Chapter 2 Collisions and Circular Motion

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Heinemann Physics 12(3e)Copyright © Pearson Education Australia 2008 Teacher’s Resource and Assessment Disk(a division of Pearson Australia Group Pty Ltd) IS BN 9781442501263

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Worked solutions

Chapter 2 Collisions and circular motion

2.1 Momentum and impulse

1 The mosquito has more momentum. Stationary objects have zero momentum.

2Momentum 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 p = mv = 9.1 ! 10 –31 ! 3.0 ! 10 7 = 2.7 ! 10 –23 kg m s –1 b p = mv = 6.0 ! 10 24 ! 3.0 ! 10 4 = 1.8 ! 10 29 kg m s –1 c p = mv = 0.10 ! 2.3 = 0.23 kg m s –1

4 D is the correct answer. If the direction of motion is different, the momentum must also bedifferent.

5 a 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 timeinterval, the force that is acting must be reduced in size.

b His momentum would change over a very short time interval. It would require a verylarge force to do this, which would be painful (or damaging) for Vijay.

6 a Change in speed: " v = v – u = 8.0 – 10 = –2 m s –1 b Change in velocity: " v = v – u = 8.0 m s –1 up – 10 m s –1 down = 18 m s –1 upwardsc " p = m" v = 0.080 ! 18 = 1.4 kg m s –1 upd 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 = 29 N upb The forces acting are gravity F g downwards and normal reaction force F N upwards.

F g = m g = 0.78 N#F = F g + F N This gives F N = 29 N up.

c As described by Newton’s third law, this is equal and opposite to the force that the floorexerts on the ball, so F = 29 N down.

8 a From the graph: F max = 500 N forwardsb As described by Newton’s third law, the maximum force that the ball exerts on the

racquet is 500 N backwards.c Net impulse = area under F – t graph = ! ! 0.020 ! 500 = 5.0 N s forwardsd Net impulse = " p = 5.0 kg m s –1forwards



Worked solutions Chapter 2 Collisions and circular motion


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e " 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 a 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 eachdummy.

b Net impulse = " p, so impulse is the same for each dummy.c The dummy in car B stops in a shorter time interval and so would experience a greater

stopping force.

d Car A: #F " t = m" v F A " 0.40 = 100 " -20and F A= -5.0 ! 10 3 NF B " 0.10 = 100 " -20F B = -2.0 ! 10 4 N

e Crumple zones reduce the magnitudes of the forces that act on the occupants of a carduring a collision. This will result in fewer or less serious injuries.

f 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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2.2 Conservation of momentum

1 a For the sports car: p = mv = 1.0 ! 10 3 " 10 = 1.0 ! 10 4 kg m s –1 east

b For the station wagon: p = mv = 2.0 ! 10 3 " 5.0 = 1.0 ! 10 4 kg m s –1 west

c Total momentum = 1.0 ! 10 4 kg m s –1 east + 1.0 ! 10 4 kg m s –1 west = 0

2 a # p i = # p f From Question 1c, # p i = 0 so # p f = 0 i.e. common velocity = 0

b It hasn’t gone anywhere. The vehicles had a total of zero momentum before the collisionand so there is zero momentum after the collision.

c The change in momentum of the sports car = " p – p f – p i = 0 – 1.0 ! 10 4 kg m s –1 east= 1.0 ! 10 4 kg m s –1 west

d The change in momentum of the station wagon " p – p f – p i = 0 – 1.0 ! 10 4 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) + (0.100 v)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) + (100 v) 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) vand 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 at5.0 m s –1. Because the water keeps its horizontal momentum, so too will the tanker retain itsmomentum. There is no transfer of momentum between the tanker and the water, so each willcontinue to travel forwards at 5.0 m s –1.

7Considering the horizontal motion only. The bag of apples falls vertically and so initially has nohorizontal 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, andone cannot be converted into the other.

9 a B is the correct answer because the trolley slows down and so has less momentum afterthe bag of apples is dropped in it.





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b A is the correct answer because the total momentum of the trolley and its contentsremains constant (conservation of momentum).

