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Edexcel International A Level Physics · Unit 1 · Mechanics and Materials
40 free, high-quality past-paper-style questions on Momentum for Edexcel IAL Physics, each with a full mark scheme. For when you have finished the past papers, or want to practise one topic at a time.IAL Physics notes and flashcards →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.
Which of the following is a correct unit for momentum?
| Answer | Mark | Notes |
|---|---|---|
| Momentum = mass × velocity, so its unit is kg × m s−1 = kg m s−1 (equivalent to N s). | ✓ 1 | Answer A |
Answer: A
Exam tip: Build a unit from the defining equation: write the units of each quantity in p = mv and multiply them.
A cyclist and her bicycle have a combined mass of 85 kg. They travel along a straight road at 7.2 m s−1.
What is their momentum?
| Answer | Mark | Notes |
|---|---|---|
| p = mv = 85 × 7.2 = 610 kg m s−1 (612). | ✓ 1 | Answer C |
Answer: C
Common error: Using ½mv2 or ½mv. Momentum is simply mass × velocity, with no factor of ½ and no square.
The total momentum of a system of interacting bodies is constant provided that
| Answer | Mark | Notes |
|---|---|---|
| The principle of conservation of linear momentum applies to any interaction, provided no resultant external force acts on the system. It does not depend on the masses, on whether the bodies stick, or on what happens to the kinetic energy. | ✓ 1 | Answer C |
Answer: C
Exam tip: When stating the principle, always include the condition: no resultant external force (a closed system).
In a rugby match a player of mass 90 kg runs at 5.0 m s−1 and tackles a stationary opponent of mass 70 kg. Immediately after the tackle the two players move together.
What is their speed immediately after the tackle?
| Answer | Mark | Notes |
|---|---|---|
| Momentum before = 90 × 5.0 = 450 kg m s−1. After: 450 = (90 + 70)v, so v = 450 ÷ 160 = 2.8 m s−1 (2.81). | ✓ 1 | Answer C |
Answer: C
Exam tip: When bodies move off together, the mass after the collision is the sum of the masses.
A bowling ball of mass 6.4 kg rolls in a straight line along a lane at 5.5 m s−1. It strikes the padded back wall of the lane head-on and rebounds along the same line at 1.5 m s−1.
Calculate the momentum of the ball before it reaches the wall.
Calculate the magnitude of the change in momentum of the ball.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| p = mv = 6.4 × 5.5 | ✓ 1 | |
| = 35 kg m s−1 (35.2) | ✓ 1 | |
| Part (b) | ||
| towards the wall positive: Δp = 6.4 × (−1.5) − 6.4 × 5.5 | ✓ 1 | |
| magnitude = 45 kg m s−1 (44.8) | ✓ 1 | |
Common error: A rebound reverses the velocity. Give the final velocity a negative sign before subtracting; 6.4 × (5.5 − 1.5) = 26 kg m s−1 is wrong.
A toy dart of mass 0.020 kg travels horizontally at 8.0 m s−1. It hits a toy car of mass 0.18 kg that is at rest on a smooth floor, and its sucker sticks to the car.
State the principle of conservation of linear momentum.
Calculate the velocity of the car and dart just after the dart sticks.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| the total momentum of a system stays constant / total momentum before = total momentum after | ✓ 1 | |
| provided no resultant external force acts on the system | ✓ 1 | |
| Part (b) | ||
| 0.020 × 8.0 = (0.020 + 0.18)v | ✓ 1 | |
| v = 0.80 m s−1 in the dart's original direction | ✓ 1 | |
Exam tip: A statement of the principle needs the condition. “Momentum is conserved” on its own earns at most half the marks.
Figure 1 shows four bodies, P, Q, R and S, moving along straight lines. The mass and the velocity of each body are labelled.
Which body has the momentum of largest magnitude?
| Answer | Mark | Notes |
|---|---|---|
| Magnitudes of p = mv: P 6.0 × 6.0 = 36, Q 0.20 × 50 = 10, R 1.0 × 25 = 25, S 90 × 0.30 = 27 kg m s−1. The largest is P. The direction of travel does not affect the magnitude. | ✓ 1 | Answer A |
Answer: A
Common error: Picking the fastest or the most massive body. Momentum depends on the product mv, so work out all four products before choosing.
A ball of mass 0.15 kg is thrown vertically upwards. Figure 1 shows the ball as it leaves the thrower's hand and as it is caught on its way down by a friend standing 1.4 m higher. Air resistance is negligible.
What is the change in momentum of the ball between leaving the hand and being caught?
| Answer | Mark | Notes |
|---|---|---|
| Taking upwards as positive: initial momentum = 0.15 × 8.0 = +1.2 kg m s−1; final momentum = 0.15 × (−6.0) = −0.90 kg m s−1. Δp = −0.90 − 1.2 = −2.1 kg m s−1, so 2.1 kg m s−1 downwards. | ✓ 1 | Answer D |
Answer: D
Exam tip: Choose a positive direction before you substitute. The ball's velocity changes sign, so the change in momentum is the sum of the two magnitudes, in the direction of the final velocity.
