Download Worked solutions Chapter 2: Collisions and

Worked solutions
Chapter 2 Collisions and circular motion
2.1 Momentum and impulse
1
The locust has more momentum. Stationary objects have zero momentum.
2
Momentum of Pavithra: p = mv = 45  3.5 = 160 kg m s–1
Momentum of Michelle: p = mv = 60  2.5 = 150 kg m s–1
Pavithra has more momentum.
3
a
b
c
4
D is the correct answer. If the direction of motion is different, the momentum must also be
different.
5
a
b
p = mv = 9.1  10–31  3.0  107 = 2.7  10–23 kg m s–1
p = mv = 6.0  1024  3.0  104 = 1.8  1029 kg m s–1
p = mv = 0.10  2.3 = 0.23 kg m s–1
This extends his stopping time and so reduces the size of the force that acts to stop him.
Using Ft = p, it can be seen that if the momentum change occurs over a long time
interval, the force that is acting must be reduced in size.
His momentum would change over a very short time interval. It would require a very
large force to do this, which would be painful (or damaging) for Vijay.
6
a
b
c
d
Change in speed: v = v – u = 8.0 – 10 = –2 m s–1
Change in velocity: v = v – u = 8.0 m s–1 up – 10 m s–1 down = 18 m s–1 upwards
p = mv = 0.080  18 = 1.4 kg m s–1 up
Impulse = p = 1.4 N s up
7
a
The net force can be found by using Ft = mv = p.
F = 1.4/0.050 = 28 N up
The forces acting are gravity F g downwards and normal reaction force F N upwards.
F g = mg = 0.78 N
F = F g + F N
This gives F N = 29 N up.
As described by Newton’s third law, this is equal and opposite to the force that the floor
exerts on the ball, so F = 29 N down.
b
c
8
a
b
c
d
From the graph: F max = 500 N forwards
As described by Newton’s third law, the maximum force that the ball exerts on the
racquet is 500 N backwards.
Net impulse = area under F–t graph = ½  0.020  500 = 5.0 N s forwards
Net impulse = p = 5.0 kg m s–1 forwards
Copyright © Pearson Australia 2012 (a division of Pearson Australia Group Pty Ltd) ISBN 978 1 4425 5459 7
Page 1
Worked solutions
e
Chapter 2 Collisions and circular motion
p = mv
5.0 = 0.10  v
v = 50 m s–1
9
The padding extends the time over which the players lose their momentum and are brought to a
stop. The force that is acting to stop them must therefore be reduced in size.
10
ai
ii
iii
b
c
d
The dummies have equal amounts of momentum before the collision. Both dummies have
zero momentum after the collision, so the change in momentum is the same for each
dummy.
Net impulse = p, so impulse is the same for each dummy.
The dummy in car B stops in a shorter time interval and so would experience a greater
stopping force.
Car A: Ft = mv
F A × 0.40 = 100 × –20
and F A = –5.0  103 N
F B × 0.10 = 100 × –20
F B = –2.0  104 N
Crumple zones reduce the magnitudes of the forces that act on the occupants of a car
during a collision. This will result in fewer or less serious injuries.
This will further extend the time taken for a passenger to come to a stop. Using Ft = p,
this will result in smaller forces acting on the person as they come to a stop.
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Worked solutions
Chapter 2 Collisions and circular motion
2.2 Conservation of momentum
1
a
b
c
2
a
b
c
d
For the sports car:
p = mv = 1.0  103 × 10 = 1.0  104 kg m s–1 east
For the station wagon:
p = mv = 2.0  103 × 5.0 = 1.0  104 kg m s–1 west
Total momentum = 1.0  104 kg m s–1 east + 1.0  104 kg m s–1 west = 0
p i = p f
From Question 1c, p i = 0 so p f = 0 i.e. common velocity = 0
It hasn’t gone anywhere. The vehicles had a total of zero momentum before the collision
and so there is zero momentum after the collision.
