Edexcel International A Level Physics · Unit 4 · Further Mechanics, Fields and Particles
Impulse and momentum in two dimensions
40 free, high-quality past-paper-style questions on Impulse and momentum in two dimensions 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.40 new this month · 146 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · every answer re-solved before publishing · How questions are checked
1IALP-U4-1-A-01Unit 4 · A2Easy1 mark
A camera records a collision between two marbles at 240 frames per second. Video-analysis software shows that one marble moves 18 mm between frame 12 and frame 18. What is the speed of the marble?
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Answer
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Notes
Frame 12 to frame 18 is 6 frame intervals, so the time is 6 ÷ 240 = 0.025 s. v = 0.018 ÷ 0.025 = 0.72 m s−1.
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Answer B
Answer: B
Common error: Counting frames instead of intervals. From frame 12 to frame 18 there are 6 intervals, not 7.
A football of mass 0.43 kg is at rest. A player kicks it and it leaves her foot at 24 m s−1. Her foot is in contact with the ball for 9.0 ms. What is the average force on the ball?
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Answer
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Notes
Δp = 0.43 × 24 = 10.3 N s. F = Δp/Δt = 10.3 ÷ 9.0 × 10−3 = 1100 N (1150).
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Answer C
Answer: C
Common error: Leaving the time in milliseconds. Convert 9.0 ms to 9.0 × 10−3 s before dividing.
An airbag reduces the risk of injury to a driver in a crash. Compared with the driver hitting the steering wheel, the airbag
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Answer
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The driver's change of momentum is the same either way (from the same speed to rest). The airbag increases the time taken for the driver to stop, so the average force F = Δp/Δt is smaller.
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Answer B
Answer: B
Exam tip: In safety questions, say what stays the same (the change of momentum) as well as what changes (the time, and so the force).
A cyclist accelerates along a straight, level road. The graph shows how the momentum of the cyclist and bicycle varies with time. What is the average resultant force on the cyclist and bicycle between t = 2 s and t = 6 s?
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Answer
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Notes
Resultant force = rate of change of momentum = gradient of the graph. Between 2 s and 6 s: F = (560 − 240) ÷ (6 − 2) = 80 N.
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Answer B
Answer: B
Exam tip: On a momentum–time graph the gradient is the resultant force. Use the change of momentum over the change of time, not a single reading divided by a single time.
A firework shell is momentarily at rest at the top of its flight when it bursts into three fragments that all move off horizontally. The figure is a top view showing the velocities of fragments 1 and 2 just after the burst. Fragment 3 has a mass of 0.25 kg. Ignore gravity and air resistance during the burst. What is the speed of fragment 3 just after the burst?
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Momentum of fragment 1 = 0.20 × 15 = 3.0 kg m s−1 east; fragment 2 = 0.10 × 40 = 4.0 kg m s−1 north. The total momentum stays zero, so fragment 3 has momentum √(3.0² + 4.0²) = 5.0 kg m s−1 (towards the south-west). Speed = 5.0 ÷ 0.25 = 20 m s−1.
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Answer C
Answer: C
Exam tip: Find the momentum of each fragment first, then combine them as vectors. The unknown fragment must cancel their resultant.
The figure shows two gliders on a level air track just before and just after they collide. How much kinetic energy is transferred to other forms in the collision?
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Answer
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Notes
Before: ½ × 0.40 × 0.80² = 0.128 J. After: ½ × 0.40 × 0.30² + ½ × 0.20 × 1.0² = 0.018 + 0.100 = 0.118 J. Transferred = 0.128 − 0.118 = 0.010 J. (Momentum is conserved: 0.32 kg m s−1 before and after.)
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Answer A
Answer: A
Exam tip: Find the total kinetic energy of every object before and after. The difference is the energy transferred to other forms, such as internal energy and sound.
A student measures the impulse of the force on a trolley with a force sensor and data logger, and the change of momentum of the trolley using a light gate, for five different collisions. The graph shows her results. Which conclusion is supported by the graph?
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Answer
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Notes
The points lie close to a straight line through the origin with gradient 1 (for example 1.0 N s of impulse for 1.0 N s change of momentum), so the impulse is equal to the change of momentum, as Newton's second law predicts.
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Answer D
Answer: D
Exam tip: A straight line through the origin shows proportionality. Equality needs the gradient as well: here it is 1, with both quantities in N s.
