(b)
Calculate:
(i) the horizontal component of the initial velocity of the
ball
(ii) the vertical component of the initial velocity of the ball.
If the time taken for the ball to reach the basket is 0.69s
calculate the height h of the top of the basket.
.
(a)
(i) uh = v cos θ
= 6.5 cos 50°
= 4.18 ms-1
s=?
uv = 4.98 ms-1
vv = ? ms-1
a = - a = -9.8 ms-2
t = 0.69s
(ii)
(b)
Gravitation
ð
ð
ð
uV = v sin θ
= 6.5 sin 50°
= 4.98 ms-1
s = ut + ½ at2
s = (4.98 x 0.69) + (½x -9.8 x 0.692)
s = 3.44 + (-2.33)
s = 1.11m
Actual height to top of basket
= 2.3 + 1.11
= 3.41m
17
3. An object thrown upwards at an angle
Example 11
A golfer on an elevated tee hits a golf ball with an initial velocity of
35.0 ms-1 at an angle of 40° to the horizontal.
The ball travels through the air and hits the ground at point R.
Point R is 12 m below the height of the tee, as shown.
The effects of air resistance can be ignored.
SQA 2003
Q21 adapted
a) Calculate:
i) The horizontal component of the initial velocity of the ball
ii) The vertical component of the initial velocity of the ball.
iii) The time taken for the ball to reach its maximum height at
point P.
b) From its maximum height at point P, the ball falls to point Q,
which is at the same height as the tee. It then takes a further
0.48s to travel from Q until it hits the ground at R.
Calculate the total horizontal distance d travelled by the ball.
(a) (i)
(iii)
uh = v cos θ
= 35 cos 40°
= 26.8 ms-1
s=?
uv = 22.5 ms-1
vv = 0 ms-1
a = - a = -9.8 ms-2
t=?
(ii)
uV = v sin θ
= 35 sin 40°
= 22.5 ms-1
t = v-u =0 – 22.5 = 2.3s
a
-9.8
(b) ttotal = 2.3 + 2.3 + 0.48
= 5.08s
sh = uh ttotal = 26.8 x 5.08 = 136.1m
Gravitation
18
ODU Problem booklet 1
P17 Q4 - 12
Previous Knowledge
This section builds on the knowledge from the following key area from
Dynamics and Space Booklet 2 - Forces
• Newton’s Laws
It also builds on the knowledge from the following key area from
Electricity and Energy Booklet 1 • Conservation of Energy
Learning Outcomes
At the end of this section you should be able to
o Analyse motion using Newton’s first and second laws (for forces
acting in one plane only)
o Recognise balanced and unbalanced forces
o Describe friction as a force acting in a direction to oppose motion
o Describe tension as a pulling force exerted by a string or cable on
another object
o Draw the velocity-time graph of a falling object when air
resistance is taken into account, including changing the surface
area of the falling object
o Analyse the motion of a rocket which may involve a constant force
on a changing mass as fuel is used up.
o Resolve forces acting at an angle into two perpendicular forces
o Calculate forces acting at an angle to the direction of movement.
o Resolve the weight of an object on a slope into a component acting
down the slope and a component acting normal to the slope.
o Combine systems of two or more forces acting in two dimensions to
obtain a resultant force.
o Carry out calculations using the following equations
o Ew = Fd
Ek=½mv2
Ep=mgh
E= Pt
o Apply conservation of energy where appropriate
Forces, Energy and Power
19
Newton’s Laws
1. An object will remain at rest or move with constant velocity
(constant speed in a straight line) unless acted on by an external
force.
F
F
m
2. An object will accelerate when acted upon by an unbalanced force.
F
m
F = ma
Acceleration =>
F = Force (newtons –N)
m = mass (kilogramme – kg)
a = acceleration (metres per second squared – ms-2)
Definition of the Newton
One newton is the force required to accelerate a mass of 1kg at a
rate of 1ms-2
An acceleration of 1ms-2 means that the velocity changes by 1ms-1
each second.
Force is a vector quantity. If several forces act on an object they
can be added together. This allows us to calculate whether they
are balanced or unbalanced.