10 a p = mv = (48 " 40) = 190 kg m s –1 rightb # p i = # p f

192 = (50 " vg)

and vg = 3.8 m s –1 rightc # p i = # p f

The girl retains her momentum as she jumps from the skateboard, so there is no transferof 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 tothe 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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2.3 Work, energy and power

1 a W = Fx = (800 " 9.8) " 90 = 7.1 ! 10 5 Jb W = Fx = (800 " 9.8) " 50 + 0.5(800 kg)(2.0 m s –1)2 = 3.9 ! 10 5 Jc P = Fv = (800 " 9.8) " 2.0 = 1568 W = 16 kW

2 a F h = F cos $ = 30cos60 ° = 15 NW = F h " x = 15 " 2.5 = 37.5 = 38 N m

b Work done by friction: W = F f " x = 10 " 2.5 = 25 N mc Work done by # F = # F " x = (15 – 10) " 2.5 = 12.5 = 13 N m

3 a " E k = work done by net force = 12.5 J = 13 Jb Energy dissipated as heat = work done by friction = 25 Jc friction

4 The vertical component of the 30 N pulling force is acting perpendicular to the dog’s directionof 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 at 2 2.5 = 0 + 0.5 " 2.5 " t 2 t = % 2.0 = 1.4 s

6 a 100 g ball: E k (i) = ! mv2 = 0.5 " 0.10 " 02 = 0200 g ball: E k (i) = ! mv2 = 0.5 " 0.20 " 9.0 2 = 8.1 J# E k (i) = 0 + 8.1 = 8.1 J

b 100 g ball: E k (f) = ! mv2 = 0.5 " 0.10 " 12 2 = 7.2 J200 g ball: E k (f) = ! mv2 = 0.5 " 0.20 " 3.0 2 = 0.9 J# E k (f) = 7.2 + 0.9 = 8.1 J

c Elastic. Kinetic energy has been conserved.d No, during the collision some energy would have been transformed into heat and sound

energy. Truly elastic collisions only occur at the atomic level.

7 a W = " U s = 0.5 kx2, where k = gradient of force–compression graph = 1.0 ! 10 3 N m –1 Then W = 0.5 " 1.0 ! 10 3 " 0.010 2 = 0.050 N m

b During compression, the girl does work to alter the shape of the ball. During release, thestored energy is released in restoring the ball to its original shape.

c P = W /t = (120 " 0.050)/60 = 0.10 W

8 Kinetic energy has been transformed into elastic potential energy0.5mv2 = 0.5 kx2

x2 = mv2/k = 0.10 " 4.0 2/1.0 ! 10 3 then x = 4.0 ! 10 –2 m = 4.0 cm

9 a Loss of gravitational potential energy = mg " h

= 80 " 9.8 " 5.0 = 3.9 ! 10 3 Jb u = 0, a = 5.0 m s –2, x = 5.0 m, v = ?

v2 = u2 + 2 ax = 0 + 2 " 5.0 " 5.0 = 50v = 7.1 m s –1

Increase in kinetic energy = 0.5 mv2





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= 0.5 " 80 ! 7.1 2 = 2.0 " 10 3 J

c This is due to energy being transformed into heat and sound due to friction.d W = F ! x = mgx = 80 " 9.8 " 5.0 = 3.9 ! 10 3 Je Heat produced = energy lost due to friction

= 3.92 ! 103

J – 2.0 ! 103

J = 1.9 ! 103

J10 a Yes, momentum is conserved in all collisions

b Inelastic; some kinetic energy has been transformed into heat and sound energyc E k (i) = ! mv2 + ! mv2

= ! " 4.0 " 3.0 2 + ! " 4.0 " 3.0 2 36 J20 J transformed into heat and sound, so E k (f) = 36 – 20 = 16 JFrom symmetry, the balls will have the same final speeds and the same kinetic energiesof 8.0 J.For each ball: E k = ! mv2 = 0.5 " 4.0 " v2 = 8.0v2 = 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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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 fromwhich the spring is made, and not the length of the spring.

