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VCE Physics
Unit 3
Motion in 1 & 2 Dimensions
Revision Questions
1
Motion - Revision Questions
Question type: Equations of Motion
In a road test, a car was uniformally accelerated from rest over a distance of
400 m in 19 sec.
The driver then applied the brakes, stopping in 5.1 sec with constant
deceleration.
Q: Calculate the acceleration of the car for the first 400 m.
A: Firstly, list information:
u=0
v=?
a=?
s = 400 m
t = 19 s
Choose the appropriate equation:
s = ut + ½at2
400 = 0 + ½a (19)2
a = 2.22 ms-2
2
Motion - Revision Questions
Question type: Sketch Graphs
In a road test, a car was uniformally accelerated from rest over a
distance of 400 m in 19 sec. The driver then applied the brakes,
stopping in 5.1 sec with constant deceleration.
The graphs A to F below should be used to answer the questions below. The
horizontal axis represents time and the vertical axis could be velocity or
distance.
A
B
C
D
E
F
Q: Which of the graphs, A to F, represents the velocity time graph for the
entire journey ?
A: Graph B
Q: Which of the graphs, A to F, best represents the distance time graph of
the car for the entire journey ?
A: Graph E
3
Motion - Revision Questions
Question type: Average Speed
In a road test, a car was uniformally accelerated from rest over a distance of 400 m in
19 sec. The driver then applied the brakes, stopping in 5.1 sec with constant
deceleration.
Q: Calculate the average speed for the entire journey, covering both the
accelerating and braking sections.
A: Average Speed = Total Distance Need to know u, the initial speed for the
braking section which equals the final
Total Time
For the accelerated part of journey: speed for the accelerating section.
For accelerating section:
u=0
v = u + at
v=?
= 0 + (2.2)(19)
-2
a = 2.2 ms
= 41.8 ms-1
s = 400 m
Now can calc s
t = 19 s
s = ut + ½at2
Need to calc
To get braking distance, Braking list
= (41.8)(5.1) + ½(-8.2)(5.1)2
acc
use Eqns of Motion
becomes
= 213.2 - 106.6
v = u + at
u = ? List does not
u = 41.8 ms-1
0 = 41.8 + a(5.1) = 106.6 m
v = 0 contain
v = 0 Still not
a = - 8.2 ms-2
a = ? enough info
a = ? enough
Total Distance = 400 + 106.6 = 506.6 m
s = ? the calculate s
s = ? info
4
Total Time = 19 + 5.1 = 24.1 s
t = 5.1s
t = 5.1s
So, Average Speed = 506.6/24.1 = 21 ms-1
Distance = 400 m
Time = 19 sec
For the braking part of journey:
Distance = needs to be calculated
Time = 5.1 sec
Motion - Revision Questions
Question type: Momentum/Impulse
In a car the driver’s head is moving horizontally at 8.0 ms-1 and collides with an
air bag as shown. The time taken for the driver’s head to come to a complete
stop is 1.6 x 10-1 s. This collision may be modelled as a simple horizontal
collision between the head of mass 7.0 kg and the air bag.
Q: Calculate the magnitude of the average contact force that the air bag exerts on the
driver’s head during this collision.
A:
Impulse = Change in Momentum
FΔt = Δ(mv)
So F = Δ(mv)/Δt
= (7.0)(8.0)/(1.6 x 10-1)
= 350 N
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Motion - Revision Questions
Question type: Air Bags/Crumple Zones
In a car the driver’s head is moving horizontally at 8.0 ms-1 and
collides with an air bag as shown. The time taken for the driver’s
head to come to a complete stop is 1.6 x 10-1 s. This collision
may be modelled as a simple horizontal collision between the
head of mass 7.0 kg and the air bag.
Q: Explain why the driver is less likely to suffer a head injury in a collision with
the air bag than if his head collided with the car dashboard, or other hard
surface.
A: The change in momentum suffered by the driver’s head is a FIXED quantity no
matter how his head is brought to rest.
• Therefore the product of F and t (ie Impulse) is also a fixed quantity.
• However the individual values of F and t may be varied as long as their
product always remains the same.
• The air bag increases the time over which the collision occurs, therefore
reducing the size of the force the head must absorb so reducing the risk of
injury.
• The air bag also spreads the force over a larger area, reducing injury risk.
• Without the air bag the driver’s head may hit a hard surface decreasing the
time to stop his head and necessarily increasing the force experienced and
thus the likelihood of injury.