The change in momentum of the sports car = p – p f – p i = 0 – 1.0  104 kg m s–1 east
= 1.0  104 kg m s–1 west
The change in momentum of the station wagon p – p f – p i = 0 – 1.0  104 kg m s–1 west
= 1.0  104 kg m s–1 east
3
Using the conservation of momentum, where the direction of positive velocity is to the right:
p i = p f
(0.200 × 9.0) + 0 = (0.200 × 3.0) + (0.100v)
and v = 12 m s–1,
i.e. the velocity of the 100 g ball after the collision is 12 m s–1 right
4
p i = p f
0 = (10.0 × 500) + (100v)
and the recoil velocity of cannon, v = 5.0 m s–1
5
Using the conservation of momentum, where the direction of positive velocity is to the right:
p i = p f
(0.100 × 40) + 0 = (0.100 + 0.080)v
and v = 22 m s–1,
i.e. 22 m s–1 right
6
Mary is correct. As the water spills from the tanker, the water continues to move forwards at
5.0 m s–1. Because the water keeps its horizontal momentum, so too will the tanker retain its
momentum. There is no transfer of momentum between the tanker and the water, so each will
continue to travel forwards at 5.0 m s–1.
7
Considering the horizontal motion only. The bag of apples falls vertically and so initially has no
horizontal momentum.
From the conservation of momentum: p i = p f
(10 × 5.0) = (20 × v) and v = 2.5 m s–1
8
C is the correct answer.
A is wrong because vertical momentum can only be transferred in a vertical direction.
B and D are wrong because momentum and energy are two completely different quantities, and
one cannot be converted into the other.
9
a
b
B is the correct answer because the trolley slows down and so has less momentum after
the bag of apples is dropped in it.
A is the correct answer because the total momentum of the trolley and its contents
remains constant (conservation of momentum).
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Worked solutions
10
a
b
c
Chapter 2 Collisions and circular motion
p = mv = (48 × 40) = 190 kg m s–1 right
p i = p f
192 = (50 × v g )
and v g = 3.8 m s–1 right
p i = p f
The girl retains her momentum as she jumps from the skateboard, so there is no transfer
of momentum to the skateboard. It keeps travelling at its speed of 3.8 m s–1.
11
The footballer’s momentum has been reduced to zero. His momentum has been transferred to
the Earth (via the fence), making it move very slightly.
12
No, the diver is not an isolated system. An unbalanced gravitational force is acting on her.
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Worked solutions
Chapter 2 Collisions and circular motion
2.3 Work, energy and power
1
a
b
c
W = Fx = (800 × 9.8) × 90 = 7.1  105 J
W = Fx = (800 × 9.8) × 50 + 0.5(800 kg)(2.0 m s–1)2 = 3.9  105 J
P = Fv = (800 × 9.8) × 2.0 = 1568 W = 16 kW
2
a
b
c
F h = Fcos = 30cos60 = 15 N
W = F h × x = 15 × 2.5 = 37.5 = 38 N m
Work done by friction: W = F f × x = 10 × 2.5 = 25 N m
Work done by F = F × x = (15 – 10) × 2.5 = 12.5 = 13 N m
3
a
b
c
E k = work done by net force = 12.5 J = 13 J
Energy dissipated as heat = work done by friction = 25 J
friction
4
The vertical component of the 30 N pulling force is acting perpendicular to the dog’s direction
of motion, so the component does zero work and does not change the energy of the dog.
5
Horizontal acceleration of dog: a = F h /m = 5.0/2.0 = 2.5 m s–2
u = 0, a = 2.5 m s–2, x = 2.5 m, t = ?
x = ut + 1/2 at2
2.5 = 0 + 0.5 × 2.5 × t2
t = 2.0 = 1.4 s
6
a
b
c
d
7
a
b
c
100 g ball: E k (i) = ½ mv2 = 0.5 × 0.10 × 02 = 0
200 g ball: E k (i) = ½ mv2 = 0.5 × 0.20 × 9.02 = 8.1 J
E k (i) = 0 + 8.1 = 8.1 J
100 g ball: E k (f) = ½ mv2 = 0.5 × 0.10 × 122 = 7.2 J
200 g ball: E k (f) = ½ mv2 = 0.5 × 0.20 × 3.02 = 0.9 J
E k (f) = 7.2 + 0.9 = 8.1 J
Elastic. Kinetic energy has been conserved.
No, during the collision some energy would have been transformed into heat and sound
energy. Truly elastic collisions only occur at the atomic level.
W = U s = 0.5 kx2, where k = gradient of force–compression graph = 1.0  103 N m–1
Then W = 0.5 × 1.0  103 × 0.0102 = 0.050 N m
During compression, the girl does work to alter the shape of the ball. During release, the
stored energy is released in restoring the ball to its original shape.
