EXAM QUESTIONS - docscientia.co.za · A. no external forces act on the system. B. no friction forces act within the system. C. no kinetic energy is lost or gained by the system

EXAM QUESTIONS

Page 313

Unit 1: Momentum and impulse

Multiple-choice questionsFour possible answers are given for the following questions. Each question has only ONE correct answer. Choose the correct answer and mark the applicable LETTER with a cross (X).1. A rocket is fired vertically. At its highest point, it explodes. Which one of the

following describes what happens to its total momentum and total kinetic energy as a result of the explosion?

Total momentum Total kinetic energyA. Unchanged IncreasedB. Unchanged UnchangedC. Increased IncreasedD. Increased Increased

Exam Questions: Unit 1

2. Which of the following quantities are conserved in an inelastic collision between two bodies?

Total linear momentum of the bodies

Total kinetic energy of the bodies

A. Yes YesB. Yes NoC. No YesD. No No


3. A ball of mass m falls from rest on to a horizontal plate and bounces off it. The magnitudes of its velocity just before and just after the bounce are v1and v2 respectively. The variation with time t of the velocity v of the ball is shown below.The magnitude of the net force on the ball is given by which one of the following?

mv1 mv1 - v2T1 t2 - t1

Mv2 mv1 + v2t3 - t2 t2 - t1

A. B.

D.C.


4. Joe is standing on the surface of a frozen pond and he throws a ball horizontally. Considering Joe and the ball together, which one of the following correctly describes the change in the magnitude of the momentum and the change in the kinetic energy of Joe and the ball immediately after the ball is thrown?

Magnitude of momentum of Joe and ball

Kinetic energy of Joe and ball

A. No change Increases B. Increases Increases C. No change No changeD. Increases No change


5. An astronaut in outer space is holding a hammer and drifting at constant velocity. The astronaut throws the hammer in the opposite direction to that in which she is drifting. What change, if any, occurs in the total kinetic energy and the total momentum of the astronaut and hammer?

Total kinetic energy Total momentumA. No change Increased B. No change No changeC. Increased Increased D. Increased No change


6. A constant force is applied to a ball of mass m. The velocity of the ball changes from v1 to v2. The impulse received by the ball is:

A. m(v2 + v1). B. m(v2 – v1).C. m(v2

2 + v12). D. m(v2

2 – v12).


7. A ball of mass 2,0 kg falls vertically and hits the ground with speed 7,0 m⋅s–1 as shown. The ball leaves the ground with a vertical speed of 3,0 m⋅s-1. The magnitude of the change in momentum of the ball is:

A. zero. C. 10 N⋅sB. 8,0 N⋅s D. 20 N⋅s


8. Which of the following quantities are conserved in an inelastic collision in an isolated system of two objects?

Linear momentum of system

Kinetic energy of system

A. Yes YesB. Yes NoC. No YesD. No No


9. Two spheres X and Y are moving towards each other along the same straight line with moments of magnitude PX and PY respectively. The spheres collide and move off with moments pX and pY respectively, as illustrated. Which one of the following is a correct statement of the law of conservation of momentum for this collision?

A. PX + PY = px + py B. PX - PY = px + pyC. PX - PY = px - py D. PX + PY = px - py


10. The momentum of a system is conserved if:A. no external forces act on the system.B. no friction forces act within the system.C. no kinetic energy is lost or gained by the system.D. the forces acting on the system are in equilibrium.

11. The velocity of a body of mass m changes by an amount Δv in a time Δt. The impulse given to the body is equal to:

A. m∆t B.

C. D. m∆v

∆t∆vm∆v

∆t


12. A ball is held at rest at point X and is then released. It drops on to a flat horizontal surface and rebounds to a maximum height at point Y. Which one of the following graphs best shows the variation with time t of the momentum p of the ball as itmoves between point X and point Y?

B.


13. A small ball P moves with speed v towards another identical ball Q along a line joining the centres of the two balls. Ball Q is at rest. Kinetic energy is conserved in the collision.

Which one of the following situations is a possible outcome of the collision between the balls?B.


14. Two spheres of masses m1 and m2 are moving towards each other along the same straight line with speeds v1 and v2 as shown:

The spheres collide. Which of the following gives the total change in linear momentum of the spheres as a result of the collision?

A. 0 B. m1v1 + m2v2C. m1v1 - m2v2 D. m2v2 - m1v1


15. An object of mass m is initially at rest. An impulse J acts on the object. The change in kinetic energy of the object is:

A.