• In addition the force will be applied over a much smaller area increasing the
likelihood of severe injury
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Motion - Revision Questions
Question type: Momentum
A railway truck (X) of mass 10 tonnes, moving at 6.0 ms-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.
10 tonnes
5 tonnes
-1
6.0 ms
Before Collision
X
Y
stationary
v ms-1
After Collision
X
Y
Q: Calculate the final speed of the joined railway trucks after collision.
A: In ALL collisions Momentum is conserved.
So Mom before collision = Mom after collision
Mom before = (10 x 103)(6.0) + (5 x 103)(0)
Mom after = (15x 103)(v)
So v = 60,000/15,000
= 4.0 ms-1
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Motion - Revision Questions
Question type: Impulse
A railway truck (X) of mass 10 tonnes, moving at 6.0 ms-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 at a speed of 4.0 ms-1.
10 tonnes
Before
Collision
6.0
5 tonnes
ms-1
X
Y
stationary
After Collision
X
4.0 ms-1
Y
Q: Calculate the magnitude of the total impulse that truck Y exerts on truck X
A: Impulse = Change in Momentum
Truck X’s change in momentum = Final momentum – Initial Momentum
= (10 x 103)(4.0) – (10 x 103)(6.0)
= -2.0 x 104 kgms-1
The mechanism for this change in momentum is the impulse supplied by Truck Y
So, I = 2.0 x 104 Ns
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Motion - Revision Questions
Question type: Inelastic Collisions
A railway truck (X) of mass 10 tonnes, moving at 6.0 ms-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 at 4.0 ms-1.
10 tonnes
5 tonnes
-1
6.0
ms
Before Collision
X
Y
stationary
After Collision
X
4.0 ms-1
Y
Q: Explain why this collision is an example of an inelastic collision. Calculate
specific numerical values to justify your answer.
A: In Inelastic collisions Momentum is conserved BUT Kinetic Energy is not.
For this collision
pBEFORE = (10 x 103)(6.0) + (5 x 103)(0.0) = 6 x 104 kgms-1
pAFTER = (15 x 103)(4.0)
= 6 x 104 kgms-1
KEBEFORE = ½mv2 = ½(10 x 103)(6.0)2 = 1.8 x 105 J
KEAFTER = ½mv2 = ½(15 x 103)(4.0)2 = 1.2 x 105 J
Thus Mom IS conserved but KE is NOT conserved, therefore this is an inelastic
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collision
Revision
Question type:
Questions
A cyclist is towing a small trailer
along a level bike track (Figure 1).
Question
The cyclist
and bike have a mass of
90 kg, and the trailer has a mass of 40
kg. type:
There are opposing constant forces
Motion - Revision Questions
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.
Question 1
What driving force is being exerted on the road by the rear tyre of the bicycle?
10
Motion - Revision Questions
Question type: Newton’s Laws
A seaplane of mass 2200 kg takes off from a smooth lake as shown. 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 graph shows
the TOTAL FORCE OPPOSING THE MOTION of the seaplane as a function of the
distance travelled.
Total Opposing
Force (N)
10000
8000
6000
A: At d = 500 m the plane is travelling at
CONSTANT VELOCITY,
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So ΣF = 0
Distance (m)
4000
2000
0
Q: What is the magnitude of the net force
acting on the seaplane after it has travelled a
distance of 500 m from the start ?
100 200 300 400 500 600
Motion - Revision Questions
Question type: Newton’s 2nd Law
Q: What is the magnitude of the seaplane’s acceleration at the 200 m mark ?
A: At d = 500 m the seaplane is subject to 0 net force (see previous question).
Thus, Driving Force = Opposing Force = 10,000 N (read from graph).
At d = 200 m the total opposing force = 2000 N (read from graph)
So ΣF = 10,000 - 2000 = 8000 N
Now, we know that ΣF = ma
So, a = ΣF/m = 8000/2200 = 3.64 ms-2
Total Opposing
Force (N)
10000
8000
6000
4000
2000
0
Distance (m)
100 200 300 400 500 600
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Motion - Revision Questions
Question type: Work
Q: Estimate the work done by the seaplane against the opposing forces in
travelling for a distance of 500 m.
A: Work = Force x Distance = Area under F vs d graph.
Area needs to be calculated by “counting squares”.
