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G481
MECHANICS MODULE 1:
MOTION
Answer Booklet
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RAB Plymstock School
Lesson 2 questions – scalars and vectors
1
The table shows vector and scalar quantities.
Speed, acceleration
Energy, power
S
Force, pressure
Velocity, displacement
V
In the blank spaces provided, label the pair of quantities that are both vectors with a V
and the pair that are both scalars with an S.
(2)
2
Explain the difference between a scalar and a vector including one example in
your answer.
Scalar……A scalar is a quantity that has magnitude only, speed……………………
…………………………………………………………………………………………..
…………………………………………………………………………………………..
…………………………………………………………………………………………..
Vector……A vector is a quantity that has magnitude and direction, velocity………...
…………………………………………………………………………………………..
…………………………………………………………………………………………..
……………………………………………………………………………………… (4)
3
Put the following quantities into a list of vectors and a list of scalars.
MASS, FORCE, SPEED, VELOCITY, WORK, DISPLACEMENT
Vector
FORCE
VELOCITY
DISPLACEMENT
Scalar
MASS
SPEED
WORK
(3)
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Lesson 3 questions – Vector addition
1 a) Explain the difference between a scalar and a vector quantity.
…………Scalar has magnitude only. A vector has magnitude and direction.……
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………… (2)
b)
Fig1 shows the path of a car as it travels around a right angles bend.
25ms-1
A
B
25ms-1
The car travels from point A to point B in 7.6 seconds at a constant speed of 25ms-1.
i)
Calculate the distance the car travels in 7.6s.
Distance = speed x time
= 25 x 7.6
Distance=……190………m (2)
ii)
Draw a line on fig1 to show the displacement of the car having travelled from
A to B.
(1)
iii)
Explain why the velocity of the car changes as it travels from A to B although
the speed remains constant.
…………Velocity is a vector and speed is a scalar. The car’s direction changes and
therefore its velocity must.……………………………………………………………..
………………………………………………………………………………………(2)
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iv)
Using a labelled vector triangle, calculate the magnitude of the change in
velocity of the car (velocity at B – velocity at A).
25m/s
25m/s
resultant V
resultant V = √(252 + 252)
= 35.4 m/s
Magnitude of velocity change = ………35.4…………….ms-1. (4)
v)
State and explain whether the car is accelerating as it travels around the bend
from A to B.
………Acceleration is change in velocity. Car has change in direction and therefore
change in velocity and is therefore accelerating.…………………………….
…………………………………………………………………………………………..
…………………………………………………………………………………………..
……………………………………………………………………………………… (2)
Total (13)
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RAB Plymstock School
Lesson 4 questions – Vector components
1a)
Fig 1.1 shows a skier being pulled up a slope at constant speed.
The tension in the wire pulls the skier with a force of 400N that acts at 40° to the
slope.
400N
40°
Fig1.1
i)
Explain with reference to the forces acting on the skier why he travels at
constant speed.
………Forces are balanced…………………………………………………………
………Component of tension parallel to slope = the friction down the slope………
…………………………………………………………………………………… (2)
ii)
Calculate the component of the tension in the wire.
1.
Parallel to the slope
400cos40=306
Component = ……306N………….
2.
Perpendicular to the slope
400cos50=257
Component = ……257N………….
(3)
b)
Describe two possible effects on the skier if the tension in the wire was
suddenly increased.
………forces no longer balanced accelerates
lift up off slope
greater air resistance
less friction
forced to let go
…………… (2)
Total (7)
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2
Fig 2.1 shows a force of 6.0 N acting at 30° to the horizontal.
6.0N
30°
horizontal
Calculate the component of the force that acts
i)
horizontally,
6cos30=5.196
horizontal component………5.2…………….N
ii)
vertically,
6cos60=3.0
vertical component……3.0……………….N
(2)
Total (2)
3
Fig 3.1 shows a ball kicked from the top of a cliff with a horizontal
velocity of 5.6ms-1. Air resistance can be neglected.
5.6ms-1
After 0.90 seconds the vertical component of the velocity is 8.8ms-1.