P = W/t = (120 × 0.050)/60 = 0.10 W
8
Kinetic energy has been transformed into elastic potential energy
0.5mv2 = 0.5kx2
x2 = mv2/k = 0.10 × 4.02/1.0  103
then x = 4.0  10–2 m = 4.0 cm
9
a
b
Loss of gravitational potential energy = mgh
= 80 × 9.8 × 5.0 = 3.9  103 J
u = 0, a = 5.0 m s–2, x = 5.0 m, v = ?
v2 = u2 + 2ax = 0 + 2 × 5.0 × 5.0 = 50
v = 7.1 m s–1
Increase in kinetic energy = 0.5mv2
= 0.5 × 80 × 7.12
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Worked solutions
c
d
e
10
a
b
c
Chapter 2 Collisions and circular motion
= 2.0 × 103 J
This is due to energy being transformed into heat and sound due to friction.
W = F  x = mgx = 80 × 9.8 × 5.0 = 3.9  103 J
Heat produced = energy lost due to friction
= 3.92  103 J – 2.0  103 J = 1.9  103 J
Yes, momentum is conserved in all collisions
Inelastic; some kinetic energy has been transformed into heat and sound energy
E k (i) = ½ mv2 + ½ mv2
= ½ × 4.0 × 3.02 + ½ × 4.0 × 3.02
36 J
20 J transformed into heat and sound, so E k (f) = 36 – 20 = 16 J
From symmetry, the balls will have the same final speeds and the same kinetic energies
of 8.0 J.
For each ball: E k = ½ mv2 = 0.5 × 4.0 × v2 = 8.0
v2 = 4.0
v = 2.0 m s–1
The balls will travel in opposite directions at 2.0 m s–1.
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Worked solutions
Chapter 2 Collisions and circular motion
2.4 Hooke’s law and elastic potential energy
1
C is the correct answer. The force constant depends only on the properties of the material from
which the spring is made, and not the length of the spring.
2
a
b
c
d
From the graph, F = 200 N
Spring constant = gradient of graph = 1.0  104 N m–1
W = U s = ½ kx2 = 0.5 × 1.0  104 × (2.0  10–2)2 = 2.0 N m
U s = ½ kx2
= 0.5 × 1.0  104 × 0.0102 = 0.50 J
e
3
a
b
c
The force constant for each spring is determined by the gradient of the force–compression
graph:
Spring A: k = 2.0  104 N m–1
Spring B: k = 1.0  104 N m–1
Spring C: k = 5.0  103 N m–1
The stiffest spring will have the highest spring constant; therefore the ranking is:
C, B, A.
Using U s = ½ kx2, where x = 0.010 m,
and using the respective spring constants for each spring:
A: U s = 1.0 J
B: U s = 0.50 J
C: U s = 0.25 J
4
½k A x2 A = ½k B x2 B
0.5 × 2.0  104 × x A 2 = 0.5 × 1.0  104 × x B 2
and x A /x B = 1/ 2 = 0.71
5
a
b
F = mg = kx
k = mg/x = 0.075 × 9.8/0.050 = 14.7 = 15 N m–1
U s = ½ kx2= 0.5 × 14.7 × 0.0502 = 1.8  10–2 J
b
x = 45 – 30 = 15 cm = 0.15 m
U s = ½ kx2= 0.5 × 14.7 × 0.152 = 0.16 J
F = mg = kx
k = mg/x,
then m = kx/g
= 14.7 × 0.15/9.8 = 0.23 kg
7
a
b
E k (i) = ½ mv2 = 0.5 × 2.5 × 1.52 = 2.8 J
Energy stored in spring is equal to initial kinetic energy of cart, i.e. 2.8 J.
8
a
U s = ½ kx2= 2.8
6
a
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Worked solutions
b
9
a
b
10
a
b
Chapter 2 Collisions and circular motion
0.5 × 400 × x2 = 2.8
x = 0.12 m
U s + E k = 2.8
½ kx2+E k = 2.8
0.5 × 400 × 0.0402 + E k = 2.8
0.32 + E k = 2.8
E k = 2.5 J
The work done by the archer on the bow is equal to the area under the force–distance
graph from XY = 0.10 m to XY = 0.30 m.
This area = 7.2 J
It is being stored as elastic potential energy in the bow.
As the string returns to its original position, all the elastic potential energy stored in the
string is transferred to the arrow.
The work done by the string on the arrow = 7.2 J
There have been two assumptions made:
i
All the work done by the archer on the bow is stored as elastic potential energy in
the string, and none is stored as strain energy in the shaft of the bow. The
assumption here is that the shaft of the bow is completely rigid.
ii
The assumption is that no energy is transformed into heat as the string returns to its
unextended position.