Balanced forces
4N
4N
Forces are balanced so object will remain at rest (or move with
constant velocity if it is moving)
Forces, Energy and Power
20
2
1
Analysing Forces
Unbalanced forces
4N
Original
diagram
6N
2kg
Equivalent
diagram
2N
2kg
If forces are positioned directly opposite one another then the resultant
force is the difference between the forces.
If the forces are not directly opposite you need to draw a vector
diagram.
4N
6N
6N
2kg
9N
4N
X
2kg
3N
2kg
θ
3N
2N
To calculate the resultant force either
a) Do a scale drawing and measure the resultant vector and
angle.
b) Use Pythagoras to calculate the magnitude of the resultant
vector and ‘sohcahtoa’ to calculate the angle.
To calculate the length of side X
X2 = 32 + 42 = 25
X= 5
To calculate the angle
Tan θ = opp = 4
Adj 3
θ = 53.1°
Forces, Energy and Power
21
Analysing Forces
Example 12
What mass is required to lift a mass of 600 kg upwards with an
acceleration of 2 ms-2. Air resistance is 800 N.
F
W = mg = 600 x 9.8 = 5880 N
F = ma = 600 x 2 = 1200 N
Total force required
= 5880 + 1200 + 800
= 7880 N upwards
W = mg
Air resistance
Example 13
Calculate the minimum force required to accelerate a mass of 400 kg
upwards with an initial acceleration of 6 ms-2, when air resistance is
250 N.
W = mg = 400 x 9.8 = 3920 N
a = 6 ms
-2
400 kg
W = mg
250 N
Forces, Energy and Power
F = ma = 400 x 6 = 2400 N
Total force required
= 2400 + 3920 + 250
= 6570 N upwards
22
ODU Problem booklet 1
P21 Q1 – 8, 11
Tension
Tension is a pulling force applied by a string or cable on another object.
a) Frictionless surface
T
2kg
1kg
F
Example 14
Calculate the tension in the rope at T if a force of 6N is applied to
the rope.
Total mass = 1 + 2 = 3kg
If a force of 6N is applied then the acceleration will be
a = F = 6 = 2ms-2
m
3
The tension in the rope at T = 2 x 1 = 2N
Example 15
T
m1= 2000kg
m2= 3000kg
An articulated lorry is returning from dropping off its load. The mass
of the lorry and the trailer are shown in the diagram.
Calculate:
a) The acceleration of the system
b) The tension in connection at T
a) a = F =
9000
= 9000 = 1.8 ms-2
m (2000+ 3000) 5000
b) F = ma = 2000 x 1.8 = 3000N
Forces, Energy and Power
23
Tension
b) Surface including friction
SQA 2011
Q3 adapted
Example 16
A car of mass 1200kg pulls a horsebox of mass 700kg along a
straight, horizontal road. They have an acceleration of 2ms-2. If the
friction between the car tyres and the road is 70 N and the horsebox
and the road is 50N, what engine force is required to maintain that
acceleration?
Total mass = 1200 + 700 = 1900 kg
F = ma = 1900 x 2 = 3800N
Total force required = 3800 + 70 + 50 = 3920N
T
m1= 6000kg
200N
m2= 20,000kg
600N
Example 17
A train pulls a carriage. The train has a mass of 20,000 kg and the
carriage has a mass of 6000 kg. The force of friction on the train is
600 N and on the carriage is 200 N. The whole system accelerates at
4 ms-2.
Calculate:
a) the force of the engine
b) the force that the coupling exerts on the carriage
a) F = ma = (20,000 + 6000) x 4 = 2-4,000 N
Total force required = 104,000 + 600 + 200 = 104,800 N
b) F = ma = 6000 x 4 = 24,000N
Total force = 224,000 + 200 = 24,200 N
Forces, Energy and Power
24
ODU Problem booklet 1
P24 Q 16 - 18
Tension
c) Mass over a pulley
Tension in rope at T = ma = 1 x 2 = 2N
m1
T
Both blocks accelerate.
Total mass = m1 + m2
Force = weight of m2 = m2g
a=F
m
m2
=
F
= m2g
m1+m2
m1+m2
Tension in string at T
= pulling force on m1
=> T = m1a
Example 18
If m1 = 2 kg and m2 = 5 kg, calculate the tension at T, assuming the
surface is frictionless.
