Momentum

Chapter 8 MCQs

For Full Scale Course: Written and Compiled By Sir Hunain Zia (AYLOTI), World Record Holder With 154 Total Personal A Grades, 11 World Records and 7 Distinctions, Educate A Change.

1

A ball of mass 0.25 kg moves at 18 m/s.

What is its momentum?

A 0.014 kg m/s
B 4.5 kg m/s
C 18.25 kg m/s
D 72 kg m/s

2

A trolley has momentum 36 kg m/s and mass 4.0 kg.

What is its velocity?

A 0.11 m/s
B 9.0 m/s
C 32 m/s
D 144 m/s

3

An object has momentum 24 kg m/s and velocity 6.0 m/s.

What is its mass?

A 0.25 kg
B 4.0 kg
C 18 kg
D 144 kg

4

Which unit is equivalent to kg m/s?

A N
B N/s
C N s
D N/m

5

A force of 12 N acts on an object for 0.50 s.

What impulse is delivered to the object?

A 0.042 N s
B 6.0 N s
C 12.5 N s
D 24 N s

6

A force acts on a trolley for 0.20 s and changes its momentum by 8.0 kg m/s.

What is the average force?

A 1.6 N
B 8.2 N
C 40 N
D 160 N

7

A ball of mass 0.40 kg moving at 15 m/s is brought to rest.

What is the magnitude of the impulse on the ball?

A 0.027 N s
B 6.0 N s
C 15 N s
D 37.5 N s

8

A 0.060 kg tennis ball moving at 30 m/s rebounds in the opposite direction at 20 m/s.

What is the magnitude of the change in momentum of the ball?

A 0.60 kg m/s
B 1.2 kg m/s
C 1.8 kg m/s
D 3.0 kg m/s

9

A ball of mass 0.20 kg moving east at 12 m/s rebounds west at 8.0 m/s.

Taking east as positive, what is the change in momentum of the ball?

A −4.0 kg m/s
B −0.80 kg m/s
C +0.80 kg m/s
D +4.0 kg m/s

10

A force of 50 N acts on an object for 0.40 s.

The object has mass 2.0 kg and is initially at rest.

What is its final speed?

A 4.0 m/s
B 10 m/s
C 20 m/s
D 25 m/s

For Full Scale Course: Written and Compiled By Sir Hunain Zia (AYLOTI), World Record Holder With 154 Total Personal A Grades, 11 World Records and 7 Distinctions, Educate A Change.

11

A 3.0 kg trolley moving at 4.0 m/s collides with a stationary 1.0 kg trolley. They stick together.

What is their common speed after collision?

A 1.0 m/s
B 3.0 m/s
C 4.0 m/s
D 12 m/s

12

A 2.0 kg trolley moving right at 6.0 m/s collides with a 4.0 kg trolley moving right at 3.0 m/s. They stick together.

What is their common speed after collision?

A 3.0 m/s
B 4.0 m/s
C 4.5 m/s
D 9.0 m/s

13

A 5.0 kg trolley moving right at 2.0 m/s collides with a 3.0 kg trolley moving left at 4.0 m/s. They stick together.

Taking right as positive, what is their velocity after collision?

A 0.25 m/s left
B 0.25 m/s right
C 2.75 m/s left
D 2.75 m/s right

14

A 0.50 kg ball moving right at 10 m/s collides with a wall and rebounds left at 6.0 m/s.

What is the magnitude of the impulse exerted by the wall on the ball?

A 2.0 N s
B 3.0 N s
C 5.0 N s
D 8.0 N s

15

A ball of mass 0.10 kg moving at 40 m/s hits a wall and rebounds at 30 m/s in the opposite direction. The contact time is 0.020 s.

What is the magnitude of the average force on the ball?

A 50 N
B 150 N
C 350 N
D 700 N

16

A student says, “Momentum is conserved in every collision because kinetic energy is always conserved.”

Which statement is correct?

A Both momentum and kinetic energy are always conserved in collisions.
B Momentum is conserved in a closed system, but kinetic energy may not be conserved.
C Kinetic energy is conserved, but momentum is never conserved.
D Momentum is only conserved if the objects stick together.