Each square has area = 2000 x 100 = 2 x 105 J
Total number of squares (up to d = 500 m) = 9 whole squares (x)
+ 6 part squares (p) = 12 whole squares.
So Work done = (12) x (2 x 105) = 2.4 x 106 J
Total Opposing
Force (N)
10000
p
8000
x
x
x
p
x
x
x
x
x
p
4000
2000
p
0
x
p
6000
p
100 200 300 400 500 600
Distance (m)
13
Motion - Revision Questions
Question type: Vectors
Adam is testing a trampoline. The diagrams show Adam at successive stages of his
downward motion.
Figure C shows Adam at a time when he is travelling DOWNWARDS and SLOWING
DOWN.
A
B
Q: What is the direction of
Adam’s acceleration at the time
shown in Figure C ? Explain your
answer.
C
D
A: Acc is UPWARD.
In order to meet the requirements set travelling downward BUT slowing
down, he must be decelerating ie.
Accelerating in a direction opposite to
his velocity. Thus acc is upward.
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Motion - Revision Questions
Question type: Force Diagrams
Q: On Figure C draw arrows that show the TWO INDIVIDUAL forces acting on Adam at
this instant. LABEL EACH ARROW with the name of the force and indicate the
RELATIVE MAGNITUDES of the forces by the LENGTHS of the arrows you draw.
A: The two forces are:
(a) Weight Force (W) acting through Adam’s centre of mass
(b) Normal Reaction Force (N) acting from the point of contact b/w
Adam’s feet and the air bag.
N.B. N >W to meet requirements that acc be upwards at this time
NOTE: The two forces are drawn offset to show the point of origin of each, they
should be coplaner ie., sit one on top of the other.
N
W
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A
B
C
D
Motion - Revision Questions
Question type: Projectile Motion
A car takes off from a ramp and the path of its centre of mass through the air is
shown below.
First model the motion of the car assuming that air resistance is small enough to
neglect.
Q: Which of the directions (A - H), best shows the VELOCITY of the car at X ?
A: Direction C.
Q: Which of the directions (A - H), best shows the VELOCITY of the car at Y ?
A: Direction D
A
B
A
B
H
G
C
D
Direction of Motion
H
G
C
X
F
D
E
Y
F
E
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Motion - Revision Questions
Question type: Projectile Motion
Q: Which of the directions (A - H), best shows the ACCELERATION of the car at X ?
A: Direction E.
Q: Which of the directions (A - H), best shows the ACCELERATION of the car at Y ?
A: Direction E.
Now suppose that AIR RESISTANCE CANNOT BE NEGLECTED.
Q: Which of the directions (A - H), best shows the ACCELERATION of the car at X ?
A: Direction F.
A
B
A
B
H
G
C
D
Direction of Motion
H
G
C
X
F
D
E
Y
F
E
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Motion - Revision Questions
Question type: Newton’s 3rd Law
a
The figure shows a cyclist with the bicycle
wheels in CONTACT with the road surface.
The cyclist is about to start accelerating
forward.
FTR
FRT
Q: Explain, with the aid of a clear force diagram, how the rotation of the wheels
result in the cyclist accelerating forwards.
A: The wheels rotate in the direction shown.
The force labelled FTR is the force the tyre exerts on the road. This force is directed
in the opposite direction to the acceleration and thus cannot be the force producing
that acceleration.
The force labelled FRT is the Newton 3 reaction force arising from the action of FTR.
It is this force (directed in the same direction as the acceleration) that actually
produces the acceleration of the bike and rider.
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Motion - Revision Questions
Question type: Centripetal Force
Mark Webber and his Formula 1 racing car are taking a corner at the
Australian Grand Prix. A camera views the racing car head on at point X on
the bend where it is travelling at constant speed. At this point the radius of
curvature is 36.0 m.
The total mass of the car and driver is 800 kg.
Q: On the diagram showing the camera’s view of
the racing car, draw an arrow to represent the
direction of the NET force acting on the racing car
at this instant.
FC
Camera's head on view
of racing car at point X
The magnitude of the horizontal force
on the car is 6400 N.
36.0 m
Q: Calculate the speed of the car.
X
A:
FC = mv2/R
So v =√FCR/m
= √(6400)(36.0)/(800)
= 17 ms-1
Camera
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Motion - Revision Questions
Question type: Centripetal Force
Q: Referring to the racing car from the previous slide, explain:
(a) Why the car needs a horizontal force to turn the corner.