B.

C. J2m

D. 2J2m

J2

2mJ2

m


16. The engine of a rocket ejects gas at high speed, as shown below.

The rocket accelerates forward because:A. the momentum of the gas is equal to the momentum of the rocket.B. the gas pushes on the air at the back of the rocket.C. the change in momentum of the gas gives rise to a force on the

rocket.D. the ejected gas creates a region of high pressure behind the rocket.


17. A stationary metal plate is hanging freely on a string. A steel ball, travelling horizontally, hits the plate. The speed of the ball after the collision is less than before, but still in a horizontal direction, as shown. Which one of the following gives a correct statement, with a valid reason, about the type of collision between the ball and the plate?


Type of collision

Reason

A. Inelastic The sphere has changed its momentum during the collision.

B. Inelastic The sphere has lost kinetic energy during the collision.C. Unknown The change in momentum of the plate during the

collision is unknown.D. Unknown The kinetic energy of the plate after the collision is

unknown.


18. The diagram shows a trolley of mass 4,0 kg moving on a frictionless horizontal table at a speed of 2,0 m·s-1. It collides with a stationary trolley also of mass 4,0 kg. Which of the following diagrams shows a possible outcome?

A.


19. A ball of mass M hits a wall at speed V normal to the wall. It rebounds with speed v normal to the wall as shown. What is the magnitude of the change in momentum of the ball and the direction of the force that the wall exerts on the ball?

Change in momentum Direction of forceA. M(V - v) To the rightB. M(V - v) To the leftC. M(V + v) To the rightD. M(V + v) To the left


20. Two trolleys P and Q, are connected by a rubber band. They are at rest on a horizontal surface. The mass of Q is twice that of P. The trolleys are pulled apart so that the band is stretched and are then released.

The ratio is

A. ¼. B. ½.C. 1. D. 2.

magnitude of initial acceleration of trolley Pmagnitude of initial acceleration of trolley Q


21. An impulse J acts on a body of mass m that is initially at rest. What is the distance moved by the body in a time t after the impulse has been delivered?

A.

B.

C.

D. Jt

Jtm

Jm

Jmt


22. A ball of mass 160 g is travelling at 25 m⋅s-1 to the right. The ball is hit horizontally by a batsman and leaves the bat at 35 m⋅s-1 to the left. The magnitude of the change in momentum of the ball is:

A. 9 600 kg⋅m⋅s-1 B. 9,6 kg⋅m⋅s-1

C. 1,6 kg⋅m⋅s-1 D. 1 600 kg⋅m⋅s-1

IEB 2019 Sup. Paper 1


23. A ball of mass m travels horizontally with speed v before colliding with a vertical wall. The ball rebounds at a speed v in a direction opposite to its initial direction. What is the magnitude of the change in momentum of the ball?

A. 0 B. ½mvC. mv D. 2mv

IEB 2018 Paper 1


24. A stationary nucleus decays and emits an alpha particle of mass m. The alpha particle is emitted with momentum p and kinetic energy E. The mass of the recoiling nucleus is 50 times greater than the mass of the alpha particle. What are the magnitudes of the momentum and kinetic energy of the recoiling nucleus?

Momentum Kinetic energyA. p EB. P E

50C. p

50E

D. p50

E50


25. The momentum of a car vs time is illustrated in the graph. The gradient of the graph represents:

A. the velocity of the car.B. the resultant force on the car.C. the kinetic energy of the car.D. the rate of change of velocity of the car

IEB 2017 Sup. Paper 1


26. A person throws an object of mass M straight up with an initial speed v. The object rises to a maximum height H above the launch point. The person now throws an object of mass ½M straight up with a speed of 2v. In terms of H, to what maximum height does the object of ½M rise above the launch point? Ignore air resistance.

A. 4H B. 2HC. 2½H D. H


27. Car R is travelling towards car P as shown in the diagram. Car R has a greater mass than car P and is moving faster. The two cars collide head on. Which statement best describes the magnitude of the forces experienced by the cars during the collision?

A. Car R experiences the greater forceB. Car P experiences the greater forceC. The cars experience equal forcesD. It depends on the ratio of the car’s masses

IEB 2017 Paper 1


28. Ball P and Q, of the same mass, are dropped onto a concrete floor. Both balls hit the concrete floor at the same speed, v. Ball P rebounds with the same vertical speed, v, but ball Q rebounds with speed ½v. Refer to the diagram. Ignore air resistance. Which one of the following statements regarding the collision of each ball with the concrete floor is correct?