Disc A slides across a horizontal air table and strikes disc B, which is at rest. The figure is a momentum vector diagram drawn to scale. It shows the momentum of A just before and just after the collision.
(a)
State the principle of conservation of linear momentum.
(1)
(b)
Use the figure to determine the components of the momentum of B after the collision, along and perpendicular to A's original direction.
(2)
(c)
The mass of B is 0.25 kg. Calculate the speed of B after the collision.
(2)
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Part (a)
the total momentum of a system remains constant provided no resultant external force acts on it
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Part (b)
A before: 0.80 kg m s−1 along; A after: 0.50 along and 0.40 perpendicular
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B: 0.80 − 0.50 = 0.30 kg m s−1 along and 0.40 kg m s−1 perpendicular, on the opposite side
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Part (c)
p = √(0.30² + 0.40²) = 0.50 kg m s−1
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v = 0.50 ÷ 0.25 = 2.0 m s−1
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Exam tip: On a momentum vector diagram, B's momentum is the vector that joins the tip of A-after to the tip of A-before: the two after-vectors must add to the before-vector.
A 1.5 kg block is at rest on a smooth horizontal surface. A horizontal force acts on it. The force–time graph is a triangle: the force rises steadily from zero to 240 N and falls steadily back to zero, the whole pulse lasting 0.050 s. What is the speed of the block after the force has acted?
Figure: force–time graph of the pulse
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Answer
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Notes
Impulse = area = ½ × 240 × 0.050 = 6.0 N s = Δp. v = 6.0 ÷ 1.5 = 4.0 m s−1.
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Answer A
Answer: A
Exam tip: The area under a force–time graph is the change of momentum. Divide by the mass to get the change of velocity.
Two toy cars move on a smooth horizontal floor. Car X (2.0 kg) moves east at 3.0 m s−1. Car Y (1.5 kg) moves north at 4.0 m s−1. They collide and lock together. What is the speed of the cars just after the collision?
Figure: the two cars just before the collision (plan view, not to scale)
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Notes
East: 2.0 × 3.0 = 6.0 kg m s−1; north: 1.5 × 4.0 = 6.0 kg m s−1. Total = √(6.0² + 6.0²) = 8.49 kg m s−1. v = 8.49 ÷ 3.5 = 2.4 m s−1 (2.42).
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Answer A
Answer: A
Common error: Adding perpendicular momenta as if they were in the same direction. Combine perpendicular components with Pythagoras.
A ball of mass 0.060 kg hits a smooth wall at 12 m s−1, travelling at 30° to the normal (the line perpendicular to the wall). It rebounds at the same speed and at 30° to the normal on the other side. What is the magnitude of the impulse on the ball?
Figure: the ball's path before and after hitting the wall (plan view)
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Answer
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Notes
The component parallel to the wall is unchanged. The perpendicular component reverses: Δp = 2 × 0.060 × 12 cos 30° = 1.2 N s (1.25), directed away from the wall.
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Answer C
Answer: C
Exam tip: Resolve along and perpendicular to the surface. Only the perpendicular component of momentum changes when a ball bounces off a smooth wall.
A proton (mass 1.67 × 10−27 kg) and an alpha particle (mass 6.64 × 10−27 kg) have momenta of the same magnitude. What is the value of (kinetic energy of the proton) ÷ (kinetic energy of the alpha particle)?
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Answer
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Ek = p2/2m, so for equal p the ratio is mα/mp = 6.64 ÷ 1.67 = 4.0 (3.98).
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Answer D
Answer: D
Common error: Inverting the ratio. For a fixed momentum, kinetic energy is inversely proportional to mass, so the lighter proton has more.
A horizontal jet of water from a hose hits a vertical wall at right angles. Water leaves the hose at a rate of 0.80 kg each second and reaches the wall at 12 m s−1. On hitting the wall the water spreads out along the wall surface and does not bounce back.
Figure: the jet of water hitting the wall (not to scale)
(a)
Explain, using Newton's laws of motion, why the water exerts a force on the wall.
(2)
(b)
Calculate the force of the water on the wall.
(2)
(c)
The jet is aimed at a different surface, from which the water bounces straight back at 4.0 m s−1. Calculate the new force on the surface.