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RAB Plymstock School
i)
Use a vector triangle to determine the resultant velocity of the ball after 0.90s.
scale must be shown if scale drawing
5.6m/s
Ө
resultant v
8.8m/s
Pythagoras: v=√(8.82 + 5.62)
=10.43m/s
cosӨ=5.6/10.43
= 57.53°
Resultant velocity: magnitude = ………10.4…………..ms-1.
Angle to the horizontal……57.5………………° (4)
ii)
Calculate
1. the vertical distance the ball falls in 0.90 s,
2. the horizontal distance the ball travels in this time.
1
Distance = speed x time
=(8.8 x 0.9)/2 (since ball is accelerating down with constant g
– but started at 0 speed)
=3.96m
2
distance = 5.6 x 0.9
=5.04m
the vertical distance = ………3.96m ……………(2)
the horizontal distance = ……5.04m ……………… (1)
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RAB Plymstock School
Lesson 7 questions – motion graphs: Displacement
1a)i) Define speed.
………the rate of change of
distance………………………………………………………
………………………………………………………………………………………
ii)
Define velocity.
………the rate of change of
displacement…………………………………………………
…………………………………………………………………………………………
iii)
State the differences between these quantities.
………speed has magnitude only as it is a scalar
quantity…………………………………
………velocity has both magnitude and direction as it is a vector quantity………
…………………………………………………………………………………………
……………………………………………………………………………………… (4)
b)
Fig 1.1 shows a fairground big wheel. The wheel is rotating in a vertical plane
and carriages travel round a circle of diameter 40m at a constant speed. The carriages
complete one revolution in 3.5 minutes.
i)
A carriage moves half a revolution from X to Y. Calculate
1.
the speed of the carriage
πr/(3.5x60/2)=1.20 ms-1
speed = ……0.6………ms-1
2.
the magnitude of the average velocity of the carriage
40/(3.5x60/2)=0.38 ms-1
magnitude of the average velocity= ……0.4………ms-1
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ii)
The carriage in (bi) returns to point X. Calculate, for the complete revolution,
1.
the speed of the carriage
2πr/3.5x60
speed = ……0.6………ms-1
2,
the average velocity of the carriage.
displacement=0
average velocity= ……0.0………ms-1
Comment on your answer.
………The carriage has ended back up in same place so displacement is zero and
therefore velocity is zero.………………………………………………………………
……The speed stays constant around the wheel……………………………
…………………………………………………………………………………………
……………………………………………………………………………………… (3)
c)
Describe how the instantaneous velocity of the carriage at Y differs from the
average velocity of the carriage after traveling from X to Y.
………The instantaneous velocity would be the same as the magnitude of the speed
whereas the average velocity shows over a period of time. ………
The velocity up would be 0.4 and the velocity down would be -0.4 so the average of
these is 0.
since velocity has direction………………………………………………………… (3)
Total [14]
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Answers and worked solutions
1.
Speed 
distance
 distance  speed time  340 m s -1  0.375 s  127.5 m
time
= 128 m to 3sf
2.
When you have read chapter 9 you will be able to do this calculation using ideas
about acceleration. There is a simpler way: the girl's average speed is
distance 15 m
average speed 

 8.57 m s 1
time
1.75s
.
As she dives from rest, her final speed must be twice this average (if we assume that
she accelerates uniformly). So, her speed on entry is
maximum speed  2  8.57 m s1  17.14 m s1  17.1 m s1 .
=17 ms-1 to 2sf
3.
The solution to this problem is similar to that for question 2. Find the average speed
and then double it. The average speed is
average speed 
distance 20.75 m

 4.15 m s 1
time
5s
.
Hence:
maximum speed  2  4.15 m s1  8.30 m s1.
4.
The velocity is calculated from the gradient of the distance–time graph. At first the
graph rises, implying a positive gradient, which gradually decreases to zero, implying
a speed of zero. The gradient then becomes negative and gradually steeper, showing
an increasing negative velocity.
time
5.
The top speed of the cyclist is 100 km h–1. If the cyclist accelerates uniformly from
rest, the average speed must be half this, i.e. 50 km h–1. To find the time taken to
travel 3 km at this speed:
speed 
distance
distance
3 km
 time 

 0.06 h  216 s
time
speed
50 km h 1
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.