A. Kinetic energy is conserved for both balls.B. The change in momentum of ball P is greater than that of ball Q.C. The contact time with the floor is the same for both balls P and Q.D. Momentum is conserved for the collision of ball P, but not for

that of ball Q.

DBE NSC November 2018


29. A trolley of mass m is moving at constant velocity v to the right on a frictionless horizontal surface. A ball of clay, also of mass m, dropped vertically, falls onto the trolley at time t, as shown in the diagram. The ball of clay sticks to the trolley. Which one of the velocity-time graphs below correctly represents the velocity of the trolley before and after time t?


C.


Contextual questions1 A car of mass 1 500 kg travels to the right at 70 km⋅h-1. It runs into the back

of a stationary van of unknown mass. The two vehicles lock together and move to the right at 7 m⋅s-1.

1.1 Define momentum.The product of the mass and velocity of the object.


1.2 Convert 70 km⋅h-1 to m⋅s-1.

= 19,44 m·s-1

1.3 Write down the law of conservation of linear momentum in words.The total linear momentum of an isolated system remains constant.

1.4 Calculate the mass of the van.Σpi = Σpf

mcvci + 0 = (mc + mv)vf(1 500)(19,44) = 7(1 500) + mv(7)

mv = 2 666,67 kg

703,6


1.5 Is this an elastic collision? Motivate your answer by means of a calculation.ƩEki = ½mv2

= ½(1 500)(19,44)2 + 0= 283 435,2 J

ƩEkf = ½mv2

= ½(1 500 + 2 666,67)(7)2

= 102 083,42 J

It is not an elastic collision as kinetic energy has been lost. (EKi > EKf)

Adapted from IEB 2019 Sup. Paper 1


2 Students are performing an experiment in the lab. A 0,9 kg trolley is travelling with a constant velocity of 2,4 m⋅s-1 on a long frictionless track when a metal cylinder of mass m is dropped vertically onto the trolley. The trolley with the metal cylinder continues to move with velocity v as shown in the diagram.


2.1 Use the graph to determine the magnitude of the velocity (v) of the trolleyafter the metal cylinder is dropped on the trolley.

v = helling van die x-t-grafiek

=

= 1,8 m·s-1

10,2 - 4,85 - 2


The students use video analysis to plotthe following position vs time graph of the trolley:

2.2 The law of conservation of momentum is a consequence of which of Newton’s laws?

Newton’s third law

2.3 Calculate the mass (m) of the metal cylinder.Ʃpi = Ʃpf

mtvti + 0 = (mt + m)vf(0,9)(2,4) + 0 = (0,9 + m)(1,8)

m = 0,3 kg


2.4 Why is the vertical velocity of the falling metal cylinder not taken into consideration?

Momentum is a vector and perpendicular directions are conserved independently.ORMomentum is a vector and the velocity of m has no horizontal component.

Adapted from IEB 2018 Paper 1


3 A small car of mass 1 084 kg was travelling east at a speed of 33 m⋅s-1. A large SUV of mass 3 447 kg was travelling west at a speed of 28 m⋅s-1. The two vehicles collided head on with each other.

Immediately after the collision, the small car was moving west at 5 m⋅s-1.3.1 Name the law you would use to calculate the velocity of the SUV

immediately after the collision.The law of conservation of linear momentum


3.2 Determine the velocity of the SUV immediately after the collision.Ʃpi = Ʃpf + east

mcvci + mSUVv(SUV)i = mcvcf + mSUVv(SUV)f(1084)(33) + (3437)(-28) = (1084)(-5) + (3437)v

v = -16,02 m·s-1

v = 16,02 m·s-1 west

3.3 Define elastic collision.A collision in which both momentum and kinetic energy are conserved.


3.4 Use a calculation to determine if the collision was an elastic collision.ƩEki = ½mv2

= ½(1084)(33)2 + ½(3437)(28)2

= 1 937 542 J

ƩEkf = ½mv2

= ½(1084)(5)2 + ½(3437)(16,02)2

= 454 587,53 J

It is not an elastic collision as kinetic energy has been lost. ƩEKi > ƩEKf

Adapted from IEB 2016 Sup. Paper 1


4 A student is experimenting with two carts on a frictionless surface. Cart X has a mass of 0,5 kg and is initially at rest. Cart Y has a mass of 1,2 kg and collides with cart X. Cart Y is initially travelling at 8 m⋅s-1 east and travels 4 m⋅s-1 east after colliding with cart X. The velocity of cart Y is represented on the velocity time graph.