Figure 1 shows a trolley carrying a spring launcher on a horizontal track. The total mass of the trolley and launcher is 2.4 kg. Everything is at rest until the launcher fires a ball of mass 0.30 kg horizontally to the left at 4.0 m s−1.
What is the speed of the trolley just after the ball is fired?
| Answer | Mark | Notes |
|---|---|---|
| Total momentum before = 0. After: 0.30 × 4.0 = 2.4 × v, so v = 1.2 ÷ 2.4 = 0.50 m s−1 to the right. | ✓ 1 | Answer B |
Answer: B
Exam tip: In a recoil from rest the two momenta are equal in size and opposite in direction. Divide the ball's momentum by the mass of the trolley alone, not by the total mass.
A rubber ball of mass 0.060 kg is dropped from rest onto a hard floor. The graph shows how its velocity v varies with time t, taking upwards as positive.
Use the graph to calculate the momentum of the ball just before it hits the floor. Give its direction.
Calculate the change in momentum of the ball during the bounce. Give its direction.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| v = −6.0 m s−1 read from the graph, so p = 0.060 × 6.0 = 0.36 kg m s−1 | ✓ 1 | |
| downwards | ✓ 1 | |
| Part (b) | ||
| Δp = 0.060 × (+4.0) − 0.060 × (−6.0) | ✓ 1 | |
| = 0.60 kg m s−1 upwards | ✓ 1 | |
Common error: Using 6.0 − 4.0 for the change in velocity. The velocity changes from −6.0 to +4.0 m s−1, a change of 10 m s−1; subtracting the speeds gives 0.12 kg m s−1, which is wrong.
Figure 1 shows a cart of mass 12 kg rolling along a level track at 1.5 m s−1. A sandbag of mass 3.0 kg is dropped vertically into the cart and stays in it. Friction is negligible.
Explain why the sandbag adds no horizontal momentum to the system.
Calculate the speed of the cart just after the sandbag lands in it.
The vertical momentum of the sandbag is not conserved when it lands. State what provides the external force that removes it.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| the sandbag moves vertically, so its velocity (and momentum) has no horizontal component | ✓ 1 | |
| Part (b) | ||
| 12 × 1.5 = (12 + 3.0) × v | ✓ 1 | |
| v = 18 ÷ 15 = 1.2 m s−1 | ✓ 1 | |
| Part (c) | ||
| the track / ground pushes up on the cart (normal contact force), which stops the sandbag's downward motion | ✓ 1 | |
Exam tip: Momentum can be conserved in one direction but not in another. Here there is no external horizontal force, so horizontal momentum is conserved even though a vertical force acts.
A football of mass 0.45 kg moves horizontally at 12 m s−1 towards a wall. It rebounds along the same line at 8.0 m s−1.
What is the magnitude of the change in momentum of the ball?
| Answer | Mark | Notes |
|---|---|---|
| Taking towards the wall as positive: Δp = 0.45 × (−8.0) − 0.45 × 12 = −9.0 kg m s−1. Magnitude 9.0 kg m s−1. | ✓ 1 | Answer D |
Answer: D
Common error: Subtracting the speeds. The ball reverses, so its initial and final momenta have opposite signs and the change is their sum in magnitude.
Two ice-hockey players skate towards each other along a straight line. Player P (mass 80 kg) moves east at 4.0 m s−1 and player Q (mass 100 kg) moves west at 3.0 m s−1. They collide and hold on to each other. Friction from the ice is negligible.
What is their speed just after the collision?
| Answer | Mark | Notes |
|---|---|---|
| East positive: total momentum = 80 × 4.0 + 100 × (−3.0) = 20 kg m s−1. 20 = 180v, so v = 0.11 m s−1 (0.111), eastwards. | ✓ 1 | Answer A |
Answer: A
Common error: Adding the two momenta as magnitudes (320 + 300). Momenta in opposite directions have opposite signs and partly cancel.
A fisherman of mass 70 kg stands in a boat of mass 50 kg. The boat is at rest on still water. He throws a 5.0 kg anchor horizontally from the boat at 4.0 m s−1 relative to the water. Water resistance is negligible.
What is the speed of the boat and fisherman just after the throw?
| Answer | Mark | Notes |
|---|---|---|
| Total momentum before = 0. After: 5.0 × 4.0 + 120v = 0, so v = −20 ÷ 120 = −0.17 m s−1: speed 0.17 m s−1 (0.167), opposite to the anchor. | ✓ 1 | Answer B |
Answer: B
Exam tip: Decide exactly which bodies move together after the event. Here the fisherman and boat share one velocity; the anchor is no longer part of that mass.