17

A 2.0 kg object moving at 5.0 m/s explodes into two parts. One part has mass 0.80 kg and moves forward at 12 m/s.

What is the velocity of the other part?

A 0.33 m/s forward
B 0.33 m/s backward
C 4.7 m/s forward
D 12 m/s backward

18

A stationary object explodes into two parts. Part X has mass 3.0 kg and moves east at 4.0 m/s. Part Y has mass 2.0 kg.

What is the velocity of part Y?

A 2.0 m/s west
B 4.0 m/s west
C 6.0 m/s west
D 12 m/s west

19

Two trolleys collide on a frictionless track.

Trolley A has mass 1.5 kg and velocity 6.0 m/s to the right. Trolley B has mass 2.5 kg and velocity 2.0 m/s to the left.

What is the total momentum before collision, taking right as positive?

A 4.0 kg m/s right
B 5.0 kg m/s right
C 9.0 kg m/s right
D 14 kg m/s right

20

A 4.0 kg trolley moving right at 3.0 m/s collides with a 2.0 kg trolley moving left at 1.0 m/s. After collision, the 4.0 kg trolley moves right at 1.0 m/s.

What is the final velocity of the 2.0 kg trolley?

A 1.0 m/s right
B 2.0 m/s right
C 3.0 m/s right
D 5.0 m/s right

21

A 0.20 kg ball moving right at 25 m/s is hit by a bat and leaves left at 35 m/s. The contact time is 0.010 s.

What is the magnitude of the average force on the ball?

A 200 N
B 500 N
C 700 N
D 1200 N

22

A force–time graph is a triangle. The force rises uniformly from 0 to 80 N in 0.20 s and then falls uniformly to 0 in the next 0.20 s.

What impulse is delivered?

A 8.0 N s
B 16 N s
C 32 N s
D 80 N s

23

A force–time graph is a rectangle of height 60 N from 0 to 0.30 s.

What is the change in momentum?

A 0.0050 kg m/s
B 18 kg m/s
C 60 kg m/s
D 200 kg m/s

24

A force–time graph consists of:

40 N for 0.20 s
then 80 N for 0.10 s

What impulse is delivered?

A 8.0 N s
B 12 N s
C 16 N s
D 120 N s

25

A 0.50 kg ball initially at rest receives an impulse of 6.0 N s.

What is its final speed?

A 3.0 m/s
B 6.0 m/s
C 12 m/s
D 24 m/s

For Full Scale Course: Written and Compiled By Sir Hunain Zia (AYLOTI), World Record Holder With 154 Total Personal A Grades, 11 World Records and 7 Distinctions, Educate A Change.

26

A 1.2 kg object moving at 8.0 m/s receives an impulse of 3.6 N s in the same direction as its motion.

What is its final speed?

A 3.0 m/s
B 8.0 m/s
C 11 m/s
D 12 m/s

27

A 2.0 kg trolley moving at 5.0 m/s receives an impulse of 14 N s opposite to its motion.

Taking the original direction as positive, what is its final velocity?

A 2.0 m/s backward
B 2.0 m/s forward
C 7.0 m/s backward
D 12 m/s forward

28

A 0.30 kg ball moving at 20 m/s is stopped by a player’s hand in 0.050 s.

What is the average force exerted on the ball?

A 6.0 N
B 12 N
C 120 N
D 400 N

29

A crumple zone increases the time taken for a car to stop during a collision.

For the same change in momentum, what happens to the average force?

A it decreases
B it increases
C it remains unchanged
D it becomes zero

30

An airbag increases the time taken for a passenger’s head to stop.

Which statement explains why this reduces injury?

A The airbag reduces the passenger’s change in momentum to zero.
B The airbag increases the impulse but decreases the time.
C The airbag keeps the impulse approximately the same but increases the stopping time, reducing force.
D The airbag makes momentum not conserved.

31

Two identical balls hit a wall with the same speed. Ball X sticks to the wall. Ball Y rebounds with the same speed.