2 a From the graph, F = 200 Nb Spring constant = gradient of graph = 1.0 ! 10 4 N m –1 c W = U s = ! kx2 = 0.5 " 1.0 ! 10 4 " (2.0 ! 10 –2)2 = 2.0 N md U s = ! kx2

= 0.5 " 1.0 ! 10 4 " 0.010 2 = 0.50 Je

3 a The force constant for each spring is determined by the gradient of the force–compressiongraph:Spring A: k = 2.0 ! 10 4 N m –1 Spring B: k = 1.0 ! 10 4 N m –1 Spring C: k = 5.0 ! 10 3 N m –1

b The stiffest spring will have the highest spring constant; therefore the ranking is:

C, B, A.c Using U s = ! kx2, where x = 0.010 m,and using the respective spring constants for each spring:A: U s = 1.0 JB: U s = 0.50 JC: U s = 0.25 J

4 ! k A x2

A = ! k B x2

B 0.5 " 2.0 ! 10 4 " xA

2 = 0.5 " 1.0 ! 10 4 " xB2

and xA/ xB = 1/ 2 = 0.71

5 a F = mg = kx k = mg / x = 0.075 " 9.8/0.050 = 14.7 = 15 N m –1

b U s = ! kx2= 0.5 " 14.7 " 0.050 2 = 1.8 ! 10 –2 J

6 a x = 45 – 30 = 15 cm = 0.15 mU s = ! kx2= 0.5 " 14.7 " 0.15 2 = 0.16 J

b F = mg = kxk = mg / x,then m = kx/ g = 14.7 " 0.15/9.8 = 0.23 kg

7 a E k (i) = ! mv2 = 0.5 " 2.5 " 1.5 2 = 2.8 J

bEnergy stored in spring is equal to initial kinetic energy of cart, i.e. 2.8 J.





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8 a U s = ! kx2= 2.80.5 " 400 " x2 = 2.8

x = 0.12 mb U s + E k = 2.8

! kx2+ E k = 2.80.5 " 400 " 0.040 2 + E k = 2.80.32 + E k = 2.8

E k = 2.5 J

9 a The work done by the archer on the bow is equal to the area under the force–distancegraph from XY = 0.10 m to XY = 0.30 m.This area = 7.2 J

b It is being stored as elastic potential energy in the bow.

10 a As the string returns to its original position, all the elastic potential energy stored in thestring is transferred to the arrow.

The work done by the string on the arrow = 7.2 Jb 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. Theassumption 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 itsun-extended position.





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2.5 Circular motion

1 a A and D are correct

b The passenger has continued straight ahead and the car has turned to the left.

2 a 8.0 m s –1

b 8.0 m s –1 southc a = v2/r = 8.0 2/9.2 = 7.0 m s –2 towards centre, i.e. east

3 a #F = ma = 1200 " 7.0 = 8.4 " 10 3 N east

b Friction between the tyres and the road surface.

4 a 8.0 m s –1 north

b towards centre i.e. west

5 The force needed to give the car a larger centripetal acceleration will eventually exceed themaximum frictional force that could act between the tyres and the road surface. At this time, thecar would skid out of its circular path.

6 a a = v2/r = 2.0 2/1.5= 2.7 m s –2 towards centre of circle

b The forces are unbalanced because the skater has an acceleration as she skates in acircular path.

c #F = ma = 50 " 2.67= 130 N towards centre of circle

dThe horizontal force exerted on the blades by the ice; i.e. friction

7 a T = 1/ f = 1/2.0 = 0.50 s

b v = 2 &r/T = 2& " 0.80/0.50 = 10 m s –1

c a = v2/r = 10 2/0.80 = 126 = 130 m s –2 towards centre of circle

d #F = ma = 2.5 " 126 = 315 = 320 N

e The force that is responsible for the centripetal acceleration of the ball is the tension inthe wire, which is directed radially inwards at all times.

f 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 /Tso T = 2 &r /v = 2 & " 62/13.9 = 28 s

9 #F = mv2 /r = 1.6 " 13.9 2/62 = 5.0 N

10 The sideways force that the air exerts on the wings of the plane.





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2.6 Aspects of circular motion

1 Velocity is tangential to path.

a westb south-west c north

2 Acceleration is towards centre of circular path.

a southb south-east

c west

3 Net force is in direction of acceleration.

a southb 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 energyis a scalar and so is constant. Answer is C.