A lorry collides head-on with a much lighter car. Which statement about the collision is correct?
| Answer | Mark | Notes |
|---|---|---|
| By Newton's third law the forces on the two vehicles are equal and opposite, and they act for the same time. So the changes in momentum are equal in size (and opposite in direction). The car, with its smaller mass, has the larger change in velocity. | ✓ 1 | Answer B |
Answer: B
Exam tip: Equal and opposite forces for the same time give equal and opposite changes in momentum — not equal changes in velocity.
In a road-safety test, a car of mass 1200 kg collides head-on with a truck of mass 6000 kg. During the collision the velocity of the car changes by 9.0 m s−1.
Explain, using Newton's laws of motion, why the change in momentum of the car is equal in size and opposite in direction to the change in momentum of the truck.
Calculate the size of the change in velocity of the truck.
Suggest why the occupants of the car are at greater risk of injury than the occupants of the truck.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| the car exerts a force on the truck and the truck exerts a force on the car that is equal in size and opposite in direction (Newton's third law) | ✓ 1 | |
| the two forces act for the same time / the contact time is the same for both | ✓ 1 | |
| force = rate of change of momentum (Newton's second law), so the changes in momentum are equal in size and opposite in direction | ✓ 1 | |
| Part (b) | ||
| change in momentum of car = 1200 × 9.0 = 10 800 kg m s−1 | ✓ 1 | |
| Δv = 10 800 ÷ 6000 = 1.8 m s−1 | ✓ 1 | |
| Part (c) | ||
| the car has a larger change in velocity in the same time, so a larger deceleration / the occupants of the car experience larger forces for their mass | ✓ 1 | |
Exam tip: An explanation using Newton's laws needs three linked ideas: equal and opposite forces, the same time of action, and therefore equal and opposite changes in momentum.
On a pool table a cue ball of mass 0.17 kg moves at 2.4 m s−1 and hits a stationary ball of mass 0.16 kg head-on. After the collision the cue ball continues in the same direction at 0.20 m s−1.
Calculate the momentum of the cue ball before the collision.
Calculate the velocity of the second ball just after the collision.
Friction from the cloth acts on both balls. Explain why it is still reasonable to apply conservation of momentum to the collision.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 0.17 × 2.4 = 0.41 kg m s−1 (0.408) | ✓ 1 | |
| Part (b) | ||
| total momentum after = 0.17 × 0.20 + 0.16v | ✓ 1 | |
| 0.408 = 0.034 + 0.16v | ✓ 1 | |
| v = 2.3 m s−1 (2.34) in the original direction of the cue ball (accept 2.4 m s−1 from using 0.41) | ✓ 1 | |
| Part (c) | ||
| friction is an external force, but it is very small compared with the forces the balls exert on each other | ✓ 1 | |
| and the collision lasts only a very short time, so friction changes the total momentum by a negligible amount during it | ✓ 1 | |
Exam tip: Carry the unrounded momentum (0.408 kg m s−1) into the next part and round only the final answer. Rounding intermediate values early can change the last significant figure.
A radium nucleus at rest decays by emitting an alpha particle. The alpha particle has mass 6.64 × 10−27 kg and moves off at 1.52 × 107 m s−1. The remaining nucleus has mass 3.69 × 10−25 kg.
Explain why the remaining nucleus moves off in the opposite direction to the alpha particle.
Calculate the speed of the remaining nucleus.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| the total momentum before the decay is zero, as the nucleus is at rest | ✓ 1 | |
| momentum is conserved, so after the decay the two momenta must be equal in size and opposite in direction to give a total of zero | ✓ 1 | |
| Part (b) | ||
| 3.69 × 10−25 × v = 6.64 × 10−27 × 1.52 × 107 | ✓ 1 | |
| v = 2.74 × 105 m s−1 | ✓ 1 | |
Exam tip: Powers of ten are easy to mistype. Estimate first: the masses differ by a factor of about 56, so the recoil speed should be about 1/56 of the alpha speed.
A student uses conservation of momentum to find the mass of trolley B. Trolley A, of mass 0.80 kg, is pushed along a runway towards B, which is at rest. A pin on A sticks into a cork on B and the trolleys move off together. A motion sensor connected to a data logger records the velocity of A. The velocity–time graph is horizontal at 0.60 m s−1 up to t = 1.20 s, drops steeply between 1.20 s and 1.26 s, and is then horizontal at 0.40 m s−1.
The runway is tilted slightly. Explain why, and describe how the student can check that the tilt is correct.
Determine the mass of trolley B.