4.1 Use the law of conservation of linear momentum to calculate themagnitude of the final velocity of cart X.

Ʃpi = Ʃpf + eastmxvxi + myvyi = mxvxf + myvyf(0) + (1,2)(8) = (0,5)vxf + (1,2)(4)

v = 9,6 m·s-1

4.2 Calculate the change in momentum of cart Y.∆py = m(vf - vi)∆py = (1,2)(4 - 8)∆py = -4,8∆py = 4,8 kg m·s-1 west


4.3 Calculate the magnitude of the force that cart X exerts on cart Y.

Fnet =

=

= 3 200 N

∆p ∆t

4,81,5 × 10-3


5 A rocket of total mass (m) is in deep space travelling at a speed of 1 000 m⋅s-1. The rocket fires its engines. The burnt fuel gas has an exhaust speed of 500 m⋅s-1 in the opposite direction to the motion of the rocket.

5.1 Explain why firing the engine changes the speed of the rocket. Use one of Newton’s laws to help you in your explanation.

The firing engine provides a resultant force on the rocket. As the firing engine pushes gas out of the exhaust and gas pushes back on the firing engine (attached to the rocket) according to Newton’s third law of motion. ORThe firing engine provides a resultant force on the rocket.According to Newton’s second law of motion, it accelerates.


5.2 Calculate the rocket’s final speed if one third of its mass is lost due to theburnt fuel being ejected in a single short burn.Ʃpi = Ʃpf

(1 000)vi = 2⁄3 mv + 1⁄3 m(-500)v = 1 750 m·s-1

Adapted from IEB 2015 Paper


6 Two different balls A and B of masses 250 g and 300 g respectively collide head-on. Before the collision, ball A is moving to the right at 4 m⋅s-1 and ball B is moving to the left at 6 m⋅s-1 before the collision. During the collision, ball A experiences an impulse of 1,5 kg⋅m⋅s-1 to the left.

6.1 Calculate the magnitude and direction of the velocity of ball A immediately after the collision.

∆py = m(vf - vi)-1,5 = (0,25)( vf - 4)

vf = -2 m·s-1

vf = 2 m·s-1 left


6.2 Calculate the magnitude and direction of the velocity of ball B immediately after the collision.

Ʃpi = ƩpfmAvAi + mBvBi = mAvAf + mBvBf

(0,25)(4) + (0,3)(-6) = (0,25)(-2) + (0,3)(vBf)vBf = -1 m·s-1

vBf = 1 m·s-1 left

IEB 2014 Paper 1


7 Two cars, A and B, collide head-on and lock together on impact, as represented in the sketch below. Ignore frictional effects during the collision but not after the collision.


7.1 Calculate the speed of car B immediately before it collided with car A.Ʃpi = Ʃpf

mAvAi + mBvBi = (mA + mB)vf(1 500)(20) + (1 000)vBi = (1 500 + 1 000)(6)

vBi = -15 m·s-1

vBi = 15 m·s-1


7.2 Use suitable calculations to determine whether or not the collision was elastic.

ƩEki = ½mv2

= ½(1 500)(20)2 + ½(1 000)(15)2

= 412 500 J

ƩEkf = ½mv2

= ½(2 500)(6)2

= 45 000 J

It is not an elastic collision as kinetic energy has been lost.ƩEKi > ƩEKf


7.3 Define impulse.The product of the net force and the contact time

7.4 Which car, if any, experiences the greater magnitude of impulse during the collision?

Neither JA = JB


7.5 Justify your answer to Question 7.4 by referring to relevant physical principles and/or supplying supporting calculations. You must refer to or make use of a suitable formula.

J = ∆p= m(vf - vi)

JA = 1 500(6 - 20)= -21 000 N·s

JB = 1 000(6 - (-15))= 21 000 N·s


ORAccording to Newton’s third law the two cars exert a force of the same magnitude but in an opposite direction on one another.

From Fnet = the change in momentum will be the same for both cars as the

contact time is also the same for both cars. Change in momentum is equal to

the impulse, thus both cars experience an impulse of the same magnitude in

opposite directions.