(b) Where this force comes from.
A: (a) Newton 1 states all objects will continue in a straight line unless acted upon by a
net force.
In order for the car to travel in a circular path a constant force directed at right angles
to the direction of motion (toward the centre of the circle) must exist.
(b) The horizontal force arises from the frictional force exerted BY THE ROAD ON
THE TYRES and is directed toward the centre of the circle the car is travelling in.
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Motion - Revision Questions
Question type: Projectile Motion
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 ms–1 at an angle of 30° to
the horizontal, as shown in Figure 1. The package lands on the ground 3.0 s after its
release. Ignore air resistance in your calculations.
Figure 1
Q: What is the value
of h in Figure 1?
A: Can approach in a
number of ways.
One way is: Find how long it takes to rise and fall back to same vertical height (h)
Then subtract from total time to give a time for vertical fall under gravity with an initial
velocity of 10 ms-1 directed at 30o below the horizontal
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Motion - Revision Questions
Question type: Projectile Motion
Figure 1
1. Time to get back to h
Since this part of the journey is symmetrical time
to reach max height = ½ time for this part of
journey.
h
2. Value of h
again analyse vertical motion
Analyse vertical motion
u = 10 Sin 30o
2
s
=
ut
+
½
at
v
=
?
o
-1
u = - 10 Sin 30 ms
+ve a = 10 ms-2
h = (5)(2) + ½ (10)(4)
v=0
v = u + at
h = 30 m
s=h
-2
+ve a = 10 ms
0
0 = -10 Sin 30 + 10t
t = 2.0 s
s=?
t = 5/10 = 0.5 sec.
t=?
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So, total time to get back to h = 1.0 s
Motion - Revision Questions
Question type: Projectile Motion
Figure 1
Q: Assuming that the helicopter continues to fly with its initial velocity,
where is it when the package lands?
Which one of the statements below is most correct?
A. It is directly above the package.
B. It is directly above a point that is 15 m beyond the package.
C. It is directly above a point that is 26 m beyond the package.
D. It is directly above a point that is 30 m from the bushwalker.
Both the package and the helicopter have the same (constant) horizontal velocity,
so the package will always be directly below the helicopter.
So, alternative A is correct
23
Motion - Revision Questions
Question type: Sketch Graphs
Q: Which of the graphs below best represents the speed of the package as a
function of time?
The key to the question is SPEED, this is the sum of both vertical and horizontal.
Speed never falls to zero (there is always a horizontal component), so C and D are out
A shows the object’s acceleration falling as it nears the ground – not so !
24
So B must be the correct answer
Motion - Revision Questions
Question type: Momentum
A car of mass 1000 kg travelling on a smooth road at 5.0 ms–1 collides
with a truck that is stationary at a set of traffic lights. After the collision
they are stuck together and move off with a speed of 2.0 ms–1
Q: How much momentum did the car transfer to the truck?
A: Mom is ALWAYS conserved.
Mom of car before = (1000)(5) = 5000 kgms-1
Mom of car after = (1000)(2) = 2000 kgms-1
Since mom is conserved
Mom loss by car = Mom gain by truck
So, Mom transferred to truck = 3000 kgms-1
Q: What is the mass of the truck?
A: p = mv
so, m = p/v
= 3000/2
= 1500 kg
25
Motion - Revision Questions
Question type: Impulse
A car of mass 1000 kg travelling on a smooth road at 5.0 ms–1 collides with a
truck that is stationary at a set of traffic lights. After the collision they are stuck
together and move off with a speed of 2.0 ms–1
Q: If the collision took place over a period of 0.3 s, what was the average
force exerted by the car on the truck?
For the truck
Impulse = Change in momentum
Ft = mv
F(0.3) = 3000
F = 10,000 N
26
Motion - Revision Questions
Question type: Relative Motion
A train is travelling at a constant velocity on a level track. Lee is standing
in the train, facing the front, and throws a ball vertically up in the air, and
observes its motion.
Q: Describe the motion of the ball as seen by Lee.
Lee sees the ball move straight up and down.
Sam, who is standing at a level crossing, sees Lee throw the ball into the air.
Q: Describe and explain the motion of the ball as seen by Sam.
From Sam’s point of view, the ball follows a parabolic path
made up of the vertical motion imparted by Lee and the
horizontal motion due to the train.