(2)
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Answer
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Notes
Part (a)
the wall changes the momentum of the water (stops it horizontally), so by Newton's second law the wall exerts a force on the water
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by Newton's third law the water exerts an equal and opposite force on the wall
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Part (b)
change of momentum each second = 0.80 × 12
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F = 9.6 N
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Part (c)
change of velocity = 12 − (−4.0) = 16 m s−1
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F = 0.80 × 16 = 13 N (12.8)
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Exam tip: When mass flows, use F = (mass per second) × (change of velocity). This is Newton's second law in momentum form.
A model hovercraft of mass 0.75 kg rests on a smooth level floor. Its fan is switched on briefly. A force sensor records the horizontal thrust: it rises steadily from zero to 1.5 N in the first 0.20 s, stays at 1.5 N until t = 1.40 s, then falls steadily to zero at t = 1.60 s. Resistive forces are negligible.
Figure: force–time graph of the thrust
(a)
State what is represented by the area under a force–time graph.
(1)
(b)
Calculate the impulse of the thrust.
(2)
(c)
Calculate the speed of the hovercraft at t = 1.60 s.
(2)
(d)
Calculate the average force on the hovercraft over the 1.60 s.
(1)
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Answer
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Notes
Part (a)
the impulse / the change of momentum
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Part (b)
area = ½ × 1.5 × 0.20 + 1.5 × 1.20 + ½ × 1.5 × 0.20
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= 2.1 N s
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Part (c)
v = 2.1 ÷ 0.75
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= 2.8 m s−1
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Part (d)
2.1 ÷ 1.60 = 1.3 N (1.31)
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Common error: Using the peak force × total time (1.5 × 1.60 = 2.4 N s). The force is smaller than its peak at the start and end, so split the area into shapes.
A student wants to show that the impulse of a force on a trolley is equal to the change of momentum of the trolley. She has a runway, a trolley carrying a card, a light gate, a force sensor, a data logger and a top-pan balance.
(a)
Describe how she could carry out the investigation. Your answer should include the measurements she makes and how she uses them.
(6)
(b)
She plots the impulse (area) on the y-axis against the change of momentum on the x-axis. State the shape and gradient of the graph that would confirm her hypothesis.
(2)
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Answer
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Notes
Part (a)
Indicative content:
—
tilt the runway until the trolley moves at constant speed after a small push, so friction is compensated
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measure the mass of the trolley and card with the balance and the length of the card (with a ruler or calipers)
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fix the force sensor at the end so the trolley collides with it; connect the sensor and light gate to the data logger
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use the light gate to find the speed before (and after, if it rebounds) the impact: speed = card length ÷ time
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record a force–time graph of the impact at a high sampling rate and find the area under it, which is the impulse
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calculate the change of momentum m(v − u) and compare with the impulse; repeat for different starting points on the runway
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Part (b)
a straight line through the origin
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with a gradient of 1
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Exam tip: In a six-mark description, cover the set-up, every measurement and the instrument used for it, how each result is calculated, and how the comparison is made.
A student uses a video camera and analysis software to study a glancing collision between a moving steel sphere and a stationary glass sphere on a smooth bench. The camera records 60 frames per second.
(a)
Explain how the student should position the camera and the scale so that distances measured from the video are accurate.
(2)
(b)
Before the collision the steel sphere moves 0.126 m in 10 frame intervals. Calculate its speed.
(2)
(c)
The total kinetic energy of the spheres calculated from the video is lower after the collision than before. Suggest one reason, other than an inelastic collision, why this could happen.
(1)
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Answer
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Notes
Part (a)
camera directly above the bench, pointing straight down / perpendicular to the plane of motion, so the image is not distorted
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a ruler or grid placed in the same plane as the spheres, to calibrate distances and avoid parallax error
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Part (b)
time = 10 ÷ 60 = 0.167 s
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v = 0.126 ÷ 0.167 = 0.76 m s−1 (0.756)
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Part (c)
the spheres are rolling, so some kinetic energy is rotational and not included in ½mv2 / rolling friction slows the spheres between the frames used
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Exam tip: When asked for a precaution, state what you do and why it improves the measurement.
A model railway wagon of mass 1.5 kg rolls at 0.60 m s−1 along a level track and couples with a stationary wagon of mass 0.50 kg. They move off together. Friction is negligible.
Figure: the wagons just before coupling (not to scale)
(a)
Calculate the velocity of the wagons just after coupling.