6 a v =T

r !2

= 2& " 0.800/1.36 = 3.70 m s –1 b a = v2/r

= 3.70 2/0.800 = 17.1 m s –2 radially inwards

c #F = ma = 0.0250 " 17.1 = 0.430 N radially inwards, so magnitude is 0.430 N

7 a

b Use a force triangle for the ball.F t = mg /sin30.0 ° = 0.0250 " 9.80/0.50 = 0.490 N

8 a r = 2.4cos60 = 1.2 m

b The forces are her weight acting vertically and the tension in the rope acting along therope towards the top of the maypole.

c She has an acceleration directed towards point B, the centre of her circular path.d Use a force triangle for the girl, showing the net force towards B.

#F = mg /tan60 = 294/1.73 = 170 N towards B.e #F = mv 2/r

170 = 30 " v2/1.2v = 2.6 m s –1

9 a #F = mv2/r = 1200 " 18 2/80 = 4860 = 4.9 kN





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b Use a force triangle for the car, showing the normal, weight and the net force actinghorizontally.tan $ = 4860/1200 " 9.8 = 0.413$ = 22 °

10The greater speed would make the car travel higher up the track. The driver would have to turnthe wheels slightly towards the bottom of the track so as to create a sideways force of frictiontowards the bottom of the track.





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2.7 Circular motion in a vertical plane

1 a It has a constant speed so its centripetal acceleration a = v2/r is also constant inmagnitude.

b At the bottom of its path, the yo-yo has an upwards acceleration and so the net force isup. This indicates that the tension force is greater than F g.

c At the top of its path, the yo-yo has an downwards acceleration and so the net force isdown. This indicates that the tension force is less than F g.

2 At this point,a = v2/r = g so v = % (r " g ) = % 1.5 " 9.80 = 3.8 m s –1

3 a The weight force from gravity and the normal force from the road.

b #F = mv2 /r = 800 " 3.8 2/10 = 1.2 " 10 3 Nc 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 feelslighter at this point.

d 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

4 a At X, mechanical energy is:

' E = E k + U g = 1/2 mv2 + mgh

= 0.5 " 500 " 2.00 2 + 500 " 9.80 " 50.0= 1000 + 245,000

= 246 000 JAt Y: U g is zero so its kinetic energy is 246 000 J1/2 mv2 = 246 0000.5 " 500 " v2 = 246 000v = % 984 = 31.4 m s –1

b At Z, mechanical energy = 246 000 J' E = E k + U g 246 000 = E k + 500 " 9.80 " 30.0246 000 = E k + 147 000

E k = 99 0000.5 " 500 v2 = 99 000v = 19.9 m s –1

c At Z, acceleration is:a = v2/r = 19.9 2/15.0 = 26.4 m s –2 down' F = ma = 500 " 26.4 = 13 200 N downBoth F N and F g are acting downwards, so:F N + F g = 13,200F N = 13 200 – 500 " 9.80 = 8300 N = 8.3 " 10 3 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





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6 If the pilot is upside-down at the top of the loop, both the normal force from the seat and gravityare acting downwards.' F = F N + F g mv2/r = F N + 80 " 9.8080 " 35 2/100 = F N + 784F

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 = 9 g so v = % (r " 9 g ) = % (400 " 88.2) = 190 m s –1

9 a F t = F g = 4.0 " 9.80 = 39 N

b First determine the speed of the ball at X. If it is released from rest, the kinetic energy atX is equal to its gravitational potential energy at Z.

E k (X) = U g (Z) = 4.0 " 9.80 " 2.0 = 78.4 J! mv2 = 78.40.5 " 4.0 " v2 = 78.4v = 6.3 m s –1 #F = mv2/r = 4.0 " 6.3 2/2.0 = 78.4 N upF t = 78.4 + mg = 78.4 + 39.2 = 120 N up

10 The wire is more likely to snap when the ball is moving through X. The bowling ball has anupwards acceleration at X and so tension is greater that the weight force. This larger tension inthe wire means it is more likely to break.