Which ball experiences the greater change in momentum?

A X, because it stops
B Y, because its velocity reverses
C both have the same change in momentum
D neither, because speed is the same before collision

32

A 0.50 kg ball moving at 6.0 m/s hits a wall and stops. Another identical ball moving at 6.0 m/s rebounds at 6.0 m/s in the opposite direction.

What is the ratio of change in momentum of the rebounding ball to the stopping ball?

A 1 : 2
B 1 : 1
C 2 : 1
D 4 : 1

33

A truck of mass 2000 kg travels at 12 m/s. A car of mass 1000 kg travels at 24 m/s in the same direction.

Which statement is correct?

A The truck has greater momentum.
B The car has greater momentum.
C They have equal momentum.
D Momentum cannot be compared without knowing acceleration.

34

Two objects have equal momentum. Object X has twice the mass of object Y.

Which statement is correct?

A X has twice the speed of Y.
B X has half the speed of Y.
C X has four times the speed of Y.
D X has the same speed as Y.

35

Two objects have equal kinetic energy. Object P has four times the mass of object Q.

Which statement about their momenta is correct?

A P has half the momentum of Q.
B P has the same momentum as Q.
C P has twice the momentum of Q.
D P has four times the momentum of Q.

36

A 1.0 kg trolley moving at 8.0 m/s collides with a stationary 3.0 kg trolley. They stick together.

What fraction of the initial kinetic energy remains after collision?

A 1/4
B 1/2
C 3/4
D all of it

37

A 2.0 kg trolley moving at 6.0 m/s collides with a stationary trolley. They stick together and move at 2.0 m/s.

What is the mass of the stationary trolley?

A 2.0 kg
B 4.0 kg
C 6.0 kg
D 8.0 kg

38

A 5.0 kg trolley moving at 4.0 m/s collides with a stationary trolley of mass 15 kg. They stick together.

What is their final speed?

A 1.0 m/s
B 2.0 m/s
C 3.0 m/s
D 4.0 m/s

39

A 3.0 kg trolley moving at 2.0 m/s collides and sticks to a 1.0 kg trolley moving in the same direction. After collision, they move at 3.5 m/s.

What was the initial speed of the 1.0 kg trolley?

A 4.0 m/s
B 6.0 m/s
C 8.0 m/s
D 12 m/s

40

A stationary cannon fires a 5.0 kg shell at 200 m/s. The cannon has mass 500 kg.

Ignoring external horizontal forces, what is the recoil speed of the cannon?

A 0.50 m/s
B 2.0 m/s
C 5.0 m/s
D 20 m/s

For Full Scale Course: Written and Compiled By Sir Hunain Zia (AYLOTI), World Record Holder With 154 Total Personal A Grades, 11 World Records and 7 Distinctions, Educate A Change.

41

A rocket ejects gas backwards.

Which statement best explains why the rocket accelerates forwards?

A The gas has no momentum, so the rocket gains all momentum.
B The rocket pushes gas backwards and the gas pushes the rocket forwards.
C The rocket becomes weightless after gas leaves it.
D Momentum is destroyed in the gas.

42

A rocket of total mass 1000 kg ejects 20 kg of gas backwards at 500 m/s relative to the ground. Initially, the rocket and gas are stationary.

What is the forward speed of the remaining rocket immediately after the gas is ejected?

A 1.0 m/s
B 10 m/s
C 10.2 m/s
D 500 m/s

43

A skater of mass 60 kg throws a 2.0 kg ball forward at 15 m/s. Initially, skater and ball are at rest.

What is the recoil speed of the skater?

A 0.25 m/s
B 0.50 m/s
C 2.0 m/s
D 15 m/s

44

A student catches a 0.40 kg ball moving at 10 m/s. The student’s hands move backwards while catching the ball, increasing the stopping time from 0.020 s to 0.10 s.

What happens to the average force on the ball?

A it becomes 1/5 of the original force
B it becomes 5 times the original force
C it remains unchanged
D it becomes zero

45

A 0.15 kg ball is moving at 24 m/s. It is hit so that it moves in the opposite direction at 16 m/s.