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6.
To convert 50 km h–1 to m s–1:
50 km h–1 = 50 km h–1 x 1000 m km–1 / 3600 s h–1 = 13.8 m s–1.
The average speed of the car is half the maximum, or 6.9 m s –1. Then:
distance = speed x time = 6.9 m s-1 x 1.4 s = 9.7 m
When the cat runs out, you first move for 0.6 s at the maximum speed of 13.8 m s –1.
Hence, the additional 'thinking distance' travelled is
distance = speed x time = 13.8 m s–1 x 0.6 s = 8.3 m
Thus, the total distance travelled is 8.3 m + 9.7 m = 18 m.
7.
4
20
130
140
time / s
The engine travels 440 m at 4 m s–1.
time = distance / speed = 440 m / 4 m s–1 = 110 s
Distance travelled = area under the speed-time graph.
total distance = (1/2 x 20 x 4) + (110 x 4) + (1/2 x 10 x 4) = 500 m
The total time for the journey is then 20 s + 110 s + 10 s = 140 s. The average speed
is then
distance 500 m
Average speed 

 3.57 m s 1
time
140 s
8.
The ball falls to the floor in 0.63 s. Its average speed during the fall is
average speed 
distance
2m

 3.17 m s 1.
time
0.63 s
Its maximum speed (the speed with which it hits the floor) is then 2 x 3.17 m s–1 =
6.35 m s–1. On the rebound, the average speed is
average speed 
distance 1.5 m

 2.73 m s 1.
time
0.55 s
The time in this equation is calculated from 1.18 s – 0.63 s = 0.55 s. The maximum
speed on the rebound must then be 2 x 2.73 m s–1 = 5.45 m s–1.
Speed /ms-1
5.45
6.35
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0.63
1.18
Time /s
RAB Plymstock School
Lesson 8 questions– acceleration
1
Fig 1.1 shows a trolley on a bench surface, connected to a mass M by a string.
The mass is released and the trolley moves along the surface. The graph shows the
variation of velocity v of the trolley with time t for the motion from A to B.
a)i)
Calculate the acceleration of the trolley between A and B.
1.8/0.8
acceleration = ……2.25…………… ms-2 (2)
ii)
Show that the distance from A to B is 0.72m.
area under graph = ½ 0.8 x 1.8 = 0.72
(2)
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b)
When the trolley reaches B the mass M has just reached the floor.
i)
Ignoring any resistive forces, calculate the time it takes the trolley to
travel from B to C.
constant speed – no acceleration. t=s/v = 1.28/1.8
time for B to C = ……0.71……………. s (3)
ii)
On the graph complete a line showing the trolley moving from B and
coming to rest at the pulley at C.
(3)
Total [10]
2
a)
Define acceleration
……The rate of change
of velocity
(change in velocity / change in time)…………………………………………… (2)
b)
Fig 2.1 shows a graph of velocity against time for a train that stops at a station.
i)
For the time interval t=40s to t=100s, calculate
1.
The acceleration of the train,
v-u/t=0-25/60
acceleration = ……-0.42..ms-2
2.
The distance travelled by the train.
½ x 60 x 25
distance ……750………. m (5)
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ii)
Calculate the distance travelled by the train during its acceleration from rest to
-1
25ms .
t=40s
½ x 40 x 25
distance…500……………..m
(2)
iii)
Calculate the journey time that would be saved if the train did not stop at the
station but continued at a constant speed of 25ms-1.
Total distance = 750 + 500 = 1250
at 25m/s So t = 1250/25 = 50s
total time to decelerate and accelerate and stop at station = 60 + 90 + 40 = 190
So time saved = 190 - 50
time saved ………140…………….s (4)
Total [13]
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Lesson 9 questions – uniform acceleration
1
In this question, two marks are available for the quality of written
communication.
An object falls vertically through a large distance from rest in air. Describe and
explain the motion of the object as it descends in terms of the forces that act and its
resulting acceleration.
……At rest acceleration = g……………………..