Weighing trolley B on a balance gives 0.38 kg. Suggest why the value from the momentum method is larger.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| so that a component of the weight balances friction / so there is no resultant external force on the trolleys | ✓ 1 | |
| a trolley given a small push moves down the runway at constant velocity / the velocity–time graph for one trolley is horizontal | ✓ 1 | |
| Part (b) | ||
| momentum before = 0.80 × 0.60 = 0.48 kg m s−1 | ✓ 1 | |
| 0.48 = (0.80 + m) × 0.40 | ✓ 1 | |
| m = 1.2 − 0.80 = 0.40 kg | ✓ 1 | |
| Part (c) | ||
| some momentum is lost to an external force such as friction / the runway is not perfectly compensated, so the velocity after the collision is measured slightly too low | ✓ 1 | |
| a lower velocity after gives a larger calculated total mass (0.80 + m) and so a larger value for m | ✓ 1 | |
Exam tip: In a “suggest” question link the error to its effect on the answer: say which measurement is affected and in which direction the result moves.
A large fish of mass 0.80 kg swims in a straight line at 1.2 m s−1. It swallows a small fish of mass 0.040 kg that is swimming directly towards it at 0.50 m s−1.
Explain why the directions of the two velocities must be taken into account when calculating the total momentum.
Calculate the velocity of the large fish just after it swallows the small fish.
State one assumption made in your calculation.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| momentum is a vector, so momenta in opposite directions have opposite signs and partly cancel | ✓ 1 | |
| Part (b) | ||
| direction of the large fish positive: total momentum = 0.80 × 1.2 + 0.040 × (−0.50) = 0.94 kg m s−1 | ✓ 1 | |
| 0.94 = (0.80 + 0.040)v | ✓ 1 | |
| v = 1.1 m s−1 (1.12) in its original direction | ✓ 1 | |
| Part (c) | ||
| the resistive force from the water is negligible during the swallow / no resultant external force acts on the two fish | ✓ 1 | |
Common error: Adding 0.020 kg m s−1 for the small fish instead of subtracting it gives 1.17 m s−1. A body moving towards the other has a velocity of the opposite sign.
Two trolleys, X and Y, collide on a horizontal runway. Friction is negligible. The graph shows how the momentum of each trolley varies with time, taking the direction of X's initial motion as positive. The trace for Y after the collision has been left off.
What is the momentum of Y after the collision?
| Answer | Mark | Notes |
|---|---|---|
| From the graph: total momentum before = 0.90 + (−0.20) = 0.70 kg m s−1. After: 0.30 + pY = 0.70, so pY = 0.40 kg m s−1 (in the positive direction). | ✓ 1 | Answer B |
Answer: B
Common error: Ignoring the negative momentum of Y before the collision. Read every value with its sign: the total before is 0.90 − 0.20, not 0.90.
Trolley A, of mass 0.90 kg, pulls a paper tape through a ticker-timer that makes 50 dots per second. A runs along a friction-compensated runway into a stationary trolley B and the two stick together. Figure 1 shows the tape, with the measured lengths marked.
What is the mass of B?
| Answer | Mark | Notes |
|---|---|---|
| Dot interval = 1/50 = 0.020 s. Before: 8.0 cm ÷ 4 = 2.0 cm per interval, so u = 0.020 m ÷ 0.020 s = 1.0 m s−1. After: 6.0 cm ÷ 5 = 1.2 cm, so v = 0.60 m s−1. 0.90 × 1.0 = (0.90 + m) × 0.60 gives 0.90 + m = 1.5 kg, so m = 0.60 kg. | ✓ 1 | Answer C |
Answer: C
Exam tip: Count the gaps between dots, not the dots: four gaps cover 8.0 cm here. Then each gap is one time interval of 1/f.
Figure 1 shows two pucks, A and B, on an air table before and after a head-on collision. Friction is negligible.
What is the velocity of B after the collision?
| Answer | Mark | Notes |
|---|---|---|
| Right positive. Before: 0.20 × 3.0 + 0.30 × (−1.0) = 0.30 kg m s−1. After: 0.20 × (−1.2) + 0.30v = 0.30, so 0.30v = 0.54 and v = +1.8 m s−1, to the right. | ✓ 1 | Answer C |
Answer: C
Common error: Giving A's final velocity a positive sign. A moves left after the collision, so its momentum is −0.24 kg m s−1 and B must carry more than the whole total.
Two sledges, X and Y, collide head-on on smooth ice. The bar chart shows the momentum of each sledge before the collision and of X after it, taking X's original direction as positive. Sledge Y and its rider have a total mass of 25 kg.
What is the speed of Y after the collision?
| Answer | Mark | Notes |
|---|---|---|
| Total momentum before = 60 + (−20) = 40 kg m s−1. After: 10 + pY = 40, so pY = 30 kg m s−1. Speed = 30 ÷ 25 = 1.2 m s−1. | ✓ 1 | Answer B |
Answer: B
Exam tip: Read the negative bar as a negative momentum. The total before is 60 − 20, and the same total must appear after the collision.