Total mass = 2 + 5 = 7kg
a=F
m
=
W = m2g = 5 x 9.8 = 49N
F
= 49 = 7ms-2
m1+m2
7
Tension in string at T
T = m1a = 2 x 7 = 14N
d) Masses suspended over a pulley
m1
m2
1. Which is the greater mass? The system
will rotate in that direction.
e.g. if m2 > m1 it will rotate clockwise
2. W = mg = m2 x 9.8
3. The whole system moves, so
a= W
m1 + m2
Example 19
If m1 = 5kg and m2 = 3kg state the direction of rotation and find the
tension in the string.
m1 > m2 so system will rotate anticlockwise.
W = mg = 5 x 9.8 = 49N
a = F = 49 = 6.125 ms-2
m 5+3
Tension in string = pulling force on m2
Forces, Energy and Power
25
F = 3 x 6.125 = 18.4N
ODU Problem booklet 1
P25 Q 20
Blocks
Calculate:
a) The force between the 6kg block and the 4 kg block
b) The force between the 4 kg block and the 2 kg block.
a) Frictionless surface
6kg
24N
4kg 2kg
Example 20
a = F = 24 = 24 = 2ms-2
m
6 +4 +2
12
Force between 6kg block and 4 kg block
F= ma = (4 +2) x 2 = 12 N
Force between 4kg block and 2 kg block
F = ma = 2 x 2 = 4 N
b) Friction between the surface and the 6kg, 4kg and 2kg blocks is
3N, 2N and 1N respectively.
30N
6kg
4kg 2kg
3N
2N
1N
Example 21
a = F = 24 = 24 = 2ms-2
m
6 +4 +2
12
Force between 6kg block and 4 kg block
F= ma = (4 +2) x 2 = 12 N
Force between 4kg block and 2 kg block
F = ma = 2 x 2 = 4 N
Forces, Energy and Power
26
ODU Problem booklet 1
P24 Q 19
Weight and Lifts
A person of mass 60kg stands on a set of scales in a lift.
The reading on the scales shows the upwards force on the person in
newtons. This is equal to their weight (Newton’s third law)
W = mg = 60 x 9.8 = 588N
a) Lift standing still OR lift travelling at constant speed.
There is no unbalanced force so the reading is the persons weight.
588N
b)
Acceleration upwards has the same effect as deceleration
downwards. You feel as if your weight has increased.
If the rate of acceleration is 2.5 ms-2
F = ma = 60 x 2.5
= 150N
c)
Reading on scales = W + F
= 588 + 150
= 738N
Acceleration downwards has the same effect as deceleration
upwards. You feel as if your weight has decreased.
If the rate of acceleration is 1.5 ms-2
F = ma = 60 x 1.5
= 90N
d)
Reading on scales = W - F
= 588 - 90
= 498N
Freefall – if the lift cable breaks…
If the lift cable breaks both the lift and the person will fall at
the same rate.
The person does not exert a force on the scales, so the reading
will be 0N.
Forces, Energy and Power
27
Weight and Lifts
Example 22
A person of mass 85kg stands on a set of bathroom scales in a lift.
What would be the reading, in newtons, if the lift was
(a) stationary
(b) accelerating upwards at 2ms-2
(c) decelerating upwards at 3ms-2
(d) accelerating downwards at 1.5ms-2
(e) moving at constant speed.
(a) W = mg = 85 x 9.8 = 833N
(b) F = ma = 85 x 2 = 170 N
Ftot = W + F = 833 + 170 = 1003 N
(c) F= ma = 85 x (-3) = -255 N
Ftot = W + F = 833 + (-255) = 578 N
W = mg
(d) F = ma = 85 x -1.5 = -127.5 N
Ftot = W + F = 833 + (-127.5) = 705.5 N
(e) W = mg = 85 x 9.8 = 833N
Example 23
A fully loaded lift has a total mass of 300kg. For safety reasons the
tension in the pulling cable must never exceed 4440N.
What is the maximum upwards acceleration of the lift?
W = mg = 300 x 9.8 = 2940N
Fun = 4440 – 2940 = 1500N
a = F = 1500 = 5 ms-2
m
300
Forces, Energy and Power
28
ODU Problem booklet 1
P22 Q12 - 15
Rockets
1. At the instant of lift-off
Fun = Thrust – (weight +
air resistance)
If Fun > 0 then the rocket
can take off.
a = Fun
m
2. As lift off continues
1. Fuel is used up => mass decreases, so weight
decreases. Fun increases so a increases.