(2)
(b)
Show that 25% of the kinetic energy is transferred to other forms in the coupling.
(3)
(c)
Explain why momentum is conserved in the coupling but kinetic energy is not.
(2)
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Answer
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Notes
Part (a)
1.5 × 0.60 = (1.5 + 0.50)v
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v = 0.45 m s−1 in the original direction
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Part (b)
before: ½ × 1.5 × 0.60² = 0.27 J
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after: ½ × 2.0 × 0.45² = 0.2025 J
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(0.27 − 0.2025) ÷ 0.27 × 100 = 25%
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Part (c)
the wagons exert equal and opposite forces on each other for the same time and there is no resultant external force, so the total momentum cannot change
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kinetic energy is not conserved because work is done deforming the coupling, transferring energy to thermal energy and sound
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Exam tip: A quick check using Ek = p2/2m: the momentum is the same before and after, so the kinetic energy falls in the ratio of the masses, 1.5 ÷ 2.0 = 0.75.
A trolley of mass 2.0 kg is at rest on a smooth horizontal track. The graph shows how the resultant force on the trolley varies with time. What is the speed of the trolley at t = 5.0 s?
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Answer
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Notes
Impulse = area under the graph, counting the area below the axis as negative: 6.0 × 2.0 − 4.0 × 2.0 = 4.0 N s. Change of momentum = 4.0 N s, so v = 4.0 ÷ 2.0 = 2.0 m s−1.
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Answer A
Answer: A
Common error: Adding the two areas as if both were positive. A force in the opposite direction gives a negative impulse, which reduces the momentum.
Ball X moves at 3.0 m s−1 towards ball Y, which is at rest on a smooth horizontal surface. After the collision X moves at 2.0 m s−1 at 35° to its original direction, as shown in the figure. Which arrow could show the direction of the velocity of Y after the collision?
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Perpendicular to the original line, the total momentum is zero, so Y must move to the opposite side from X (downwards in the figure). Along the line, X's component falls from 3.0 to 2.0 cos 35° = 1.6 m s−1, so Y must gain forward momentum. Only S is both forward and on the opposite side.
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Answer D
Answer: D
Exam tip: Treat the two directions separately: the momentum components across the original line must cancel, and the components along it must add up to the original momentum.
A sphere rolls across a horizontal table during a collision experiment. A camera above the table records 50 frames per second. The figure combines the images of the sphere from six consecutive frames on a grid of 1.0 cm squares. What is the average speed of the sphere?
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Answer
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From image 1 to image 6 the centre moves 6.0 cm across and 8.0 cm up: distance = √(6.0² + 8.0²) = 10 cm. Six images span 5 frame intervals: time = 5 ÷ 50 = 0.10 s. v = 0.10 ÷ 0.10 = 1.0 m s−1.
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Answer C
Answer: C
Common error: Counting images instead of intervals. Six consecutive frames contain five intervals of 1/50 s.
A hockey ball of mass 0.16 kg moves east across the pitch. A player's stick strikes it and it moves off north. The figure shows the velocities just before and just after the hit. The stick is in contact with the ball for 4.0 ms.
(a)
Explain why the change of momentum of the ball is not equal to 0.16 × (16 − 12).
(1)
(b)
Calculate the magnitude of the change of momentum of the ball.
(3)
(c)
Determine the direction of the impulse on the ball.
(1)
(d)
Calculate the average force of the stick on the ball.
(1)
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Answer
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Notes
Part (a)
momentum is a vector and the two velocities are in different directions, so the change must be found by vector subtraction
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Part (b)
initial momentum 0.16 × 12 = 1.92 kg m s−1 east; final momentum 0.16 × 16 = 2.56 kg m s−1 north
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Δp = final − initial: 1.92 kg m s−1 west and 2.56 kg m s−1 north
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|Δp| = √(1.92² + 2.56²) = 3.2 N s
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Part (c)
tan−1(2.56 ÷ 1.92) = 53° north of west (53.1°)
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Part (d)
F = 3.2 ÷ 4.0 × 10−3 = 8.0 × 102 N, in the direction of the impulse
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Common error: Subtracting the speeds. The change of momentum is the final momentum minus the initial momentum as vectors: reverse the initial vector and add it to the final one.