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Chapter review

1 a v2 = u2 + 2 ax = 0 + 2 " 9.8 " 10

v = 14 m s –1

p = mv = 0.200 " 14= 2.8 kg m s –1 down

b p = mv = 0.200 " 10= 2.0 kg m s –1 up so C is correct.

c 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.

d # F " t = " p #F = = 4.8/1.0 ! 10 –3 = 4.8 ! 10 3 N up

2 a " E k = E k (f) – E k (i) = 0.5 " 50.0 " 3.00 2 – 0 = 225 Jb u = 0, t = 10 s, v = 3.0 m s –1, x = ?

x = t (u + v)/2 = 10 " 3.0/2 = 15 mW = ( F cos $) " x = (200 cos 60 ° ) " 15= 1500 J = 1.50 " 10 3 J

c The amount of kinetic energy converted into heat = 1.5 ! 10 3 J – 225 J = 1.28 " 10 3 J

3 a P = work done by student/time taken = 1500/10.0 = 150 Wb P = work done by friction/time taken = 1275/10.0 = 128 Wc 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 doneThen 2.0 – 5.0 = F f " 3.0Hence 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 p (i) = mv = 250 " 0 = 0b p (f) = mv = 50 " 4.0 = 200 kg m s –1 eastc Since momentum is conserved, the momentum of sled = 200 kg m s –1 west

8 p(sled) = mv = 200 v = 200 kg m s –1 westand v = 1.0 m s –1 west

9 # p (i) = # p (f)50 " 4.4 west + 200 kg m s –1 west = 250 v 420 = 250 v v = 1.7 m s –1 west

10 a 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





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b 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 westThen " p of boy = 84 kg m s –1 west – 220 kg m s –1 west = 136 kg m s –1 east. Notice thatthis is equal and opposite to the change in momentum of the sled i .e. momentum isconserved.

11 # E k (i) = ! mv2 + 0 = 0.5 " 0.250 " 10 2 = 12.5 Jthen # 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 " vv = 1.0 m s –1 east

13 % energy efficiency = ( # E k after collision/ # E k before collision) ! 100/1= [0.5 " 0.400 " 1.0 2]/[0.5 " 0.300 " 2.0 2 + 0.5 " 0.100 " 2.0 2] ! 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 = ! kx2= = 0.5 " 2000 " 0.10 2 = 10 J

E k of ball = ! mv2 = 0.5 " 0.050 " 8.0 2 = 1.6 JThen 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 asthe ball travels inside the tube. Therefore more of the spring energy will be available to the ballas kinetic energy. Increasing the spring constant will result in more elastic potential with agreater E K and hence greater velocity.Using a ball with a lower mass will result in the ball being launched with a greater velocityconsequential on the formula ! mv2 = 0.5 kx2.B is incorrect. Increasing the length of the tube will result in greater energy loss due to frictioninside the tube, and consequently there will be less kinetic energy available to the ball onlaunching.

17 a 10 m s –1 north so A.b 10 m s –1 west so D.c 10 m s

–1

south so C.18 v = 2 &r /T

so T = 2 &r /v = 2 " & " 20/10 = 12.6 = 13 s

19 a Centripetal acceleration a = v2/r = 10 2/20 = 5.0 m s –2 west

b #F = ma = 1500 " 5.0 = 7.5 ! 10 3 N south

c F = ma = 1500 ! 5.0= 7.5 ! 10 3 N eastThis unbalanced force is friction, i.e. F f = 7.5 ! 10 3 N east.





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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 constantlychanging.

21 Use a force triangle with weight, normal and net force (acting horizontally).

#F = mg tan33 ° = 55 " 9.8 " 0.6494 = 350 N#F = m v2/r = 35055 " v2/28 = 350v2 = 178v = 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 NThis normal force is significantly larger than the normal force of 540 N ( F N = m g ) that wouldact if they were riding on a horizontal track.

23 a i At top: #F = mv2/r = 50 " 5.0

2/10 = 125 N down

F N = F g – 125 = 490 – 125 = 370 N upii At bottom: #F = mv2 /r = 50 " 5.0 2/10 = 125 N up

F N = F g +125 = 620 N upb D. At the top of the ride, F N < F g so he feels lighter than usual.

Chapter 2 Collisions and Circular Motion

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