2. Air resistance decreases (as air gets thinner)
3. Gravitational field strength decreases (further from
planet).
Example 24
A rocket produces a thrust of 400,000 N it has a mass of 60,000 kg.
Calculate the acceleration on take off.
W = mg = 6000 x 9.8
= 58,800 N
Fun = 400,000 – 58,800
= 341,200 N
a = F = 341,200 = 56.9 ms-2
m
6000
Forces, Energy and Power
29
ODU Problem booklet 1
P22 Q9 - 10
Resolving Forces
F
Fv
θ
Fh
Force is a vector quantity.
This means a force can be
resolved into two
component forces, which
are perpendicular to one
another.
cos θ = adj = Fh
hyp F
sin θ = opp = FV
hyp F
Fh = F cos θ
FV = F sin θ
Example 25
A box of weight 120N is placed on a smooth horizontal surface. A
force of 20 N is applied to the box as shown in the diagram.
Calculate the work done in pulling the box a distance of 50m along the
surface.
Component of weight acting parallel to the surface
Fparallel = F cos 30°
= 20 cos 30°
= 17.3 N
Work done = Force x distance
= 17.3 x 50
= 866J
SQA 2009 Q3 adapted
Forces, Energy and Power
30
ODU Problem booklet 1
P26 Q1 - 4
Forces in Two Dimensions
Example 26
Find the acceleration of
the mass shown in the
diagram on the right.
110N
25°
60 kg
25°
Starting point –
110N
You have two choices
1. Draw a vector diagram to scale and measure the resultant force
OR
2. The forces are symmetrical so you can work out the horizontal
component for one force, then double it.
Horizontal component of 110N at 25° = 110 cos 25° = 99.7 N
Total force = 2 x 99.7 = 199.4 N
a = F = 199.4 = 3.3 ms-2
m
60
Example 27
a)
Calculate the
vertical component of the
force in each cord if the
tension in the elastic cord
is 4.5 x 103 N.
b) If the mass of the
capsule and occupants is
236 kg, calculate the
initial acceleration
a) F = 4.5 x 103 x cos 21° = 4.2 x 103 N
b) a = Ftot = 2 x 4.2 x 103 = 35.6 ms-2
m
236
Forces, Energy and Power
31
SQA 2005 Q22
adapted
ODU Problem booklet 1
P27 Q5
Forces on a Slope
F parallel
θ
θ
F perpendicular
W = mg
Example 28
4kg
20º
The trolley moves down the slope with constant speed. Calculate the
component of the weight acting parallel with the slope.
F = mg sin θ
= 9.8 x 4 x sin 20
= 13.4 N
Forces, Energy and Power
32
Forces on a Slope
Example 29
800 kg
Friction = 250N
25º
A car of mass 800 kg rolls down a hill. Assuming friction of 250 N
calculate the acceleration of the car down the slope.
F = mg sin θ
=800 x 9.8 x sin 25
= 3313.3 N
Unbalanced force down slope = 3313.3 – 250
= 3063.3 N
a = F = 3063.3 = 3.8 ms-2
m
800
Example 30
The force of friction acting on the crate is 44N. The rope snaps and
the acceleration of the crate is -6ms-2. It comes to a halt, then
accelerates down the slope. Why is the acceleration down the slope
less than the acceleration up the slope?
Going up the slope the component of weight and friction act in the
same direction.
Going down the slope the component of weight and friction act in
opposite directions, so the unbalanced force is less.
Forces, Energy and Power
33
ODU Problem booklet 1
P27 Q6 - 9
Conservation of Energy
We can use the idea of conservation
of energy to allow us to predict what
happens in certain situations.
The pendulum swings between A and
C, passing through point B. At points
A and C it has potential energy. At
point B it has kinetic energy.
Using conservation of energy we can
predict that all the potential energy
at A is converted to kinetic energy at
B.
Ep = mgh
A
Ek = ½ mv2
c
B
Ep = Ek
mgh = ½ mv2
mgh = ½ mv2
v = !2𝑔ℎ
Example 31
A pendulum of mass 0.6kg is
released from a height of
0.2m.