∆p ∆t


The drivers of both cars have the same mass. Both drivers are wearing seatbelts which stretch slightly to bring the drivers to rest, relative to the car, in the same time period. Neither car has air bags.7.6 Which driver is likely to experience the greater force during the collision?

Driver B

Fnet =

=

∆p ∆t

m(v∆)∆t


∆vA = 6 - 20

= -14 m·s-1

∆vB = 6 - (-15) = 2 m·s-1

The change in velocity of driver B will be greater and therefore he experiences a greater force because each have the same mass and stopping time is the same.


Explain your answer with reference to one or more suitable formulae.7.7 Using principles of physics, explain how the airbag protect the driver during a

collision.

The change in momentum is the same with or without airbags ∴ airbags increase the time during which momentum changes, thereby decreasing the force:

Fnet = ; Fnet ∝

IEB 2013 Paper 1

∆p ∆t

1 ∆t


8 Rob and Kate design an experiment to determine how the hardness of the barrier into which a car crashes affects the maximum force exerted on the car during the collision. They use a toy car which moves along a frictionless track at constant speed and crashes into different barriers attached to a force sensor as shown below.

8.1 Name the independent variable in this experiment.The hardness/type of material of the crash surface/barrier.


8.2 State TWO variables that would need to be kept constant in order to ensure a fair test.Simplified sketch graphs to show the relationship between force and time for two different barriers:

Mass of carSpeed of carType of material used to make carMass/size of barrierAngle of attack


OR momemtum of car

8.3 Rob and Kate make the following conclusions:“The collision of the car with the steel barrier results in the greatest change in momentum since the car experiences the greatest maximum force.” (Rob)“The collision of the car with the foam rubber barrier results in the greatest change in momentum since the force acts on the car for a longer period of time.” (Kate)

By means of suitable calculations prove who is correct (Rob, Kate or neither). Clearly show the method that you use to arrive at your answer.


Neither are correct since the change in momentum is the same with both barriers.

J = F∆t = “area under curve” = change in momentumJmetal = (0,5)(0,04)(12,6)

= 0,252 N·sJrubber = (0,5)(0,12)(4,2)

= 0,252 N·s


8.4 “The car bumper is designed to prevent or reduce physical damage to the front and rear ends of passenger motor vehicles in low-speed collisions.” Bumpers on older cars are made from steel. Newer models of cars have plastic bumpers lined with a foam cushioning material. Refer to the results of the experiment, as shown in the graphs, to explain how plastic-foam bumpers protect the car from damage more than steelbumpers do.

Plastic foam bumper increases the time over which the momentum changes, therefore decreasing the force acting on the car since,

Fnet = ; Fnet ∝ and change in momentum is the same for steel and foam.∆p ∆t

1 ∆t


8.5 A car of mass 1 200 kg crashes into a lamp post while moving at 20 m·s-1 and it rebounds from the lamp post at 8 m·s-1. The net force exerted by the lamp post on the car is 2,1 × 105 N backwards. Calculate the time that the car was in contact with the lamp post.

Fnet =

1 200(-8 - 20)-2,1 × 105

= 0,16 s

IEB 2012 Paper 1

∆p ∆t

∆t =


9 In the diagram, railway truck A is moving along a horizontal track. It collides with a stationary truck B and on colliding, the two join together. Immediately before the collision, truck A is moving with speed 5,0 m⋅s-1. Immediately after collision, the speed of the trucks is v. The mass of truck A is 800 kg and the mass of truck B is 1 200 kg.


9.1 Define linear momentum.The product of the mass and velocity of the object.

9.2 Calculate the speed v immediately after the collision.Ʃpi = Ʃpf

mAvAi + mBvBi = (mA + mB)vf(800)(5) + 0 = (800 + 1 200)vf

vf = 2 m·s-1


9.3 Calculate the total kinetic energy lost during the collision.ƩEk = ½mv2

= ½(800)(5)2 + ½(1 200)(0)2

= 10 000 J

ƩEkf = ½mv2

= ½(2 000)(2)2

= 4 000 J

Ek, lost = 10 000 - 4 000 = 6 000 J

9.4 Suggest what has happened to the lost kinetic energy.Transformed into heat and sound energy.


10 A ball of mass 2,0 kg falls vertically and hits the ground with speed 7,0 m·s–1

as shown below.

The ball leaves the ground with a vertical speed 3,0 m·s-1.10.1 Write down Newton’s second law in terms of momentum.The net force acting on an object is equal to the rate of change of momentum.