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Motion - Revision Questions
Question type: Newtons 2nd Law
In Figure 2, a car of mass 1000 kg is being towed on a level road by a van of
mass 2000 kg. There is a constant retarding force, due to air resistance and
friction, of 500 N on the van, and 300 N on the car. The vehicles are travelling at
a constant speed.
Figure 2
Q: What is the magnitude of the force driving the van?
As the vehicles are travelling at constant speed, a = 0 and thus ΣF = 0
Thus driving force = total retarding force
Thus driving Force = 800 N
Q: What is the value of the tension, T, in the towbar?
Looking at the car alone the forces acting are the tension in the
towbar and the retarding force.
Since the car is travelling at constant velocity, ΣF = 0, so tension =
retarding force = 300 N
28
Motion - Revision Questions
Question type: Equations of Motion
When travelling at a speed of 15.0 ms–1 the van driver stops the engine, and
the van and car slow down at a constant rate due to the constant retarding
forces acting on the vehicles.
Q: How far will the van and car travel before coming to rest?
1.
Determine deceleration
ΣF = ma
so a = ΣF/m
= 800/3000
= 0.266 ms-2
2. Use equations of motion to find distance
u = 15.0 ms-1
v=0
a = - 0.266 ms-2
s=?
t=?
Another approach:
Change in Kinetic Energy = work done
mathematically ½ mv2 = Fd
v2 = u2 + 2as
0 = (15)2 +2(-0.266)s
s = 422 m
so d = 0.5(3000)(15)2
800
= 422 m
29
Motion - Revision Questions
Question type: Work and Energy
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 in Figure 4. 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
Figure 4
Q: What is the height, h, between the floors?
At the second floor the box has Potential Energy = mgh
On reaching the ground floor this has been converted to Kinetic Energy
(½mv2) plus the work done against friction in moving down the slope
thus,
mgh = ½ mv2 + Fd
(30.0)(10)h = ½ (30)(8.0)2 + 50(6.0)
h = 4.2 m
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Motion - Revision Questions
Question type: Elastic Potential Energy
Figure 4
The box then slides along the frictionless floor, and is momentarily
stopped by a spring of stiffness 30 000 N m–1
Q: How far has the spring compressed when the box has come to rest?
KE of box is converted to Elastic Potential Energy in the spring
Thus ½ mv2 = ½ kx2
½ (30.0)(8.0)2 = ½ (30,000)(x)2
x = 0.25 m
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Motion - Revision Questions
Question type: Work and Energy
A model rocket of mass 0.20 kg is launched by means of a spring, as shown
in Figure 1. 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 in Figure 2.
Q: How much energy is stored in the spring when it is compressed?
A: Compressing the spring requires work to be done on it. (this work is stored
as elastic potential energy in the spring)
Work done = area under graph up to a compression of 0.2 m
= ½ (0.2)(1000)
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= 100 J
Motion - Revision Questions
Question type: Energy Conversion
Q: What is the speed of the rocket as it leaves the spring?
A: All the elastic potential energy stored at max compression
(100 J) will be converted to Kinetic Energy at release. So
100 = ½ mv2
v = √1000
= 31.6 ms-1
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Motion - Revision Questions
Question type: Equations of Motion
Q: What is the maximum height, above the
spring, reached by the rocket? You should ignore
air resistance on the way up since the rocket is
very narrow.
+ve
u = 31.6 ms-1
v=0
a = -10 ms-2
s=?
t=?
v2 = u2 +2as
0 = (31.6)2 + 2(-10)s
s = 1000/20
= 50 m
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Motion - Revision Questions
Question type: Newtons 2nd Law
Retarding
Force (R)
(Newtons)
time (s)
Q: What is the acceleration of the rocket
at a time 5 s after the parachute opens?
+ve
A: At t = 5.0 s, R = 1.8 N
Weight of Rocket W = mg
= (0.2)(10)
= 2.0 N
R
W
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 as R in Figure 3,
and its variation as a
function of time after the
parachute opened is shown
in Figure 4.
ΣF = ma
a = ΣF/m = (2.0 – 1.8)/ 0.2
= 1.0ms-2
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Motion - Revision Questions
Question type: Centripetal Force
Fc
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.
Figure 5 shows a train at position P taking a
curve on horizontal ground, at a constant
speed, in the direction shown by the arrow.
Q: At point P shown on the figure,
draw an arrow that shows the direction
of the force exerted by the rails on the
wheels of the train.