……resultant force = W or air resistance R = 0
……
Air resistance force acts up / opposes weight
resultant force = W – R
acceleration is < g
…………………………………………………………………………………………..
R increases with v
eventually R = W / resultant force = 0
acceleration = 0 /constant(terminal) velocity / speed
…………………………………………………………………………………… (6)
1 for spelling and punctuation
1 mark for technical terms
b)
Explain how a free-fall diver can increase the rate at which she descends
through the air.
…reduce the amount of air resistance
dive head first / bring arms in / streamlining
…………………………………………………………………………………… (2)
Total [10]
2
a)i)
Define speed……rate of change of distance / distance travelled over
unit time……………………………………………………(1)
ii)
Distinguish between speed and velocity…speed has magnitude
only……………………………..
……velocity has magnitude and direction (2)
b)
Use the equations given below, which represent uniformly accelerated motion
in a straight line, to obtain an expression for v in terms of u, a and s only.
v = u + at
t=(v-u)/a
sub t into: s = (u + v)t/2
s=(u+v)(v-u)/2a
2as= uv –uv + v2 - u2
2as = v2 - u2
v2 = u2 + 2as
(2)
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Fig 2.1 shows a ball kicked from the top of a cliff with a horizontal velocity of 5.6ms1
. Air resistance can be neglected.
5.6ms-1
i)
Show that after 0.90s the vertical component of the velocity is 8.8ms-1.
v=u+at
v=0+9.81x0.9
v=8. 8ms-1.
ii)
(2)
Use a vector triangle to determine the resultant velocity of the ball after 0.90s.
scale must be shown if scale drawing
5.6m/s
Ө
resultant v
8.8m/s
Pythagoras: v=√(8.82 + 5.62)
=10.43m/s
cosӨ=5.6/10.43
= 57.53°
Resultant velocity: magnitude = ……10.4……………..ms-1.
Angle to the horizontal……57.5………………° (4)
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iii)
Calculate
3. the vertical distance the ball falls in 0.90 s,
4. the horizontal distance the ball travels in this time.
1
Distance = speed x time
=(8.8 x 0.9)/2 (since ball is accelerating down with constant g
– but started at 0 speed)
=3.96m
3
distance = 5.6 x 0.9
=5.04m
the vertical distance = 3.96m ……………(2)
the horizontal distance = ……5.04m ……………… (3)
Total (14)
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RAB Plymstock School
Lesson 11 questions – projectile motion 1
Take g = 9.8 ms-2
1
A skier moves at 11.0 ms-1 down a 16º slope. What is the skier’s
a)
vertical velocity
vy=11sin16
…3.03……… ms-1 (2)
b)
horizontal velocity
vx=11cos16
2
…10.6……… ms-1 (2)
Total [4]
-1
A ball is thrown at 20º to the horizontal at 20.0 ms , What is the ball’s
a)
vertical velocity
vy=20sin20
…6.84……… ms-1 (2)
b)
horizontal velocity
vy=20cos20
…18.8……… ms-1 (2)
c)
Describe the shape of the trajectory that the ball follows.
…………………Parabolic……………………………………………………………
……………………………………………………………………………………… (1)
Total [5]
-1
3
A ball is thrown vertically upwards at 19.6 ms .
a)
Fill in the table of the velocity of the ball at different times below:
Time t/s
Velocity v/ ms-1
v=u + at
1.0
v= 19.6 +((-9.8)x1)
v=9.8
v= 19.6 +((-9.8)x2)
2.0
v=0
v= 19.6 +((-9.8)x3)
3.0
v=-9.8
v= 19.6 +((-9.8)x3)
4.0
v=-19.6
(8)
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c) What is the displacement after
i)
2.0s
s=ut + ½ at2
s = 19.6x2 + ½ ((-9.8)x22)
s=19.6m………………………………… (2)
ii)
4.0s
s=ut + ½ at2
s = 19.6x4 + ½ ((-9.8)x42)
s=0m ………………………… (2)
c)
How far does it travel in the first 4.0 seconds?