A student investigates the impulse of a force and the change of momentum it produces. A trolley of mass 0.620 kg carries a card of length 0.0500 m. It passes through a light gate, collides with a buffer fixed to a force sensor, and rebounds back through the same light gate. The card blocks the gate for 0.0400 s before the collision and for 0.0625 s after it. The graph shows the force recorded by the sensor during the collision.
(a)
Calculate the magnitude of the change of momentum of the trolley.
(3)
(b)
Use the graph to estimate the impulse of the force on the trolley.
(2)
(c)
The buffer is replaced by a stiffer one. The trolley reaches the buffer and leaves it with the same speeds as before. Describe and explain how the force–time graph would differ from the one shown.
(2)
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Answer
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Notes
Part (a)
speed before = 0.0500 ÷ 0.0400 = 1.25 m s−1
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speed after = 0.0500 ÷ 0.0625 = 0.800 m s−1, in the opposite direction
✓ 1
Δp = 0.620 × (1.25 + 0.800) = 1.27 N s
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Part (b)
each square represents 10 × 10−3 s × 10 N = 0.10 N s, and the area is about 12 squares
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impulse ≈ 1.2 N s (allow 1.1–1.3 N s)
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Part (c)
the area under the graph is unchanged, because the change of momentum, and so the impulse, is unchanged
✓ 1
the contact time is shorter, so the peak force (and the average force) is greater: a narrower, taller pulse
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Common error: Using only the speed before the collision. The trolley rebounds, so the velocity reverses and the two speeds add when you calculate the change of momentum.
The graph shows how the kinetic energy Ek of each of two trolleys, X and Y, varies with the square of its momentum, p2.
(a)
Use the graph and the equation Ek = p2/2m to determine the mass of each trolley.
(2)
(b)
Trolley X, with momentum 1.2 kg m s−1, collides with trolley Y, which is at rest. The trolleys couple together. Calculate the kinetic energy transferred to other forms in the collision.
(3)
(c)
Calculate the fraction of the original kinetic energy that is transferred.
(1)
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Answer
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Notes
Part (a)
X: gradient = 3.0 ÷ 3.0 = 1.0 = 1/2m, so m = 0.50 kg
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Y: gradient = 1.0 ÷ 3.0 = 1/2m, so m = 1.5 kg
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Part (b)
before: Ek = 1.2² ÷ (2 × 0.50) = 1.44 J
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after: the momentum is still 1.2 kg m s−1 and the mass is 2.0 kg, so Ek = 1.2² ÷ (2 × 2.0) = 0.36 J
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transferred = 1.44 − 0.36 = 1.1 J (1.08)
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Part (c)
1.08 ÷ 1.44 = 0.75 (equal to 1.5 ÷ 2.0)
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Exam tip: When the momentum is the same before and after, Ek = p2/2m is the quickest route: the kinetic energy simply scales with 1/m.
A gymnast of mass 60 kg jumps down from a platform and reaches the floor at 5.0 m s−1. The graph shows the resultant force on her as she is brought to rest, for two landings: A, with her legs kept straight, and B, bending her knees as she lands.
(a)
Calculate the change of momentum of the gymnast during the landing.
(1)
(b)
Explain, with reference to the graph, why bending the knees reduces the risk of injury. Your answer should refer to impulse and to Newton's second law.
(6)
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Answer
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Notes
Part (a)
Δp = 60 × 5.0 = 3.0 × 102 N s (300 kg m s−1)
✓ 1
Part (b)
Indicative content:
—
she has the same change of momentum (300 N s) in both landings, from 5.0 m s−1 to rest
✓ 1
impulse = change of momentum, so the areas under curves A and B are equal
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Newton's second law: force = rate of change of momentum, F = Δp/Δt
✓ 1
bending the knees increases the stopping time from about 0.050 s to about 0.30 s
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so the average force falls from about 6 kN to about 1 kN / the peak force falls from 12 kN to 2.0 kN
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smaller forces on her bones and joints mean less risk of injury
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Exam tip: In safety explanations, say what stays the same (the change of momentum, the area under the graph) before saying what changes (the time, and so the force).
Sphere A, of mass 0.064 kg, rolls along a straight horizontal groove and hits sphere B, of mass 0.032 kg, which is at rest. A video camera records 120 frames per second, and tracking software gives the position x of the centre of each sphere in every frame. The graph shows the results. Ignore friction and rotation.
(a)
Calculate the speed of A before the collision.