What is the maximum velocity
of the pendulum?
v = !2𝑔ℎ
ð v = √2𝑥 9.8 𝑥 0.2
ð v = 1.98ms-1
Forces, Energy and Power
C
c
Example 32
A pendulum travels at 4.5 ms-1 when
it is at the bottom of its swing.
a) What is the maximum height
gain?
v = !2𝑔ℎ
ð h = v2 = 4.52 = 1.03m
2g 2 x 9.8
b) If the mass is 0.1kg what is the
potential energy at this point?
Ep = mgh
= 0.1 x 9.8 x 1.03
= 1.01J
34
Conservation of Energy
Example 33
SQA 2014 Q23 adapted
A block of wood of mass 0.20kg is suspended from the ceiling by
thin cords of negligible mass. A dart of mass 0.050g is thrown at
the stationary block of wood. Just before the dart hits the block
it is travelling horizontally at a velocity v. The dart sticks into
the block. The dart and block then swing to a maximum height of
0.15m as shown.
Use conservation of energy to show that the velocity of the dart
and block just after the collision is 1.7 ms-1.
ð
ð
ð
Ep = Ek
mgh = ½ mv2
v = 2𝑔ℎ
v = 2𝑥 9.8 𝑥 0.15
v = 2.94
v = 1.7 ms-1 as required
Forces, Energy and Power
35
ODU Problem booklet 1
P29 Q1 - 6
Momentum
Learning Outcomes
At the end of this section you should be able to
o State that momentum can be calculated using p=mv
o State the law of conservation of momentum – When two objects
collide (or explode) the total momentum before the collision (or
explosion) is equal to the total momentum after the collision (or
explosion) in the absence of net external forces.
o State that momentum is always conserved in inelastic collisions,
elastic collisions and explosions
o State that total energy is always conserved in inelastic collisions,
elastic collisions and explosions
o State that kinetic energy is conserved in elastic collisions but not
in inelastic collisions.
o Use the law of conservation of momentum to calculate unknown
quantities in collisions or explosions in one dimension and in cases
where the objects may move in opposite directions.
o State that impulse= change in momentum
o State that impulse = area under a force-time graph
o Use change in momentum = area under force-time graph to
calculate unknown quantities
o Explain crumple zones, air bags, crash helmets etc. In terms of
impulse and change in momentum.
o Describe the relationship between explosions and Newton’s third
law
Collisions, Explosions and Impulse
36
Momentum
Momentum = mass x velocity
p = mv
p = momentum
m = mass
v = velocity
** CARE = don’t mix up p for momentum with P for pressure or ρ for
density
Law of conservation of momentum
The total momentum before a collision or explosion is equal to the
total momentum after a collision or explosion in the absence of net
external forces.
Any momentum question should always start off with a statement of
the law of conservation of momentum – it qualifies what you do next.
Remember that velocity is a vector quantity so direction is important.
Collisions, Explosions and Impulse
37
Collisions where the objects stick together
Before collision
11ms-1
m1= 1000kg
•
•
•
After collision
V=?
Oms-1
m2= 1200kg
m1= 1000kg
m2= 1200kg
Always start of with a statement of the law of conservation of
momentum
Label the objects and velocities from left to right.
o Use u for velocities before collision
o Use v for velocities after the collision
Substitute in the numbers and work out the answer.
Total momentum before collision
= total momentum after collision
m1u1 + m2u2
= ( m1+m2) v
(11 x 1000) + (0 x 1200) = (1000+ 1200) v
11,000 + 0 = 2200 v
v = 11,000 = 5ms-1 to the right
2200
Example 34
1.2 ms-1
0.65 kg
v
0 ms-1
0.9
kg
0.9
0.65 kg
0.9 kg
The two vehicles collide and stick together. What will the velocity of the
vehicles be after the collision?
Total momentum before collision
= total momentum after collision
m1u1 + m2u2
= ( m1+m2) v
(0.65 x 1.2) + (0 x 0.9) = (0.65 + 0.9) v
0.78 + 0 = 1.55v
v = 0.78 = 0.5 ms-1 to the right
1.55
Collisions, Explosions and Impulse
38
ODU Problem booklet 1
P31 Q1 -4, 6, 9
Collisions where the objects stick together
Example 35
1.2 ms-1
0.7 ms-1
3.0 ms-1
2 kg
? kg
2 kg
? kg
After the collision the trolleys stick together and travel to the left at
0.7 ms-1.