What impulse is delivered to the ball?

A 1.2 N s
B 3.6 N s
C 6.0 N s
D 40 N s

46

A 0.80 kg object moving at 5.0 m/s receives a force that acts opposite to its motion for 0.25 s. The object stops.

What is the average force?

A 1.0 N
B 4.0 N
C 16 N
D 20 N

47

A 1200 kg car travelling at 15 m/s comes to rest in 0.30 s during a collision.

What is the average force on the car?

A 54 N
B 600 N
C 60 000 N
D 180 000 N

48

A constant resultant force acts on an object.

Which statement is correct?

A Its momentum changes by equal amounts in equal times.
B Its momentum remains constant.
C Its momentum changes only if its mass changes.
D Its momentum changes by equal amounts in equal distances.

49

A moving object has momentum p and kinetic energy E. Its speed is doubled while its mass remains constant.

What are its new momentum and kinetic energy?

	momentum	kinetic energy
A	2p	2E
B	2p	4E
C	4p	2E
D	4p	4E

50

Two trolleys move towards each other and stick together after collision.

Trolley X: mass 2.0 kg, speed 5.0 m/s to the right
Trolley Y: mass 3.0 kg, speed 4.0 m/s to the left

Taking right as positive, what is the final velocity of the combined trolleys?

A 0.40 m/s left
B 0.40 m/s right
C 2.0 m/s left
D 2.0 m/s right

For Full Scale Course: Written and Compiled By Sir Hunain Zia (AYLOTI), World Record Holder With 154 Total Personal A Grades, 11 World Records and 7 Distinctions, Educate A Change.

Answer Key

Q	Ans	Q	Ans	Q	Ans	Q	Ans	Q	Ans
1	B	11	B	21	D	31	B	41	B
2	B	12	B	22	B	32	C	42	C
3	B	13	A	23	B	33	C	43	B
4	C	14	D	24	C	34	B	44	A
5	B	15	C	25	C	35	C	45	C
6	C	16	B	26	C	36	A	46	C
7	B	17	A	27	A	37	B	47	C
8	D	18	C	28	C	38	A	48	A
9	A	19	A	29	A	39	C	49	B
10	B	20	C	30	C	40	B	50	A

For Full Scale Course: Written and Compiled By Sir Hunain Zia (AYLOTI), World Record Holder With 154 Total Personal A Grades, 11 World Records and 7 Distinctions, Educate A Change.

Detailed Explanations

1. B

Momentum = mass × velocity
p = mv
p = 0.25 × 18
p = 4.5 kg m/s

2. B

p = mv
v = p / m
v = 36 / 4.0
v = 9.0 m/s

3. B

p = mv
m = p / v
m = 24 / 6.0
m = 4.0 kg

4. C

Force = change in momentum / time
N = kg m/s²
N s = kg m/s² × s
N s = kg m/s
So kg m/s is equivalent to N s

5. B

Impulse = force × time
Impulse = 12 × 0.50
Impulse = 6.0 N s

6. C

Impulse = change in momentum
Force = change in momentum / time
F = 8.0 / 0.20
F = 40 N

7. B

Initial momentum = 0.40 × 15 = 6.0 kg m/s
Final momentum = 0
Change in momentum = 6.0 kg m/s
Impulse magnitude = 6.0 N s

8. D

Take original direction as positive.
Initial velocity = +30 m/s
Final velocity = −20 m/s
Change in velocity = −20 − 30 = −50 m/s
Change in momentum = 0.060 × −50 = −3.0 kg m/s
Magnitude = 3.0 kg m/s
Trap: rebounding means velocity reverses, so you add the speed changes.

9. A

East is positive.
Initial velocity = +12 m/s
Final velocity = −8.0 m/s
Change in momentum = m(v − u)
Change in momentum = 0.20(−8.0 − 12)
Change in momentum = 0.20 × −20
Change in momentum = −4.0 kg m/s

10. B

Impulse = force × time
Impulse = 50 × 0.40 = 20 N s
Impulse = change in momentum
Final momentum = 20 kg m/s
v = p / m
v = 20 / 2.0
v = 10 m/s

For Full Scale Course: Written and Compiled By Sir Hunain Zia (AYLOTI), World Record Holder With 154 Total Personal A Grades, 11 World Records and 7 Distinctions, Educate A Change.