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Worked solutions
Chapter 2 Collisions and circular motion
2.5 Circular motion
1
a
b
A and D are correct
The passenger has continued straight ahead and the car has turned to the left.
2
a
b
c
8.0 m s–1
8.0 m s–1 south
a = v2/r = 8.02/9.2 = 7.0 m s–2 towards centre, i.e. west
3
a
b
F = ma = 1200 × 7.0 = 8.4 × 103 N east
Friction between the tyres and the road surface.
4
a
b
8.0 m s–1 north
towards centre, i.e. east
5
The force needed to give the car a larger centripetal acceleration will eventually exceed the
maximum frictional force that could act between the tyres and the road surface. At this time, the
car would skid out of its circular path.
6
a
b
c
d
7
a
b
c
d
e
f
a = v2/r
= 2.02/1.5
= 2.7 m s–2 towards centre of circle
The forces are unbalanced because the skater has an acceleration as she skates in a
circular path.
F = ma
= 50 × 2.67
= 130 N towards centre of circle
The horizontal force exerted on the blades by the ice; i.e. friction
T = 1/f
= 1/2.0 = 0.50 s
v = 2r/T
= 2 × 0.80/0.50 = 10 m s–1
a = v2/r
= 102/0.80 = 126 = 130 m s–2 towards centre of circle
F = ma
= 2.5 × 126 = 315 = 320 N
The force that is responsible for the centripetal acceleration of the ball is the tension in
the wire, which is directed radially inwards at all times.
If the wire breaks, the ball will move off at a tangent to the circle with speed 10 m s–1.
8
v = 50/3.6 = 13.9 m s–1
v = 2r/T
so T = 2r/v = 2 × 62/13.9 = 28 s
9
F = mv2/r = 1.6 × 13.92/62 = 5.0 N
10
The sideways force that the air exerts on the wings of the plane.
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Worked solutions
Chapter 2 Collisions and circular motion
2.6 Aspects of circular motion
1
Velocity is tangential to path.
a
west
b
south-west
c
north
2
Acceleration is towards centre of circular path.
a
south
b
south-east
c
west
3
Net force is in direction of acceleration.
a
south
b
north-east
c
north-west
4
The ball will travel east at a tangent to its path at point E.
5
Momentum is a vector so momentum changes as direction of velocity changes. Kinetic energy
is a scalar and so is constant. Answer is C.
6
a
b
c
7
8
a
b
Use a force triangle for the ball.
F t = mg/sin30.0 = 0.0250 × 9.80/0.50 = 0.490 N
a
b
r = 2.4cos60 = 1.2 m
The forces are her weight acting vertically and the tension in the rope acting along the
rope towards the top of the maypole.
She has an acceleration directed towards point B, the centre of her circular path.
Use a force triangle for the girl, showing the net force towards B.
F = mg/tan60 = 294/1.73 = 170 N towards B.
F = mv2/r
170 = 30 × v2/1.2
v = 2.6 m s–1
c
d
e
9
2r
T
= 2 × 0.800/1.36 = 3.70 m s–1
a = v2/r
= 3.702/0.800 = 17.1 m s–2 radially inwards
F = ma
= 0.0250 × 17.1 = 0.430 N radially inwards, so magnitude is 0.430 N
v=
a
b
F = mv2/r = 1200 × 182/80 = 4860 = 4.9 kN
Use a force triangle for the car, showing the normal, weight and the net force acting
horizontally.
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Worked solutions
Chapter 2 Collisions and circular motion
tan = 4860/1200 × 9.8 = 0.413
 = 22
10
The greater speed would make the car travel higher up the track. The driver would have to turn
the wheels slightly towards the bottom of the track so as to create a sideways force of friction
towards the bottom of the track.
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Worked solutions
Chapter 2 Collisions and circular motion
2.7 Circular motion in a vertical plane
1
a
b
c
d
It has a constant speed so its centripetal acceleration a = v2/r is also constant in
magnitude.
At the bottom of its path, the yo-yo has an upwards acceleration and so the net force is
up. This indicates that the tension force is greater than F g .
At the top of its path, the yo-yo has an downwards acceleration and so the net force is
down. This indicates that the tension force is less than F g .
At the bottom of its circular path.
2
At this point,
a = v2/r = g
so v = (r × g) = 1.5 × 9.80 = 3.8 m s–1
3
a
b
c
d
4
a
b
c
The weight force from gravity and the normal force from the road.
F = mv2/r = 800 × 342/10 = 1.3 × 103 N down
Yes. When the driver is moving over a hump, the normal force is less than her weight mg.