Trolley X moves along a level, friction-free runway and collides with trolley Y, which is at rest. The four sketch graphs show possible variations of the momentum p of each trolley with time t.
Which graph is consistent with Newton's third law and the conservation of momentum?
| Answer | Mark | Notes |
|---|---|---|
| Newton's third law: the trolleys exert equal and opposite forces on each other for the same time, so Y gains momentum at exactly the rate X loses it. Y's trace must rise while X's falls, and by the same amount, so the total stays constant. Only Graph D shows this. | ✓ 1 | Answer D |
Answer: D
Exam tip: On a momentum–time graph for a closed system, the two traces must change at the same time and by equal and opposite amounts, so their sum is a horizontal line.
Two pucks, A of mass 0.12 kg and B of mass 0.16 kg, move along a straight line on an air table. A camera records an image of each puck every 0.10 s. Figure 1 shows the positions of the pucks against a centimetre scale before and after A collides with B, which is initially at rest.
Use Figure 1 to determine the velocity of A before the collision.
Determine the velocities of A and B after the collision.
Show that these results are consistent with conservation of momentum.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| images of A are 6.0 cm apart (allow 5.8–6.2 cm) | ✓ 1 | |
| u = 0.060 m ÷ 0.10 s = 0.60 m s−1 | ✓ 1 | |
| Part (b) | ||
| A: 2.0 cm per 0.10 s = 0.20 m s−1 | ✓ 1 | |
| B: 3.0 cm per 0.10 s = 0.30 m s−1, both in A's original direction | ✓ 1 | |
| Part (c) | ||
| before: 0.12 × 0.60 = 0.072 kg m s−1 | ✓ 1 | |
| after: 0.12 × 0.20 + 0.16 × 0.30 = 0.024 + 0.048 = 0.072 kg m s−1, the same as before | ✓ 1 | |
Exam tip: Measure across several gaps and divide, rather than measuring one gap. For B after the collision, four gaps span 12 cm, which is easier to read accurately than one gap of 3 cm.
Trolleys A and B move towards each other on a level, friction-compensated runway and collide. Two motion sensors record their velocities. The graph shows the velocity v of each trolley against time t, taking A's initial direction as positive. The mass of A is 0.50 kg.
Calculate the change in momentum of A.
Determine the mass of B.
Show that the total momentum of the two trolleys before the collision is zero.
The experiment is repeated with the same starting velocities, but this time the trolleys stick together. State and explain the velocity of the trolleys after the collision.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| A: u = +0.80 m s−1, v = −0.20 m s−1 read from the graph | ✓ 1 | |
| Δp = 0.50 × (−0.20 − 0.80) = −0.50 kg m s−1 (0.50 kg m s−1 in the negative direction) | ✓ 1 | |
| Part (b) | ||
| change in momentum of B = +0.50 kg m s−1 and its change in velocity = 0.10 − (−0.40) = 0.50 m s−1 | ✓ 1 | |
| m = 0.50 ÷ 0.50 = 1.0 kg | ✓ 1 | |
| Part (c) | ||
| 0.50 × 0.80 + 1.0 × (−0.40) = 0.40 − 0.40 = 0 | ✓ 1 | |
| Part (d) | ||
| they stop / velocity = 0 | ✓ 1 | |
| the total momentum is zero before, so it must be zero afterwards; with one combined mass this means zero velocity | ✓ 1 | |
Common error: Reading the change in velocity of B as 0.10 − 0.40. B was moving in the negative direction, so its change in velocity is 0.10 − (−0.40) = 0.50 m s−1.
Figure 1 shows two trolleys, P of mass 0.40 kg and Q of unknown mass, held at rest on a level track with a compressed spring between them. A buffer is fixed at each end of the track. When the spring is released, the trolleys move apart at constant speed and hit their buffers at the same instant.
State the total momentum of the trolleys just after the spring is released. Explain your answer.
Explain why the ratio of the speeds of P and Q equals the ratio of the distances in Figure 1.
Determine the mass of Q.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| zero | ✓ 1 | |
| the trolleys were at rest before, and no resultant external force acts, so the total momentum stays zero / the two momenta are equal and opposite | ✓ 1 | |
| Part (b) | ||
| both travel at constant speed for the same time, so speed = distance ÷ time gives vP : vQ = 0.90 : 0.60 | ✓ 1 | |
| Part (c) | ||
| 0.40 × vP = mQ × vQ, so mQ = 0.40 × 0.90 ÷ 0.60 | ✓ 1 | |
| = 0.60 kg | ✓ 1 | |
Exam tip: You do not need the actual speeds. Because the time is the same for both trolleys, the distances are in the same ratio as the speeds.
Figure 1 shows a student sitting on trolley X; the student and trolley have a total mass of 60 kg. The student holds a rope tied to trolley Y, of mass 40 kg. Both trolleys start at rest on a level floor, with a gap of 4.0 m between them. The student pulls the rope in, hand over hand. Friction is negligible.