11. B

Momentum before = 3.0 × 4.0 = 12 kg m/s
Total mass after = 3.0 + 1.0 = 4.0 kg
Common speed = 12 / 4.0
Common speed = 3.0 m/s

12. B

Momentum before:
- 2.0 × 6.0 = 12 kg m/s
- 4.0 × 3.0 = 12 kg m/s
Total momentum = 24 kg m/s
Total mass = 6.0 kg
Common speed = 24 / 6.0
Common speed = 4.0 m/s

13. A

Take right as positive.
Momentum before:
- 5.0 × 2.0 = +10 kg m/s
- 3.0 × −4.0 = −12 kg m/s
Total momentum = −2 kg m/s
Total mass after = 8.0 kg
Final velocity = −2 / 8.0 = −0.25 m/s
Answer = 0.25 m/s left

14. D

Take right as positive.
Initial velocity = +10 m/s
Final velocity = −6.0 m/s
Change in momentum = 0.50(−6.0 − 10)
Change in momentum = 0.50 × −16
Change in momentum = −8.0 kg m/s
Magnitude of impulse = 8.0 N s

15. C

Initial velocity = +40 m/s
Final velocity = −30 m/s
Change in velocity = −30 − 40 = −70 m/s
Change in momentum magnitude = 0.10 × 70 = 7.0 N s
Force = impulse / time
F = 7.0 / 0.020
F = 350 N

16. B

Momentum is conserved in a closed system.
Kinetic energy is only conserved in elastic collisions.
In inelastic collisions, kinetic energy is transferred to thermal energy, sound, deformation, etc.
So momentum can be conserved even when kinetic energy is not.

17. A

Initial momentum = 2.0 × 5.0 = 10 kg m/s
Momentum of first part = 0.80 × 12 = 9.6 kg m/s
Remaining momentum = 10 − 9.6 = 0.4 kg m/s forward
Mass of other part = 2.0 − 0.80 = 1.2 kg
Velocity = 0.4 / 1.2 = 0.33 m/s forward

18. C

Initial momentum = 0 because object is stationary.
Momentum of X = 3.0 × 4.0 = 12 kg m/s east
Therefore Y must have momentum 12 kg m/s west.
Velocity of Y = 12 / 2.0
Velocity = 6.0 m/s west

19. A

Take right as positive.
Momentum of A = 1.5 × 6.0 = +9.0 kg m/s
Momentum of B = 2.5 × −2.0 = −5.0 kg m/s
Total momentum = +4.0 kg m/s
Answer = 4.0 kg m/s right

20. C

Initial momentum:
- 4.0 × 3.0 = 12 kg m/s
- 2.0 × −1.0 = −2 kg m/s
Total initial momentum = 10 kg m/s
Final momentum of 4.0 kg trolley = 4.0 × 1.0 = 4 kg m/s
Let final velocity of 2.0 kg trolley = v
4 + 2v = 10
2v = 6
v = 3.0 m/s right

21. D

Initial velocity = +25 m/s
Final velocity = −35 m/s
Change in velocity = −35 − 25 = −60 m/s
Change in momentum magnitude = 0.20 × 60 = 12 N s
Force = impulse / time
F = 12 / 0.010
F = 1200 N

22. B

Impulse = area under force–time graph
Triangle base = 0.20 + 0.20 = 0.40 s
Height = 80 N
Area = 1/2 × 0.40 × 80
Impulse = 16 N s

23. B

Impulse = area under force–time graph
Rectangle area = force × time
Impulse = 60 × 0.30
Impulse = 18 N s
Change in momentum = impulse = 18 kg m/s

24. C

Total impulse = area under force–time graph
First part = 40 × 0.20 = 8.0 N s
Second part = 80 × 0.10 = 8.0 N s
Total impulse = 16 N s

25. C

Impulse = change in momentum
Initial momentum = 0
Final momentum = 6.0 kg m/s
v = p / m
v = 6.0 / 0.50
v = 12 m/s

For Full Scale Course: Written and Compiled By Sir Hunain Zia (AYLOTI), World Record Holder With 154 Total Personal A Grades, 11 World Records and 7 Distinctions, Educate A Change.