10.2 Draw a vector diagram to determine the change in momentum of the ball.


10.3 If the ball is in contact with the ground for 0,02 s. Determine the force exerted by the ground on the ball.

Fnet = + up from ground

m(vf - vi)∆t

(2,0)(3,0 - (-7,0))0,02

= 1 000 N up

∆p ∆t

=

=


11.1 The diagram below shows two identical balls A and B on a horizontal surface. Ball B is at rest and ball A is moving with speed v along a line joining the centres of the balls. The mass of each ball is M.


During the collision of the balls, the magnitude of the force that ball A exerts on ball B is FAB and the magnitude of the force that ball B exerts on ball A is FBA.

11.1 On the diagram below, add labelled arrows to represent the magnitude and direction of the forces FAB and FBA.


The balls are in contact for a time Δt. After the collision, the speed of ball A is +vAand the speed of ball B is +vB in the directions shown.

As a result of the collision, there is a change in momentum of ball A and of ball B.


11.2 Use Newton’s second law of motion to deduce an expression relating the forces acting during the collision to the change in momentum of:11.2.1 ball B; 11.2.2 ball A.

Fnet = FBA∆t = M(vAf - v)m(vf - vi)

∆tM(vBf )

∆tFAB∆t = M(vBf) (vi = 0 m·s-1)

∆p ∆t

FAB=

=


11.3 Apply Newton’s third law and your answers to Question 11.2, to deduce that the change in momentum of the system (ball A and ball B) as a result of this collision, is zero.

FAB = -FBA (according to third law)M(vBf) = M(vAf - v) (∆t is the same for A and B.)

Mv = M(vBf) + M(vAf)0 = (mvAf + mvBf) - Mv

= pf - pi0 = ∆p

This shows that the net change in momentum is zero.


11.4 Deduce, that if kinetic energy is conserved in the collision, then after the collision, ball A will come to rest and ball B will move with speed v.

From:Mv = M(vAf) + M(vBf)

v = (vAf) + (vBf)

Since: EK = ½mv2

v2 = (vBf)2 + (vAf)2 because ½m is constant.

If vAf = 0 thenv = vBf


12 Large metal bars can be driven into the ground using heavy falling object. In the situation shown, the object has a mass 2,0 × 103 kg and the metal bar has a mass of 400 kg. The object strikes the bar at a speed of 6,0 m.s-1. It comes to rest on the bar without bouncing. As a result of the collision, the bar is driven into the ground to a depth of 0,75 m.


12.1 Determine the speed of the bar immediately after the object strikes it.Ʃpi = Ʃpf

mv(object)i + mBvBi= (mobject + mB)vf(2 × 103)(6,0) + 0 = (2 × 103 + 400)vf

vf = 5 m·s-1


12.2 Determine the average frictional force exerted by the ground on the bar.vf

2 = vi2 + 2a∆x ↓+

0 = (5)2 + 2a(0,75)a = -16,67 m·s-2

Fnet = ma-f + mg = ma

-f + 2 400(+9,8) = 2 400(-16,67)f = 63 528 N upwards


13 A toy rocket of mass 0,12 kg contains 0,59 kg of water as shown in the diagram. The space above the water contains high-pressure air. The nozzle of the rocket has a circular cross-section of radius 1,4 mm. When the nozzle is opened, water emerges from the nozzle at a constant speed of 18 m·s-1.The density of water is 1 000 kg·m-3.The volume of water ejected per second through the nozzle is 1,1 × 10-4 m3.1 m3 of water = 1000 kg


13.1 Write down the law of conservation of linear momentum.The total linear momentum of an isolated system remains constant.

13.2 Write down Newton’s second law in terms of momentum.The net force acting on an object is equal to the rate of change of momentum.

13.3 Deduce that the upward force that the ejected water exerts on the rocketis 1,98 N. Explain your working by reference to Newton’s laws of motion.


Mass of water ejected per second = 1,1 × 10-4(1 000)

= 0,11 kg·s-1

Fnet =

= m(vf - vi)∆t

Fnet =

m∆t0,11(18-0)1

= 1,98 N

∆p ∆t


∆p ∆t

=

=

This is the change in momentum per second or the resultant force. According to Newton’s second law this is the force on the ejected water and according to Newton’s third law there is an equal force in magnitude but opposite in direction that acts upwards on the rocket, approximately 2,0 N.