36
Motion - Revision Questions
Question type: Centripetal Force
The radius of curvature of a track that is safe
at 60 km/h is approximately 200 m.
A: Centripetal Force
Fc = mv2
R
Q: What is the radius of curvature of a track
that would be safe at a speed of 120 km/h, This can be solved as a ratio question
assuming that the track is constructed to the
same strength as for a 60 km/h curve?
m(60)2 = m(120)2
200
x
x = 800 m
A: B
37
Motion - Revision Questions
Question type: Vector Addition
Q: At point Q the driver applies the brakes
to slow down the train on the curve.
Which of the arrows (A to D) indicates the
direction of the net force exerted on the
wheels by the rails?
A: B
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Motion - Revision Questions
Question type: Momentum
A small truck of mass 3.0 tonne collides
with a stationary car of mass 1.0 tonne.
They remain locked together as they move
off.
The speed immediately after the collision
was known to be 7.0 ms-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.
Q: What value did Robin obtain?
A: PBEFORE = PAFTER
PBEFORE = (3000)(x)
PAFTER = (3000 + 1000)(7.0)
So, 3000x = 28,000
x = 9.3 ms-1
The calculated value is questioned by the other
investigator, Chris, who believes that
conservation of momentum only applies in
elastic collisions.
A: Momentum is conserved in ALL
Q: Explain why Chris’s comment is wrong.
types of collisions whether they be
elastic or inelastic.
KE is not conserved in this type
(inelastic) collision.
39
Motion - Revision Questions
Question type: Elastic/Inelastic Collisions
Q: Use a calculation to show whether the
collision was elastic or inelastic.
A: Total KE before collision = ½ mv2 = ½ (3000)(9.3)2
= 129735 J
Total KE after collision = ½ mv2 = ½ (4000)(7.0)2
= 98,000 J
KE is NOT conserved
So collision is INELASTIC
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Motion - Revision Questions
Question type: Projectile Motion
Fred is playing tennis on the deck
of a moving ship.
He serves the ball so that it leaves
the racket 3.0 m above the deck
and travels perpendicular to the
direction of motion of the ship.
The ball leaves the racket at an
angle of 8° to the horizontal.
At its maximum height it has a
speed of 30.0 ms-1. You may ignore
air resistance in the following
questions.
Q: With what speed, relative to the
deck, did the ball leave Fred’s racket?
Give your answer to three significant
figures.
A: Assuming projectile behaviour:
VHORIZONTAL = 30 ms-1 throughout the flight
30 ms-1
x = initial speed
x
8o
30 ms-1
thus, Cos 8 = 30/x
x = 30/Cos 8
= 30.3 ms-1
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Motion - Revision Questions
Question type: Projectile Motion
Q: At its highest point, how far was the
ball above the deck?
30 ms-1
x
8o
VVERT
30 ms-1
A: Using the equations of motion for the balls flight from racquet to point
of max height, where VVERT = 0
+ve
u = VVERT = 30.3 Sin 8
v=0
a = -10 ms-2
s=?
t=?
v2 = u2 + 2as
0 = (30.3 Sin 8)2 + 2(-10)s
s = 0.89 m
So Height above the DECK = 3.0 + 0.89 = 3.89
42 m
Motion - Revision Questions
Question type: Relative Motion
The ship is travelling straight ahead at a
velocity of 10 ms-1
Q: When the ball is at its highest point at what
speed is it moving relative to the ocean?
30 ms-1
10 ms-1
A: V BALL REL OCEAN = VSHIP + VBALL
VSHIP
=
+
VBALL
30
10
θ
VBALL REL OCEAN
VBALL REL OCEAN = √(30)2 + (10)2
= 31.6 ms-1
Q: at what angle is the ball travelling relative
to the direction of the ship’s travel?
Tan θ = 30/10 = 3.0
θ = 71.60
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Motion - Revision Questions
Question type: Universal Gravitation
Newton was the first person to quantify the gravitational force
between two masses M and m, with their centres of mass
separated by a distance R as
F= GMm
R2
where G is the universal gravitational constant, and has a
value of 6.67 × 10-11 N m2 kg2.
For a mass m on the surface of Earth (mass M) this
becomes F = gm, where g = GM
R2
Q: Which one of the expressions (A to 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.
A: B
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Motion - Revision Questions
Question type: Gravitational Force
Q: What is the magnitude of the force exerted by Earth on a
water molecule of mass 3.0 × 10-26 kg at the surface of Earth?