26. C

Initial momentum = 1.2 × 8.0 = 9.6 kg m/s
Impulse acts in same direction, so add it.
Final momentum = 9.6 + 3.6 = 13.2 kg m/s
Final speed = 13.2 / 1.2
Final speed = 11 m/s

27. A

Initial momentum = 2.0 × 5.0 = 10 kg m/s
Impulse acts opposite to motion, so it is negative.
Final momentum = 10 − 14 = −4 kg m/s
Final velocity = −4 / 2.0
Final velocity = −2.0 m/s
Answer = 2.0 m/s backward

28. C

Initial momentum = 0.30 × 20 = 6.0 kg m/s
Final momentum = 0
Change in momentum magnitude = 6.0 kg m/s
Force = change in momentum / time
F = 6.0 / 0.050
F = 120 N

29. A

Average force = change in momentum / time
For the same change in momentum:
- increasing stopping time decreases average force
That is why crumple zones reduce collision forces.

30. C

The passenger’s change in momentum is approximately the same.
The airbag increases the time over which the head stops.
Force = change in momentum / time
Larger time means smaller average force.
So injury risk decreases.

31. B

Ball X sticks:
- change in momentum = mv
Ball Y rebounds with same speed:
- velocity changes from +v to −v
- change in momentum = 2mv
Ball Y has greater change in momentum because velocity reverses.

32. C

Stopping ball:
- Δp = mv = 0.50 × 6.0 = 3.0 kg m/s
Rebounding ball:
- Δp = m(−6 − 6) = −6.0 kg m/s
- magnitude = 6.0 kg m/s
Ratio rebounding : stopping = 6 : 3
Ratio = 2 : 1

33. C

Truck momentum = 2000 × 12 = 24 000 kg m/s
Car momentum = 1000 × 24 = 24 000 kg m/s
They have equal momentum
Acceleration is irrelevant.

34. B

Momentum = mass × velocity
Equal momentum means mv is the same.
If X has twice the mass, it must have half the speed.
Answer = X has half the speed of Y

35. C

Kinetic energy = p² / 2m
If kinetic energy is the same, then p² is proportional to m.
So momentum is proportional to √m.
P has 4 times the mass of Q.
Therefore P has √4 = 2 times the momentum.
Answer = P has twice the momentum of Q

36. A

Initial momentum = 1.0 × 8.0 = 8.0 kg m/s
Total mass after collision = 1.0 + 3.0 = 4.0 kg
Final speed = 8.0 / 4.0 = 2.0 m/s
Initial kinetic energy:
- Ek = 1/2 × 1.0 × 8.0² = 32 J
Final kinetic energy:
- Ek = 1/2 × 4.0 × 2.0² = 8 J
Fraction remaining = 8 / 32 = 1/4

37. B

Initial momentum = 2.0 × 6.0 = 12 kg m/s
Final speed = 2.0 m/s
Let total mass after collision = M
M × 2.0 = 12
M = 6.0 kg
Stationary trolley mass = 6.0 − 2.0
Mass = 4.0 kg

38. A

Initial momentum = 5.0 × 4.0 = 20 kg m/s
Total mass after collision = 5.0 + 15 = 20 kg
Final speed = 20 / 20
Final speed = 1.0 m/s

39. C

Final momentum = total mass × final velocity
Final momentum = (3.0 + 1.0) × 3.5 = 14 kg m/s
Momentum of 3.0 kg trolley initially:
- 3.0 × 2.0 = 6.0 kg m/s
Momentum of 1.0 kg trolley initially:
- 14 − 6 = 8.0 kg m/s
Speed = 8.0 / 1.0
Speed = 8.0 m/s

40. B

Initial total momentum = 0
Shell momentum = 5.0 × 200 = 1000 kg m/s forward
Cannon momentum must be 1000 kg m/s backward.
Recoil speed = 1000 / 500
Recoil speed = 2.0 m/s

For Full Scale Course: Written and Compiled By Sir Hunain Zia (AYLOTI), World Record Holder With 154 Total Personal A Grades, 11 World Records and 7 Distinctions, Educate A Change.