14 A ride in a pleasure park has a carriage of mass 450 kg, which is travelling horizontally at a speed of 18 m·s-1. It passes through a shallow tank containing stationary water. The tank is 9,3 m in length. The carriage leaves the tank at a speed of 13 m·s-1.

As the carriage passes through the tank, the carriage loses momentum and causes some water to be pushed forwards with a speed of 19 m·s-1 in the direction of motion of the carriage.


14.1 For the carriage passing through the water-tank, deduce that the magnitude of its total change in momentum is 2 250 kg·m·s-1.

Δp = mΔv= 450(18 - 13)= 2250 kg·m·s-1

14.2 Use the answer in Question 14.1 to deduce that the mass of water moved in the direction of motion of the carriage is 118 kg.

Δpcarriage = Δpwater2 250 = m(19)

m = 118,42 kg


14.3 Calculate the magnitude of the average acceleration of the carriage in the water

vave =

= 15,5 m·s-1

= 0,60 s

F = ma

= (450)a

a = 8,3 m·s-2

(18 + 13)2

t =

=

2 2500,60


Dv9,3

15,5

14.4 Determine the total loss in kinetic energy of the carriage when passingthrough the water-tank.

Ek lost = ½m(vf2 - vi

2)= ½(450)(132 - 182)= -34 875 J

14.5 Determine the gain in kinetic energy of the water that is moved in thedirection of motion of the carriage.

Ek = ½mv2

= ½(118,42)(19)2

= 21 374,81 J


14.6 By reference to the principles of conservation of momentum and of energy, explain your answers in Questions 14.4 and 14.5.

Some water will be thrown “sideways”This accounts for the difference in the kinetic energies as this water will not have any momentum in the forward direction. Momentum is conserved. Not all the water moves in the direction of the carriage. Some water splashed up, using up the EK that was transferred for movement.


15 Initially a girl on roller skates is at rest on a smooth horizontal pavement. The girl throws a parcel, of mass 8 kg, horizontally to the right at a speed of 4 m·s-1. Immediately after the parcel has been thrown, the girl-roller-skate combination moves at a speed of 0,6 m·s-1. Ignore the effects of friction and rotation.


15.1 Will the girl-roller-skate combination move TO THE RIGHT or TO THE LEFT after the parcel is thrown? NAME the law in physics that can beused to explain your choice of direction.

To the left according to Newton’s third law.

The total mass of the roller skates is 2 kg.

15.2 Calculate the mass of the girl.Σpi = Σpf

mvgirl(i) + mvparcel(i) = mvgirl(t) + mvparcel(f)0 = (m+2)(-0,6) + (8)(4)

m = 51,33 kg


15.3 Calculate the magnitude of the impulse that the girl-roller-skate combinationis experiencing while the parcel is being thrown.

J = ∆p= m∆v= (51,33 + 2)(-0,6 - 0)= -32 N·s

15.4 Without any further calculation, write down the change in momentumexperienced by the parcel while it is being thrown.

32 kg m·s-1 to the right

DBE NSC November 2018 Paper 1


16 A 2 kg block is at rest on a smooth, frictionless, horizontal table. The length of the block is x. A bullet of mass 0,015 kg, travelling east at 400 m·s-1, strikes the block and passes straight through it with constant acceleration. Refer to the diagram. Ignore any loss of mass of the bullet and the block.

16.1 State the law of conservation of linear momentum in words.The total linear momentum of an isolated system remains constant.


The block moves eastwards at 0,7 m·s-1 after the bullet has emerged from it.

16.2 Calculate the magnitude of the velocity of the bullet immediately after itemerges from the block.

Σpi = Σpfmv(bullet)i + mv(block)i = mv(bullet)f + mv(block)f

(0,015)(400) + 0 = (0,015)v(bullet)f + (2)(0,7)v(bullet)f = 306,67 m·s-1


16.3 If the bullet takes 0,002 s to travel through the block, calculate the length, x, of the block.

∆p OR v1 + vf∆t 2m(vf - vi) 400 + 306,67

∆t 2(0,015)(306,666 - 400) = 0,71 m

0,002= 700 N

Fnet =

=

=

∆x = t

= 0,002


Fnet = ma ∆x = vi∆t + ½a∆t2

700 = (0,015)a = (400)(0,002) + ½(-46 666,67)(0,002)2

a = 46 666,67 m·s-2 = 0,71 m



17 The graph alongside shows how the momentum of car A changes with time just before and just after a head-on collision with car B.Car A has a mass of 1 500 kg, while the mass of car B is 900 kg.Car B was travelling at a constant velocity of 15 m·s-1 west before the collision. Take east as positive and consider the system as isolated.