Her apparent weight is given by the normal force that is acting and so the driver feels
lighter at this point.
At point of lift-off, F N = 0 and a = g
a = v2/r = g
so v = (r × g) = 10 × 9.80 = 9.9 m s–1 = 36 km h–1
At X, mechanical energy is:
E = E k + U g
= ½ mv2 + mgh
= 0.5 × 500 × 2.002 + 500 × 9.80 × 50.0
= 1000 + 245 000
= 246 000 J
At Y: U g is zero so its kinetic energy is 246 000 J
½ mv2 = 246 000
0.5 × 500 × v2 = 246 000
v = 984 = 31.4 m s–1
At Z, mechanical energy = 246 000 J
E = E k + U g
246 000 = E k + 500 × 9.80 × 30.0
246 000 = E k + 147 000
E k = 99 000
0.5 × 500v2 = 99 000
v = 19.9 m s–1
At Z, acceleration is:
a = v2/r = 19.92/15.0 = 26.4 m s–2 down
F = ma = 500 × 26.4 = 13 200 N down
Both F N and F g are acting downwards, so:
F N + F g = 13,200
F N = 13 200 – 500 × 9.80 = 8300 N = 8.3 × 103 N down
5
For the cart to just lose contact at Z, F N = 0
a = v2/r = g
so v = (r × g) = (15.0 × 9.80) = 12.1 m s–1
6
If the pilot is upside-down at the top of the loop, both the normal force from the seat and gravity
are acting downwards.
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Worked solutions
Chapter 2 Collisions and circular motion
F = F N + F g
mv2/r = F N + 80 × 9.80
80 × 352/100 = F N + 784
F N = 980 – 784 = 200 N down
7
If the normal force is zero, then:
a = v2/r = g
so v = (r × g) = (100 × 9.80) = 31 m s–1
8
a = v2/r = 9g
so v = (r × 9g) = (400 × 88.2) = 190 m s–1
9
a
b
10
The wire is more likely to snap when the ball is moving through X. The bowling ball has an
upwards acceleration at X and so tension is greater that the weight force. This larger tension in
the wire means it is more likely to break.
F t = F g = 4.0 × 9.80 = 39 N
First determine the speed of the ball at X. If it is released from rest, the kinetic energy at
X is equal to its gravitational potential energy at Z.
E k (X) = U g (Z) = 4.0 × 9.80 × 2.0 = 78.4 J
½ mv2 = 78.4
0.5 × 4.0 × v2 = 78.4
v = 6.3 m s–1
F = mv2/r = 4.0 × 6.32/2.0 = 78.4 N up
F t = 78.4 + mg = 78.4 + 39.2 = 120 N up
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Worked solutions
Chapter 2 Collisions and circular motion
Chapter review
1
a
b
c
d
v2 = u2 + 2ax
= 0 + 2 × 9.8 × 10
v = 14 m s–1
p = mv
= 0.200 × 14
= 2.8 kg m s–1 down
p = mv
= 0.200 × 10
= 2.0 kg m s–1 up so C is correct.
Impulse = p = p f – p i
= 2.0 kg m s–1 up – 2.8 kg m s–1 down = 4.8 kg m s–1 up so D is correct.
Ft = p
F = = 4.8/1.0  10–3 = 4.8  103 N up
c
E k = E k (f) – E k (i) = 0.5 × 50.0 × 3.002 – 0 = 225 J
u = 0, t = 10 s, v = 3.0 m s–1, x = ?
x = t(u + v)/2 = 10 × 3.0/2 = 15 m
W = (F cos ) × x = (200 cos60) × 15 = 1500 J = 1.50 kJ
The amount of kinetic energy converted into heat = 1.5  103 J – 225 J = 1.28 × 103 J
3
a
b
c
P = work done by student/time taken = 1500/10.0 = 150 W
P = work done by friction/time taken = 1275/10.0 = 128 W
P = work done by net force/time taken = 225/10.0 = 22.5 W
4
The section between x = 2.0 m and x = 5.0 m. It is during this interval that the puck loses kinetic
energy due to friction.