………19.6 x 2 = 39.2m………………………………………… (2)
d)
Draw a sketch velocity-time graph for motion of the ball the instant after it has
left the hand until it is caught again. Take the upward direction to be positive and
assume no air resistance.
v / ms-1
19.6
2
4
t/s
-19.6
Any 4 – line counts as 2, going +ve to –ve and straight. Count axes labels and units
separately. (x and y both correct for 1)(4)
e)
Explain how the graph can be used to find the distance and displacement.
………Area under graph
………distance add up (don’t worry about negative area)
………displacement add up taking negative area into account (2)
Total [20]
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Lesson 13 answers – projectile motion
1
Here you can work on predicting the positions of objects that are
accelerated. One archery competition requires archers to fire a total of 90
arrows for a maximum possible score of 900. The targets are 1.22 m in
diameter at 60, 50 and 40 metres distance. You can model the motion of the
arrow to find out what problems the archer faces.
The arrow leaves the bow at 60 m s–1 and travels at almost this speed
horizontally for the whole of its flight.
The arrow, of course, falls because of the acceleration due to gravity. You can
find its position at any moment by working out how far it has moved
horizontally and how far it has fallen vertically.
a.
The archer shoots the arrow horizontally at the 40 m target. How far
does it drop over this range?
Time of flight
40 m

 0.67 s
60 m s –1
Vertical drop
2
1
1
2 
h  gt 2   9.81 m s – 2   s   2.2 m
2
2
3 
b.
How would the archer make allowance for this fall? Think carefully
before committing yourself.
………… Aim above the target.…………………………………………..
c
Now try to calculate the fall at 50 m and 60 m.
Similar calculations
2
h50 
h60 
1
5 
 9.81 m s – 2   s   3.4 m
2
6 
1
2
 9.81 m s – 2  1 s  4.9 m
2
d.
50 years ago the release speed of an arrow was about 30 m s –1. What
effect would this have on the vertical distance the arrow fell? Calculate the
drop for a range of 60 m to check your answer.
Twice the trip time so four times fall, at 60 m now drops 20 m
e.
We have ignored the effect of air resistance in these calculations. How
could you take account of it?
… You can explore it most easily by making a computer model of the motion.
……………………………………………………………………………………..
2
A body is projected horizontally from ground level with a speed of 24ms-1 at
an angle of 30 degrees above the horizontal. Neglect air resistance and calculate:
a)
The vertical part of the velocity
……12m/s………….. (2)
b)
The time to reach its highest point
……1.2s………….. (2)
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c)
The greatest height reached
…7.3m…………….. (2)
d)
The horizontal range of the body
……51m………….. (2)
Total [8]
3
fig 3.1 shows the trajectory of a projectile fired from the ground if air
resistance is neglected.
parabola
a)
Describe the changes in energy of the projectile that occur throughout its
flight.
… A projectile has maximum kinetic energy the instant it leaves the force
projecting it.
It loses kinetic energy as it travels upwards as this energy is being
transformed into gravitational energy and so has minimum kinetic energy at
the top of the trajectory (it is not zero since it still has the same amount of
horizontal speed as it started with).
It therefore has maximum gravitational potential energy at the top of the
trajectory.
The reverse happens on the way down as the gravitational energy is
transformed into kinetic energy on the way down back to the maximum as it
hits its target,
(this is the same amount as at the beginning of the flight because of the
conservation of energy.) (5)
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b)
Describe the changes in speed of the projectile that occur throughout its flight.
……Horizontal speed stays constant
since air resistance neglected
vertical velocity max at zero height
This decreases as since gravity is acting in the downward direction
To zero at top of flight
Projectile accelerates with gravity on the way down
To maximum speed at instant before hitting ground
This is the same as the speed it started with………………………………… (5)
c)
Explain qualitatively what would happen if the projectile experienced air
resistance. Use a diagram to help your explanation if necessary.
(The red line shows the projection of a projectile without the affects of air
resistance taken into account.
The blue line shows the projection of a projectile with the affects of air
resistance taken into account.)
You can see that a projectile loses vertical and horizontal speed
and because of this doesn’t go as high
or as far as it would without air resistance… (3)
Total [13]
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