What is the mass of the second trolley?
Total momentum before collision
= total momentum after collision
m1u1 + m2u2
= ( m1+m2) v
(2.0 x 1.2) + (-3.0 x m2) = (2.0 + m2) x -0.7
2.4 + (-3 m2) = -1.4 + (-0.7 m2)
2.4 + 1.4 = (-0.7 m2) + 3 m2
3.8 = 2.3 m2
m2 = 3.8 = 1.65 kg
2.3
Prove that the collision is inelastic.
If the collision is inelastic then kinetic energy will not be conserved.
Ek before ≠ Ek after
Ek before
½ m1u1
+ ½ m2u22
[½ x 2 x 1.22] + [½ x 1.65 x 32]
1.44
+ 7.43
Total Ek before = 8.87J
2
Ek after
½ (m1+m2) v2
[½ x (2+1.65) x 0.72]
Total Ek after = 0.89J
Since total Ek before ≠ total Ek after this is an inelastic collision
Collisions, Explosions and Impulse
39
ODU Problem booklet 1
P31 Q1 -4, 6, 9
Collisions where the objects do not stick together
Before collision
10ms-1
m1= 2kg
•
•
•
After collision
1ms-1
5 ms-1
V2
m2= 10kg
m1= 2kg
m2= 10kg
Always start of with a statement of the law of conservation of
momentum
Label the objects and velocities from left to right.
o Use u for velocities before collision
o Use v for velocities after the collision
Substitute in the numbers and work out the answer.
Total momentum before collision
= total momentum after collision
m1u1 + m2u2
= m1v1 +m2v2
(2 x 10) + (1 x 10) = (-5 x 2) + 10v2
20 + 10 = -10 + 10v2
40 = 10 v2
Example 36
6 ms-1
5kg
0 ms-1
4 ms-1
2 kg
5kg
? ms-1
2 kg
Calculate the velocity of the right hand vehicle after the collision.
Total momentum before collision
= total momentum after collision
m1u1 + m2u2
= m1v1 +m2v2
(6 x 5) + (2 x 0) = (4 x 5) + 2v2
30 + 0 = 20 + 2v2
30 – 20 = 2v2
v2 = 10 = 5ms-1 to the right
2
Collisions, Explosions and Impulse
40
Collisions where the objects do not stick together
Example 37
Before collision
20 ms-1
2 ms-1
After collision
10 ms-1
2 kg
0.4 kg
0.4 kg
V2
2 kg
Trolley A hits trolley B and bounces back with a velocity of 10ms-1 to the
left. Find the velocity of trolley B after the collision and prove that the
collision was elastic.
Total momentum before collision
= total momentum after collision
m1u1 + m2u2
= m1v1 +m2v2
(0.4 x 20) + (2 x 2) = (0.4 x -10) + 2v2
8 + 4 = -4 + 2v2
16 = 2v2
v2 = 16 = 8ms-1 to the right
2
If the collision is elastic then kinetic energy will be conserved.
Ek before = Ek after
Ek before
½ m1u1
+ ½ m2u22
[½ x 0.4 x 202] + [½ x 2 x 22]
80
+
4
Total Ek before = 84J
Ek after
½ m1v12 + ½ m2v22
[½ x 0.4 x 102] + [½ x 2 x 82]
20
+
64
Total Ek after = 84J
2
Since total Ek before = total Ek after this is an elastic collision
Collisions, Explosions and Impulse
41
ODU Problem booklet 1
P31 Q 5,7,8
Types of Collision
There are two types of collision
In both types of collision the total energy and total momentum are always
conserved.
1. Elastic collisions
In elastic collisions the total kinetic energy is conserved. Practical
collisions are never 100% efficient, but some collisions can appear to be
very close to 100% efficient such as the collision between two snooker
balls.
2. Inelastic collisions
In inelastic collisions the total kinetic energy is not conserved.
The only way to be sure which type of collision has taken place is to check
the values for the kinetic energy before and after the collision and see
whether or not it has been conserved.
Summary of momentum
Inelastic
Collision
Elastic
Collision
Explosion
Total Energy
Conserved
Kinetic Energy
Not Conserved
Total momentum
Conserved
Conserved
Conserved
Conserved
Conserved
Not Conserved
Conserved
Collisions, Explosions and Impulse
42
Explosions
Before explosion
After explosion
The total momentum before an explosion is zero.