17.1 What do you understand by the term isolated system as used in physics?A system on which the resultant external force is zero.

Use the information in the graph to answer the following questions.17.2 Determine the magnitude of the velocity of car A just before the collision

p = mv30 000 = (1500)v

v = 20 m·s-1


17.3 Calculate the velocity of car B just after the collision.Σpi = Σpf + east

mAvAi + mBvBi = mAvAf + mBvBf30 000+ (900)(-15) = 14 000 + 900vf

vf = 2,78 m·s-1 east

17.4 Calculate the magnitude of the net average force acting on car A during the collision.Δp Δt14 000 - 30 000

0,1= -160 000 N DBE NSC November 2016 Paper 1

Fnet =

=


18 Dancers have to learn many skills, including how to land correctly. A dancer of mass 50 kg leaps into the air and lands feet first on the ground. She lands on the ground with a velocity of 5 m·s-1. As she lands, she bends her knees and comes to a complete stop in 0,2 seconds.

18.1 Calculate the momentum with which the dancer reaches the groundp = mv

= 50(5)= 250 kg·m·s-1


18.2 Calculate the magnitude of the net force acting on the dancer as she lands.Assume that the dancer performs the same jump as before but lands without bending her knees.

ΔpΔt

0 - 2500,2

= 1 250 N

18.3 Will the force now be GREATER THAN, SMALLER THAN or EQUAL TO the force calculated in Question 18.2?

Greater than

Fnet =

=


18.4 Explain the answer to Question 18.3.

For the same momentum change, the contact time will be less.

Fnet ∝ ⇒ if Δt decreases Fnet will increase.


1Δt


19 A teacher demonstrates the principle of conservation of linear momentum using two trolleys. The teacher first places the trolleys, A and B, some distance apart on a flat frictionless horizontal surface, as shown in the diagram. The mass of trolley A is 3,5 kg and that of trolley B is 6,0 kg. Trolley A moves towards trolley B at constant velocity. The table below shows the position of trolley A for time intervals of 0,4 s before it collides with trolley B.


Relationship between position and time for trolley APosition of trolley A (m) 0 0,2 0,4 0,6Time (s) 0 0,4 0,8 1,2


19.1 Use the table above to prove that trolley A is moving at constant velocitybefore it collides with trolley B.

ΔxΔt0,20,40,40,80,61,2

= 0,5 m·s-1

v =

=

=

=


At time t = 1,2 s, trolley A collides with stationary trolley B. The collision time is 0,5 s after which the two trolleys move off together.19.2 Calculate the magnitude of the average net force exerted on trolley B by

trolley A.Σpi = Σpf Δp

mAvAi + mBvBi = (mA + mB)vf Δt(3,5)(0,5) + 0 = (3,5 + 6)vf m(vf - vi)

vf = 0,184 m·s-1 Δt(6)(0,184 - 0)

0,5= 2,21 N; A on B

Fnet =

DBE NSC May-June 2017 Paper 1

=

=


20 The diagram below shows two skateboards, A and B, initially at rest, with a cat standing on skate board A. The skateboards are in a straight line, one in front of the other and a short distance apart. The surface is flat, frictionless and horizontal.

EACH skateboard has a mass of 3,5 kg. The cat, of mass 2,6 kg, jumps from skateboard A with a horizontal velocity of 3 m·s-1 and lands on skateboard B with the same velocity of 3 m·s-1. Refer to the diagram below:


20.1 Calculate the velocity of skateboard A just after the cat has jumped fromit.

Σpi = Σpfmcvci + mBvBi = mcvcf + mSBvSBf

0 + 0 = (2,6)(3) + (3,5) vSBfvSBf = 2,23 m·s-1 to the left


Immediately after the cat has landed on skateboard B, the cat and skateboard B move horizontally to the right at 1,28 m·s-1.

20.2 Calculate the magnitude of the impulse on skateboard B as a result of the cat’s landing.J = Δp

= mΔv= (2,6)(1,28 - 3) OR 3,5(1,28 - 0)= -4,48 N·s = -4,48 N·s

DBE NSC Feb-March 2018 Paper 1


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EXAM QUESTIONS - docscientia.co.za · A. no external forces act on the system. B. no friction forces act within the system. C. no kinetic energy is lost or gained by the system