Explain why the total momentum of the two trolleys remains zero while the student pulls.
At one instant X is moving towards Y at 0.30 m s−1. Calculate the velocity of Y at that instant.
Determine the distance moved by X by the time the trolleys meet.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| the tension in the rope pulls X and Y with equal and opposite forces / the forces are internal to the system | ✓ 1 | |
| there is no resultant external force on the system, so its total momentum stays at its initial value of zero | ✓ 1 | |
| Part (b) | ||
| 0 = 60 × 0.30 + 40 × vY | ✓ 1 | |
| vY = −0.45 m s−1, i.e. 0.45 m s−1 towards X | ✓ 1 | |
| Part (c) | ||
| at every instant vX : vY = 40 : 60, so the distances moved are in the ratio 2 : 3 | ✓ 1 | |
| X moves 4.0 × 2/5 = 1.6 m | ✓ 1 | |
Exam tip: If the total momentum is zero at every instant, the speeds are always in the inverse ratio of the masses. So the distances moved in the same time are in that ratio too.
The graph shows how the momentum p of each of two trolleys, X and Y, depends on its velocity v.
State what the gradient of each line represents.
Use the graph to determine the mass of X and the mass of Y.
X moves at 2.0 m s−1 along a friction-free track and collides with Y, which is at rest. They stick together. Use the graph to find the momentum of X, and calculate the velocity of X and Y after the collision.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| the mass of the trolley (since p = mv) | ✓ 1 | |
| Part (b) | ||
| X: 6.0 ÷ 4.0 = 1.5 kg | ✓ 1 | |
| Y: 2.0 ÷ 4.0 = 0.50 kg | ✓ 1 | |
| Part (c) | ||
| momentum of X at 2.0 m s−1 = 3.0 kg m s−1 from the graph | ✓ 1 | |
| 3.0 = (1.5 + 0.50) × v, so v = 1.5 m s−1 | ✓ 1 | |
Exam tip: Take a gradient from a large triangle that ends on gridline crossings, such as (4.0, 6.0) for X, rather than from a small section of the line.
Two trolleys, X of mass 0.50 kg and Y of mass 1.50 kg, are held together by a compressed spring and move together along a straight runway at 0.40 m s−1. The spring is released. Afterwards X moves at 0.20 m s−1 in the opposite direction to the original motion. Friction is negligible.
What is the speed of Y after the spring is released?
| Answer | Mark | Notes |
|---|---|---|
| Original direction positive. Before: 2.00 × 0.40 = 0.80 kg m s−1. After: 0.50 × (−0.20) + 1.50v = 0.80, so v = 0.90 ÷ 1.50 = 0.60 m s−1. | ✓ 1 | Answer D |
Answer: D
Common error: Treating a separation as if it started from rest. Here the trolleys were already moving, so the total momentum before is not zero.
A motion sensor records the velocity of trolley P (mass 0.60 kg) as it collides with a stationary trolley Q on a friction-compensated runway. The velocity–time graph for P is horizontal at 0.50 m s−1 until t = 1.0 s, then drops steeply and is horizontal at 0.10 m s−1 from t = 1.1 s. After the collision Q moves at 0.40 m s−1 in the same direction as P.
What is the mass of Q?
| Answer | Mark | Notes |
|---|---|---|
| Momentum lost by P = 0.60 × (0.50 − 0.10) = 0.24 kg m s−1. This is the momentum gained by Q: m × 0.40 = 0.24, so m = 0.60 kg. | ✓ 1 | Answer B |
Answer: B
Exam tip: Read both velocities of P from the graph. The momentum P loses is the momentum Q gains.
A pair of figure skaters glide together across the ice at 2.0 m s−1. The female skater has mass 50 kg and the male skater has mass 75 kg. He pushes her forwards, and immediately afterwards she moves at 3.5 m s−1 in the original direction. Friction from the ice is negligible.
Calculate the velocity of the male skater immediately after the push.
Show that the changes in momentum of the two skaters are equal in size and opposite in direction.
Determine the speed the female skater would need after the push for the male skater to be left at rest.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| total momentum before = (50 + 75) × 2.0 = 250 kg m s−1 | ✓ 1 | |
| 250 = 50 × 3.5 + 75v | ✓ 1 | |
| v = 1.0 m s−1 in the original direction | ✓ 1 | |
| Part (b) | ||
| female skater: 50 × (3.5 − 2.0) = +75 kg m s−1 | ✓ 1 | |
| male skater: 75 × (1.0 − 2.0) = −75 kg m s−1 | ✓ 1 | |
| Part (c) | ||
| 250 = 50v + 75 × 0 | ✓ 1 | |
| v = 5.0 m s−1 | ✓ 1 | |
Common error: Starting with zero total momentum. The skaters were already moving, so the total before is (50 + 75) × 2.0, not zero.