In these questions there will only be two objects moving in opposite
directions since the force exerted by object A on object B is equal
but opposite to the force exerted by object B on object A. (Newton’s
Third Law)
Total momentum before explosion = total momentum after explosion
0 = m 1 v 1 + m2 v 2
Example 38
A cannon of mass 1200 kg fires a cannonball of mass 15 kg at a
velocity of 60 ms-1 East. Assuming the force of friction is negligible,
calculate the velocity of the cannon just after firing.
Total momentum before explosion = total momentum after explosion
0 = m1v1 + m2v2
0 = (1200 x v1) + (15 x 60)
- 1200 v1 = 900
v1 = -900
1200
v1 = 0.75 ms-1 West
SQA -2004 Q4
Example 39
Two people stand together on an ice rink. One has a mass of 55kg and
the other has a mass of 45 kg. They push apart and the lighter one
moves off with a velocity of 0.7 ms-1. Calculate the velocity of the
other skater.
Total momentum before explosion = total momentum after explosion
0 = m1v1 + m2v2
0 = (45 x -0.7) + (55 x v2)
0 = -31.5 + 55 v2
55 v2 = 31.5
v2 = 0.6 ms-1 to the right
Collisions, Explosions and Impulse
43
ODU Problem booklet 1
P32 Q10 - 13
Momentum and Impulse
Starting from Newton’s second law
F = ma
F=𝑚!
Not
examinable
𝑣−𝑢
𝑡 !
Ft = mv - mu
Impulse = force x time
Change in momentum = mv – mu
During a collision there is a change in momentum caused by the impulse.
If the time of contact is short the applied force must be large. The
force can be decreased if the contact time is increased. This is the
science behind crumple zones, air bags and soft surfaces in playgrounds.
We can illustrate the impulse by drawing a graph.
F
0
F
t
0
t
Impulse = area under a force-time graph.
If the graph is curved you would be given the area. If the graph is
triangular you could be expected to calculate the area.
If you are showing the comparison between two surfaces you must make
the differences clear – use the same graph, but label it clearly.
Example 40
Two samples of foam for the inside of crash helmets are tested. The
force time graph for sample 1 is given. If the time of impact is
doubled using sample 2, show how this alters the applied force if the
change in momentum is the same.
F
Sample 1
Sample 2
0
Collisions, Explosions and Impulse
44
t
Momentum and Impulse
Example 41
A golf club hits a stationary ball of mass 0.4 kg. The ball moves off at
4.2 ms-1 and the time of contact is 0.025s. Calculate the average force
exerted on the ball by the club.
Ft = mv – mu
F x 0.025 = 0.4 (4.2 – 0)
F = 1.68 = 67.2N
0.025
Example 42
A cue exerts a force of 40N on a stationary ball of mass 0.3 kg. If the
impact lasts for 60 ms find the speed the ball leaves the cue.
Ft = mv – mu
40 x 60 x 10-3 = 0.3 (v – 0)
2.4 = 0.3v
v = 8 ms-1
Example 43
SQA
2005
Q5
A car is designed with a “crumple-zone” so that the front of the car
collapses during impact. The purpose of the crumple-zone is to
A decrease the driver’s change in momentum per second
B increase the driver’s change in momentum per second
C decrease the driver’s final velocity
D increase the driver’s total change in momentum
E decrease the driver’s total change in momentum
Collisions, Explosions and Impulse
45
Momentum and Impulse
Example 44
A gymnast of mass 40kg is practising on a trampoline
SQA 2010 Q 23
adapted
The speed of the gymnast, as she lands on the trampoline, is 6.3 ms-1.
She rebounds with a speed of 5.7 ms-1.
a) Calculate the change in momentum of the gymnast.
b) The gymnast was in contact with the trampoline for 0.50s.
Calculate the average force exerted by the trampoline on the
gymnast.
a) Change in momentum = m(v-u)
b)
= 40 x ((-5.7) -6.3)
= 40 x -12
= -480 kg ms-1
Ft = change in momentum
F x 0.5 = -480
F = -960N
Collisions, Explosions and Impulse
46
ODU Problem booklet 1
P33 Q14 - 26