(2)
(b)
Determine the speeds of A and B after the collision.
(2)
(c)
Show that the collision is elastic.
(2)
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Answer
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Notes
Part (a)
A moves 10 cm in 10 frames, i.e. 1.0 cm per frame
✓ 1
v = 0.010 m × 120 s−1 = 1.2 m s−1
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Part (b)
A: 2.0 cm in 6 frames (0.050 s), so 0.40 m s−1
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B: 8.0 cm in 6 frames, so 1.6 m s−1
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Part (c)
momentum: before 0.064 × 1.2 = 0.0768 kg m s−1; after 0.064 × 0.40 + 0.032 × 1.6 = 0.0256 + 0.0512 = 0.0768 kg m s−1, so the speeds are consistent
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kinetic energy: before ½ × 0.064 × 1.2² = 0.0461 J; after ½ × 0.064 × 0.40² + ½ × 0.032 × 1.6² = 0.00512 + 0.0410 = 0.0461 J, unchanged, so elastic
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Exam tip: With frame data, time = number of frame intervals ÷ frame rate. Read the gradient over as many frames as possible to reduce the uncertainty.
A stationary object in deep space breaks into exactly three fragments. Which statement must be true?
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Answer
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Notes
The total momentum stays zero, so the three momentum vectors add to zero. Placed head to tail they form a closed triangle, which lies in a plane: the velocities all lie in one plane. Their speeds depend on their masses, the kinetic energy increases, and no two need be opposite.
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Answer B
Answer: B
Exam tip: Three vectors with a zero sum always form a closed triangle — a useful picture for any explosion into three pieces.
A 2.0 kg trolley moving at 3.0 m s−1 collides head-on with a stationary 1.0 kg trolley. Friction is negligible. Which pair of velocities after the collision, both in the original direction, describes an elastic collision?
Figure: the trolleys before the collision (not to scale)
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Answer
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Notes
All four pairs conserve momentum (6.0 kg m s−1). Kinetic energy before = ½ × 2.0 × 3.0² = 9.0 J. Only 1.0 and 4.0 m s−1 give 1.0 + 8.0 = 9.0 J after. The pairs 2.0 and 2.0 m s−1 and 1.5 and 3.0 m s−1 lose kinetic energy (6.0 J and 6.75 J); 0.5 and 5.0 m s−1 would gain it (12.75 J), which is impossible without an energy source.
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Answer D
Answer: D
Exam tip: Momentum conservation alone cannot tell you whether a collision is elastic. Check the total kinetic energy too.
In an investigation of impulse, a trolley of mass (0.850 ± 0.005) kg carries a card of length (0.0400 ± 0.0005) m. It runs through a light gate into a spring buffer attached to a force sensor and rebounds through the same gate. The card takes (0.0500 ± 0.0010) s to pass through the gate before the impact and 0.0750 s afterwards. The data logger gives the area under the force–time graph as 1.10 N s.
(a)
Calculate the magnitude of the change of momentum of the trolley.
(3)
(b)
Calculate the percentage uncertainty in the momentum of the trolley before the impact.
(2)
(c)
The percentage uncertainty in the change of momentum is 4%. Deduce whether the results support the idea that impulse equals change of momentum.
(2)
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Answer
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Notes
Part (a)
u = 0.0400 ÷ 0.0500 = 0.800 m s−1
✓ 1
v = −0.0400 ÷ 0.0750 = −0.533 m s−1
✓ 1
Δp = 0.850 × (0.800 + 0.533) = 1.13 N s
✓ 1
Part (b)
percentage uncertainties: mass 0.59%, length 1.25%, time 2.0%
✓ 1
total = 3.8% (3.84)
✓ 1
Part (c)
4% of 1.13 = 0.045, so Δp lies between 1.09 and 1.18 N s
✓ 1
1.10 N s lies within this range, so the results support the idea
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Common error: Forgetting that the trolley rebounds. The velocity after the impact is negative, so the speeds add in the change of momentum.
On a smooth ice surface, an ice-hockey puck A moving at 2.2 m s−1 strikes an identical stationary puck B (same mass). After the collision A moves at 1.1 m s−1 at 60.0° to its original direction.
Figure: the pucks before and after the collision (plan view, not to scale)
(a)
Show that the components of B's velocity are 1.65 m s−1 along A's original direction and 0.953 m s−1 perpendicular to it.