41. B

Rocket pushes gas backwards.
Gas pushes rocket forwards.
This is Newton’s third law.
Momentum is conserved:
- gas gains backward momentum
- rocket gains forward momentum

42. C

Initial momentum = 0
Gas mass = 20 kg
Gas velocity = −500 m/s
Remaining rocket mass = 1000 − 20 = 980 kg
Momentum conservation:
- 980v + 20(−500) = 0
- 980v = 10 000
- v = 10 000 / 980
- v = 10.2 m/s

43. B

Initial momentum = 0
Ball momentum = 2.0 × 15 = 30 kg m/s forward
Skater momentum = 30 kg m/s backward
Skater speed = 30 / 60
Skater speed = 0.50 m/s

44. A

Stopping time increases from 0.020 s to 0.10 s.
0.10 / 0.020 = 5
Stopping time becomes 5 times larger.
Force = change in momentum / time
So average force becomes 1/5 of the original force

45. C

Take original direction as positive.
Initial velocity = +24 m/s
Final velocity = −16 m/s
Change in velocity = −16 − 24 = −40 m/s
Change in momentum magnitude = 0.15 × 40
Impulse = 6.0 N s

46. C

Initial momentum = 0.80 × 5.0 = 4.0 kg m/s
Final momentum = 0
Change in momentum magnitude = 4.0 kg m/s
Force = change in momentum / time
F = 4.0 / 0.25
F = 16 N

47. C

Initial momentum = 1200 × 15 = 18 000 kg m/s
Final momentum = 0
Change in momentum magnitude = 18 000 kg m/s
Force = change in momentum / time
F = 18 000 / 0.30
F = 60 000 N

48. A

Resultant force = rate of change of momentum.
If resultant force is constant, momentum changes at a constant rate.
Therefore momentum changes by equal amounts in equal times.

49. B

Momentum = mv
If speed doubles:
- momentum doubles
Kinetic energy = 1/2 mv²
If speed doubles:
- kinetic energy becomes 2² = 4 times larger
Answer = 2p and 4E

50. A

Take right as positive.
Momentum of X:
- 2.0 × 5.0 = +10 kg m/s
Momentum of Y:
- 3.0 × −4.0 = −12 kg m/s
Total momentum = −2 kg m/s
Total mass after collision = 2.0 + 3.0 = 5.0 kg
Final velocity = −2 / 5.0
Final velocity = −0.40 m/s
Answer = 0.40 m/s left

For Full Scale Course: Written and Compiled By Sir Hunain Zia (AYLOTI), World Record Holder With 154 Total Personal A Grades, 11 World Records and 7 Distinctions, Educate A Change.

Common Traps From This Chapter

Trap	Correct Rule
Momentum formula	p = mv
Impulse formula	impulse = force × time
Impulse meaning	impulse = change in momentum
Rebound questions	velocity changes sign
Object stops	final momentum = 0
Force from collision	F = Δp / Δt
Longer collision time	smaller average force
Force–time graph	area = impulse
Momentum unit	kg m/s or N s
Conservation of momentum	total momentum before = total momentum after
Sticking together	common final velocity
Opposite directions	use positive and negative signs
Stationary explosion	total momentum remains zero
Momentum conserved	only if system is closed
Kinetic energy in collisions	not always conserved
Rebounding ball	greater Δp than stopping ball
Equal momentum	larger mass means smaller speed
Equal kinetic energy	momentum depends on √mass
Rocket motion	gas backward, rocket forward
Constant force	equal momentum change in equal times

EAC

Education

Support

Other Projects

Want To Teach Online?