5
The loss in kinetic energy due to friction = work done on puck by the frictional force F f
E k = work done
Then 2.0 – 5.0 = F f × 3.0
Hence F f = –1.0 N so magnitude of frictional force is 1.0 N
6
A, the loss in kinetic energy of the puck = heat produced by the frictional force = 3.0 J
7
a
b
c
8
p(sled) = mv = 200v = 200 kg m s–1 west
and v = 1.0 m s–1 west
9
p(i) = p(f)
50 × 4.4 + 200 kg m s–1 = 250v
420 = 250v
v = 1.7 m s–1
10
a
2
a
b
b
p(i) = mv = 250 × 0 = 0
p(f) = mv = 50 × 4.0 = 200 kg m s–1 east
Since momentum is conserved, the momentum of sled = 200 kg m s–1 west
p(i) of sled = 200 kg m s–1 west
p(f) of sled = 200 × 1.68 = 336 kg m s–1 west
p of sled = 336 west – 200 west = 136 kg m s–1 west
p(i) of boy = 50 × 4.4 = 220 kg m s–1 west
p(f) of boy = 50 × 1.68 = 84 kg m s–1 west
Copyright © Pearson Australia 2012 (a division of Pearson Australia Group Pty Ltd) ISBN 978 1 4425 5459 7 Page 14
Worked solutions
Chapter 2 Collisions and circular motion
Then p of boy = 84 kg m s–1 west – 220 kg m s–1 west = 136 kg m s–1 east. Notice that
this is equal and opposite to the change in momentum of the sled i.e. momentum is
conserved.
11
E k (i) = ½ mv2 + 0 = 0.5 × 0.250 × 102 = 12.5 J
then E k (f) = 12.5 × 0.95 = 11.9 = 12 J so D is correct.
12
Assume that the direction of positive velocity is east and use conservation of momentum.
p(i) = p(f)
(0.300 × 2.0) + (0.100 × –2.0) = 0.400 × v
v = 1.0 m s–1 east
13
% energy efficiency = (E k after collision/E k before collision)  100/1
= [0.5 × 0.400 × 1.02]/[0.5 × 0.300 × 2.02 + 0.5 × 0.100 × 2.02]  100/1
= (0.20/0.80) × 100/1
= 25% so C is correct.
14
B, 75% of the initial kinetic energy has been transformed into heat and sound energy.
15
The elastic potential energy stored in the spring during compression:
U s = ½ kx2= = 0.5 × 2000 × 0.102 = 10 J
E k of ball = ½ mv2 = 0.5 × 0.050 × 8.02 = 1.6 J
Then energy transformed into heat = 10 J – 1.6 J = 8.4 J
16
A, C and D are all correct.
Reducing the length of the tube will reduce the distance over which the frictional force acts as
the ball travels inside the tube. Therefore more of the spring energy will be available to the ball
as kinetic energy. Increasing the spring constant will result in more elastic potential with a
greater E K and hence greater velocity.
Using a ball with a lower mass will result in the ball being launched with a greater velocity
consequential on the formula ½ mv2 = 0.5kx2.
B is incorrect. Increasing the length of the tube will result in greater energy loss due to friction
inside the tube, and consequently there will be less kinetic energy available to the ball on
launching.
17
a
b
c
18
v = 2r/T
so T = 2r/v = 2 ×  × 20/10 = 12.6 = 13 s
19
a
b
c
10 m s–1 north so A.
10 m s–1 west so D.
10 m s–1 south so C.
Centripetal acceleration a = v2/r
= 102/20 = 5.0 m s–2 west
F = ma
= 1500 × 5.0 = 7.5  103 N south
ΣF = ma
= 1500  5.0
= 7.5  103 N east
This unbalanced force is friction, i.e. F f = 7.5  103 N east.
Copyright © Pearson Australia 2012 (a division of Pearson Australia Group Pty Ltd) ISBN 978 1 4425 5459 7 Page 15
Worked solutions
Chapter 2 Collisions and circular motion
20
C is correct because the speed of the car remains constant.
A, B and D are incorrect because all these quantities depend on direction, which is constantly
changing.
21
Use a force triangle with weight, normal and net force (acting horizontally).
F = mg tan33 = 55 × 9.8 × 0.6494 = 350 N
F = mv2/r = 350
55 × v2/28 = 350
v2 = 178
v = 13.3 m s–1 = 48 km h–1
22
Use a force triangle with weight, normal and net force (acting horizontally).
F N = mg/cos33 = 643 = 640 N
This normal force is significantly larger than the normal force of 540 N (F N = mg) that would
act if they were riding on a horizontal track.
23
a
b
At top: F = mv2/r = 50 × 5.02/10 = 125 N down
F N = F g – 125 = 490 – 125 = 370 N up
ii
At bottom: F = mv2/r = 50 × 5.02/10 = 125 N up
F N = F g +125 = 620 N up
D. At the top of the ride, F N  F g so he feels lighter than usual.
i
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