During a training session an athlete of mass 60 kg stands at rest on a skateboard of mass 4.0 kg on a smooth level floor. A coach throws a medicine ball of mass 5.0 kg horizontally to her at 6.0 m s−1 and she catches it. Resistive forces are negligible.
Calculate the speed of the athlete, skateboard and ball just after the catch.
Moving at this speed, she immediately throws the ball back towards the coach. It leaves her hands at 6.0 m s−1 relative to the floor. Calculate her speed on the skateboard after the throw.
Explain why her speed increases when she throws the ball back.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 5.0 × 6.0 = (60 + 4.0 + 5.0)v | ✓ 1 | |
| v = 30 ÷ 69 = 0.43 m s−1 (0.435) | ✓ 1 | |
| Part (b) | ||
| taking the ball's original direction as positive, total momentum is still 30 kg m s−1 | ✓ 1 | |
| 30 = 64v + 5.0 × (−6.0) | ✓ 1 | |
| v = 60 ÷ 64 = 0.94 m s−1 (0.938) | ✓ 1 | |
| Part (c) | ||
| she pushes the ball backwards, so the ball pushes her forwards with an equal and opposite force (Newton's third law) | ✓ 1 | |
| the ball's momentum changes from about +2.2 kg m s−1 to −30 kg m s−1, so she and the skateboard gain an equal amount (about 32 kg m s−1) of forward momentum / the total momentum stays 30 kg m s−1, so the ball's negative momentum must be balanced by more forward momentum for her | ✓ 1 | |
Exam tip: In a two-stage problem the total momentum of the system stays the same through every stage. Work it out once and reuse it.
Two gliders move on a level air track. Each glider carries a card of length 0.100 m. Glider A (mass 0.300 kg) passes through light gate 1 in 0.0625 s and then collides with glider B (mass 0.500 kg), which is at rest. After the collision A rebounds and passes back through light gate 1 in 0.250 s, while B passes through light gate 2 in 0.0862 s.
Describe how a light gate and data logger are used to find the velocity of a glider.
Calculate the momentum of glider A before the collision.
Calculate the total momentum after the collision. Take the original direction of A as positive.
The percentage uncertainty in each total momentum is ±3%. Deduce whether these results are consistent with conservation of momentum.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| the card on the glider interrupts the light beam, and the logger records the time for which the beam is blocked | ✓ 1 | |
| velocity = length of card ÷ time for the card to pass through the gate | ✓ 1 | |
| Part (b) | ||
| 0.300 × (0.100 ÷ 0.0625) = 0.300 × 1.60 = 0.480 kg m s−1 | ✓ 1 | |
| Part (c) | ||
| velocity of A after = −0.100 ÷ 0.250 = −0.400 m s−1 | ✓ 1 | |
| velocity of B after = 0.100 ÷ 0.0862 = 1.16 m s−1 | ✓ 1 | |
| total = 0.300 × (−0.400) + 0.500 × 1.16 = 0.460 kg m s−1 | ✓ 1 | |
| Part (d) | ||
| absolute uncertainties = 3% of 0.480 and 3% of 0.460 ≈ 0.014 kg m s−1 each, giving ranges 0.466–0.494 and 0.446–0.474 kg m s−1 | ✓ 1 | |
| the ranges overlap, so the results are consistent with conservation of momentum | ✓ 1 | |
Common error: Forgetting that A rebounds. Its velocity after the collision is negative, so its momentum subtracts from B's.
Trolley A, of mass 0.50 kg, moves along a friction-compensated runway towards trolley B, which is at rest. They stick together on impact. The graph shows the displacement x of each trolley from A's starting point against time t.
What is the mass of B?
| Answer | Mark | Notes |
|---|---|---|
| Gradients: before, A moves at 0.60 m ÷ 1.0 s = 0.60 m s−1. After, the pair moves at (1.00 − 0.60) m ÷ (3.0 − 1.0) s = 0.20 m s−1. 0.50 × 0.60 = (0.50 + m) × 0.20 gives 0.50 + m = 1.5 kg, so m = 1.0 kg. | ✓ 1 | Answer C |
Answer: C
Common error: Dividing 1.00 m by 3.0 s. Velocity is the gradient of the section after the collision, so use the change in displacement over that section only.
Figure 1 shows three trolleys on a straight, level track. A moves towards B, which is at rest, while C moves towards them. A hits B and they stick together; the pair then hits C and all three stick together. Friction is negligible.
What is the final velocity of the three trolleys?
| Answer | Mark | Notes |
|---|---|---|
| Momentum is conserved through both collisions, so only the start and the end matter. Right positive: 2.0 × 1.2 + 0 + 3.0 × (−0.20) = 1.8 kg m s−1. Final mass 6.0 kg, so v = 1.8 ÷ 6.0 = 0.30 m s−1 to the right. | ✓ 1 | Answer A |
Answer: A
Exam tip: With no external force, the total momentum is the same at every stage. You do not need the velocity after the first collision to find the final velocity.