A: F = GMm
R2
= (6.67 x 10-11)( 5.98 x 1024)(3.0 x 10-26)
(6.37 x 106)2
= 2.95 x 10-25 N
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Motion - Revision Questions
Question type: Kepler’s 3rd Law
A satellite in a circular orbit of radius 3.8 × 108 m
around Earth has a period of 2.36 × 106 s.
Q: Calculate the mass of Earth. You must show
your working.
A:
R3 = GMe
T2 4π2
Me =
(4π2) (3.8 x 108)3
(6.67 x 10-11) (2.36 x 106)2
= 5.83 x 1024 kg
46
Motion - Revision Questions
Question type: Universal Gravitation
The radius of the orbit of Earth in its circular motion around the
Sun is 1.5 × 1011 m
Q: Indicate on the diagram, with
an arrow, the direction of the
acceleration of Earth.
Q: Calculate the mass of the Sun.
Take the value of the gravitational
constant G = 6.67 × 10–11 N m2 kg–2.
(Figure 3).
Question assumes you know the period of rotation of the earth around the sun
ie. 365 days = 365 x 24 x 60 x 60 = 31536000 s (3.15 x 107 s)
F C = Fg
me4π2 R/T2 = Gmems/R2
ms = 4π2R3
GT2
ms =
4(3.14)2(1.5 x 1011)3
6.67 x 10-11(3.15 x 107)2
ms = 2 x 1030 kg
47
Motion - Revision Questions
Question type: Gravitational Field Strength
Nato III is a communication satellite that has a mass of 310 kg and orbits Earth at a
constant speed at a radius R = 4.22 x 107 m from the centre of Earth.
Q: Calculate the magnitude of the Earth’s gravitational
field at the orbit radius, R = 4.22 x 107 m, of Nato III.
Give your answer to 3 sig figs. You MUST show your
working. G = 6.67 x 10-11 Nm2kg-2 Me = 5.98 x 1024 kg.
R
Earth
Nato III
A:
g = GMe/R2
= (6.67 x 10-11)(5.98 x 1024)
-----------------------------------------(4.22 x 107)2
= 0.224 Nkg-1
48
Motion - Revision Questions
Question type: Universal Gravitation
Q: What is the speed of Nato III in its orbit ?
A: The centripetal force (FC) required by Nato III to
complete its circular orbit is supplied by the force of
gravitational attraction (Fg) between Earth and Nato III.
Thus Fg = FC
Thus GMem/R2 = mv2/R
So
v = √GMe/R
= √(6.67 x 10-11)(5.98 x 1024)
-------------------------------------(4.22 x 107)
= 3074 ms-1
= 3.1 x 103 ms-1
R
Earth
Nato III
49
Motion - Revision Questions
Question type: Circular Motion
Q: Which ONE of the following statements (A - D) about Nato III is correct ?
A: The net force acting on Nato III is zero and therefore it does not accelerate.
B: The speed is constant and therefore the net force acting on Nato III is zero.
C: The is a net force acting on Nato III and therefore it is accelerating.
D: There is a net force acting on Nato III, but it has zero acceleration.
A: Alt C is correct
R
Earth
Nato III
50
Motion - Revision Questions
Question type: Weight & Weightlessness
The Russian space station MIR (Russian meaning - peace) is in a circular
orbit around the Earth at a height where the Gravitational Field Strength is
8.7 Nkg-1
Q: Calculate the magnitude of the gravitational force exerted by Earth on
the astronaut of mass 68 kg on MIR
A:
W = mg
= (68)(8.7)
= 592 N
When the astronaut wishes to rest he has to lie down and strap himself into bed.
Q: What is the magnitude of the force that the bed exerts on the astronaut before he
begins to fasten the strap ?
A: 0 N
51
Motion - Revision Questions
Question type: Weightlessness
Newspaper articles about astronauts in orbit sometimes speak about zero
gravity when describing weightlessness.
Q: Explain why the astronaut in the orbiting MIR is not really weightless.
A: Weight is the action of a gravitational field acting on a mass.
For true weightlessness to exist the gravitational field strength needs to be
zero.
There is no place in the universe where the gravitational field strength is zero,
so nothing is truly “weightless”.
The astronaut actually does have weight, as calculated earlier.
The astronaut’s “apparent weight” is zero because he is not subject to a
reaction force inside the spacecraft since both he and the spacecraft are in a
state of constant free fall.
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