(3)
(b)
Calculate the speed of B and the angle its path makes with A's original direction.
(2)
(c)
Show that the collision is elastic.
(2)
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Answer
Mark
Notes
Part (a)
along: 2.2m = 1.1m cos 60.0° + mvx
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perpendicular: 0 = 1.1m sin 60.0° + mvy
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vx = 2.2 − 0.55 = 1.65 m s−1 and |vy| = 1.1 × sin 60.0° = 0.953 m s−1
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Part (b)
v = √(1.65² + 0.953²) = 1.9 m s−1 (1.91)
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tan−1(0.953 ÷ 1.65) = 30.0°, on the opposite side of the line from A
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Part (c)
kinetic energy before ∝ 2.2² = 4.84 (½m × 4.84)
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after ∝ 1.1² + (1.65² + 0.953²) = 1.21 + 3.63 = 4.84, the same, so kinetic energy is conserved
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Exam tip: With equal masses the mass cancels from every equation, so you can work with velocities. An elastic collision between equal masses, one at rest, sends them off at 90° to each other.
A space probe of total mass 600 kg moves at 200 m s−1 in the +x direction, relative to a distant star. Springs push a 150 kg module away from it. Just afterwards the module moves at 250 m s−1 at 20.0° to the +x direction. Ignore external forces.
Figure: the probe before and just after the module separates (not to scale)
(a)
Calculate the total momentum of the probe before the separation.
(1)
(b)
Calculate the components of the module's momentum after the separation.
(2)
(c)
Determine the velocity of the remaining 450 kg of the probe after the separation.
(3)
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Answer
Mark
Notes
Part (a)
600 × 200 = 1.20 × 105 kg m s−1 in the +x direction
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Part (b)
x: 150 × 250 cos 20.0° = 3.52 × 104 kg m s−1
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y: 150 × 250 sin 20.0° = 1.28 × 104 kg m s−1
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Part (c)
vx = (1.20 × 105 − 3.52 × 104) ÷ 450 = 188 m s−1 and vy = −1.28 × 104 ÷ 450 = −28.5 m s−1
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v = √(188² + 28.5²) = 191 m s−1
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at tan−1(28.5 ÷ 188) = 8.60° to the +x direction, on the opposite side from the module
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Common error: Taking the total momentum before as zero. The probe was already moving, so each component after must add up to the momentum it had before.
A tennis ball of mass 0.060 kg travels horizontally at 15 m s−1 towards a racket. The graph shows how the force of the racket on the ball varies with time during the impact. The ball leaves the racket horizontally in the opposite direction. Ignore the weight of the ball. What is the speed of the ball as it leaves the racket?
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Answer
Mark
Notes
Impulse = area of the triangle = ½ × 800 × 5.0 × 10−3 = 2.0 N s. Take the final direction as positive: 2.0 = 0.060v − 0.060 × (−15), so 0.060v = 2.0 − 0.90 = 1.1 and v = 18 m s−1 (18.3).
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Answer A
Answer: A
Common error: Forgetting the momentum the ball already had. Part of the impulse is used to stop the ball before it can speed up in the opposite direction.
Trolley P, of mass 0.60 kg, moves along a straight, level track towards trolley Q, which is at rest. They collide at t = 2.0 s. The graph shows the displacement of each trolley from its own starting position. Friction is negligible. Which row gives the mass of Q and the type of collision?
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Answer
Mark
Notes
Gradients: P before 0.40 m s−1, P after 0.20 m s−1, Q after 0.40 m s−1. Momentum: 0.60 × 0.40 = 0.60 × 0.20 + m × 0.40, so m = 0.30 kg. Kinetic energy before = 0.048 J; after = 0.012 + 0.024 = 0.036 J, so kinetic energy is lost: 0.30 kg; inelastic.
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Answer B
Answer: B
Exam tip: On a displacement–time graph, read each velocity as a gradient over a long section of the line. Then apply conservation of momentum before testing kinetic energy.
A puck glides east across a horizontal air table. A compressed spring inside it is released and the puck separates into two parts, A and B. The figure shows the motion before the separation and the velocity of A just after it. Friction is negligible.
(a)
Show that, just after the separation, the velocity of B has components of 0.461 m s−1 east and 0.600 m s−1 south.
(3)
(b)
Calculate the speed of B and the angle its path makes with east.