Figure 1 shows a foam block of mass 0.475 kg hanging at rest from light strings. A dart of mass 0.025 kg is fired horizontally into the block and sticks in it. The block and dart swing upwards until their centre of mass has risen through a vertical height of 5.0 cm. Air resistance is negligible.
Calculate the speed of the block and dart just after the impact.
Calculate the speed of the dart just before it hits the block.
Explain why momentum, but not kinetic energy, can be assumed to be conserved during the impact.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| kinetic energy just after impact = gravitational potential energy gained, ½mv2 = mgh | ✓ 1 | |
| v = √(2 × 9.81 × 0.050) | ✓ 1 | |
| = 0.99 m s−1 (0.990) | ✓ 1 | |
| Part (b) | ||
| 0.025 × u = (0.025 + 0.475) × 0.990 | ✓ 1 | |
| u = 19.8 m s−1 (accept 20 m s−1) | ✓ 1 | |
| Part (c) | ||
| no resultant horizontal external force acts during the very short impact, so momentum is conserved | ✓ 1 | |
| kinetic energy is not conserved because work is done deforming the foam / energy is transferred to internal (thermal) energy and sound as the dart embeds | ✓ 1 | |
Exam tip: Split the event into stages and use the right principle for each: momentum for the impact, then energy conservation for the swing. Do not use energy conservation across the impact.
A student investigates collisions in which trolley A, of unknown mass mA, is released from the same point on a ramp so that it always reaches a friction-compensated runway at the same speed u. On the runway A hits a stationary trolley B and they stick together. The student varies the mass mB of B by adding masses and measures the common velocity v after each collision. The graph shows 1/v plotted against mB, with a line of best fit.
Show that 1/v = 1/u + mB/(mAu).
Use the graph to determine u.
Use the graph to determine mA.
Explain why the runway is friction-compensated and why A is released from the same point every time.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| conservation of momentum: mAu = (mA + mB)v | ✓ 1 | |
| 1/v = (mA + mB)/(mAu) = 1/u + mB/(mAu) | ✓ 1 | |
| Part (b) | ||
| intercept on the 1/v axis = 1.25 s m−1 = 1/u | ✓ 1 | |
| u = 0.80 m s−1 (allow 0.78–0.82) | ✓ 1 | |
| Part (c) | ||
| gradient = (3.75 − 1.25) ÷ 1.00 = 2.5 s m−1 kg−1 (allow 2.4–2.6) | ✓ 1 | |
| gradient = 1/(mAu), so mA = 1 ÷ (2.5 × 0.80) = 0.50 kg (allow 0.48–0.52) | ✓ 1 | |
| Part (d) | ||
| friction-compensated so that there is no resultant external force and momentum is conserved / A does not slow down before or after the collision | ✓ 1 | |
| same release point so that u is the same in every run, which the analysis assumes (it is a constant in the equation) | ✓ 1 | |
Exam tip: To turn an equation into a straight-line graph, match it to y = mx + c: here y = 1/v, x = mB, gradient 1/(mAu) and intercept 1/u.
A student forgets to compensate a runway for friction. Trolley A, of mass 0.60 kg, is pushed along the runway and collides with a stationary trolley B; they stick together. The graph shows the velocity of A recorded by a motion sensor. The dashed lines continue the two sloping sections to the moment of the collision at t = 1.0 s.
Explain why the graph slopes downwards before and after the collision.
Use the graph to find the velocity of A just before the collision and the velocity of the joined trolleys just after it.
Determine the mass of B.
Another student uses the velocity at t = 0 and the velocity at t = 3.0 s instead. Calculate the mass of B this would give, and explain why it is wrong.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| friction (an external force) acts against the motion, so the trolleys decelerate | ✓ 1 | |
| Part (b) | ||
| just before: 0.80 m s−1 | ✓ 1 | |
| just after: 0.50 m s−1 (allow 0.49–0.51) | ✓ 1 | |
| Part (c) | ||
| 0.60 × 0.80 = (0.60 + m) × 0.50 | ✓ 1 | |
| m = 0.96 − 0.60 = 0.36 kg | ✓ 1 | |
| Part (d) | ||
| 0.60 × 0.90 = (0.60 + m) × 0.30 gives m = 1.2 kg | ✓ 1 | |
| friction removes momentum during the 3.0 s, so momentum is conserved only across the brief collision itself; velocities must be taken just before and just after it | ✓ 1 | |
Common error: Using velocities taken long before or long after the collision when an external force acts. Extrapolate the graph to the instant of the collision, where friction has had no time to change the momentum.
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