(2)
(c)
Calculate the energy transferred from the spring to kinetic energy.
vE = (0.150 − 0.104) ÷ 0.10 = 0.461 m s−1 and vN = −0.060 ÷ 0.10 = −0.600 m s−1, i.e. 0.600 m s−1 south
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Part (b)
v = √(0.461² + 0.600²) = 0.757 m s−1
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tan−1(0.600 ÷ 0.461) = 52.5° south of east
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Part (c)
kinetic energy before = ½ × 0.30 × 0.50² = 0.0375 J; after = ½ × 0.20 × 0.60² + ½ × 0.10 × 0.757² = 0.036 + 0.0286 = 0.0646 J
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energy from spring = 0.0646 − 0.0375 = 0.027 J (0.0271)
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Exam tip: The puck was already moving, so the momentum before is not zero. Resolve every momentum along east and north and balance each direction separately.
A student uses video analysis to study a glancing collision. Sphere A, of mass 0.30 kg, rolls towards sphere B, of mass 0.20 kg, which is at rest on a horizontal surface. Tracking software marks the centre of each sphere every 0.10 s; the figure shows the positions on a grid of 2.0 cm squares. A collides with B when A's centre is at (24, 0) cm. Assume friction is negligible and ignore any rotation of the spheres.
(a)
Show that the speed of A before the collision is 0.60 m s−1.
(1)
(b)
Determine the x and y components of the velocities of A and B after the collision.
(2)
(c)
Show that momentum is conserved in both the x and the y directions.
(2)
(d)
Determine whether the collision is elastic.
(2)
(e)
Explain why the student measures displacements over four intervals instead of between two neighbouring images.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
A moves 24 cm in 4 intervals: 0.24 m ÷ 0.40 s = 0.60 m s−1
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Part (b)
A: 12 cm and 8.0 cm in 0.40 s, so vx = 0.30 m s−1, vy = 0.20 m s−1
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B: 18 cm and −12 cm in 0.40 s, so vx = 0.45 m s−1, vy = −0.30 m s−1
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Part (c)
x: before 0.30 × 0.60 = 0.18 kg m s−1; after 0.30 × 0.30 + 0.20 × 0.45 = 0.09 + 0.09 = 0.18 kg m s−1
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y: before 0; after 0.30 × 0.20 + 0.20 × (−0.30) = 0.06 − 0.06 = 0
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Part (d)
kinetic energy before = ½ × 0.30 × 0.60² = 0.054 J; after = ½ × 0.30 × (0.30² + 0.20²) + ½ × 0.20 × (0.45² + 0.30²) = 0.0195 + 0.02925 = 0.04875 J ≈ 0.049 J
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kinetic energy decreases (by about 10%), so the collision is inelastic
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Part (e)
the displacement measured is larger, so the percentage uncertainty in reading the positions (and in the velocity) is smaller
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Exam tip: From a tracking plot, read the change in x and the change in y over the same number of intervals and divide each by the same time. Then conserve momentum in each direction separately.
On a smooth, level rink, bumper car X (total mass 200 kg) moves at 3.0 m s−1 and strikes bumper car Y (total mass 300 kg), which is at rest. The figure shows the directions in which the cars move just after the collision. Ignore friction during the collision.
(a)
Write down equations for the conservation of momentum along, and perpendicular to, the original direction of X. Use vX and vY for the speeds after the collision.
(2)
(b)
Show that vX is about 1.5 m s−1 and calculate vY.
(3)
(c)
Determine the percentage of the original kinetic energy that is transferred to other forms in the collision.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
along: 200 × 3.0 = 200vX cos 50.0° + 300vY cos 30.0°
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perpendicular: 200vX sin 50.0° = 300vY sin 30.0°
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Part (b)
from the perpendicular equation, vY = 200 sin 50.0° vX ÷ (300 sin 30.0°) = 1.02vX
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substituting: 600 = (128.6 + 265.4)vX, so vX = 1.52 m s−1
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vY = 1.56 m s−1
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Part (c)
before: ½ × 200 × 3.0² = 900 J; after: ½ × 200 × 1.52² + ½ × 300 × 1.56² = 232 + 363 = 595 J
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(900 − 595) ÷ 900 × 100 = 34%
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Common error: Using the same angle function for both cars. Resolve each velocity separately: the cosine of each car's own angle gives its component along the original line, the sine gives the component across it.
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