Practise Conservation of momentum. 14 exam-style questions plus unlimited generated ones on this subtopic, at up to four difficulty levels, with full mark schemes and a progress tracker. Free, no account needed.
In a closed system, the total momentum before an event equals the total momentum after it. This is Higher tier content. You need to use conservation of momentum to calculate velocities after collisions and explosions, and to explain events such as recoil.
Grade by grade
What you need to be able to do, from the first marks up to the top grade.
6
State the law of conservation of momentumIn a closed system, the total momentum before an event equals the total momentum after it.
6
Calculate the total momentum of a systemAdd the momentum of each object, taking direction into account.
7
Calculate the velocity after objects joinTotal momentum before = (m1 + m2) × v after.
7
Explain explosions and recoil using momentumThe total momentum is zero before, so the parts move apart with equal and opposite momentum.
7
Explain what a closed system isNo external forces act on the system, so its total momentum cannot change.
8
Solve collisions with objects moving in opposite directionsUse + and − for direction; the sign of the answer gives the direction of motion.
Notes
The law
In a closed system, the total momentum before an event is equal to the total momentum after the event. This is called conservation of momentum.
A closed system is one on which no external forces act.
Events include collisions (objects hit each other and bounce apart or stick together) and explosions (objects that start together push apart).
Collisions
Total momentum before = total momentum after: m1u1 + m2u2 = m1v1 + m2v2.
If the objects stick together, they move off with the same velocity: m1u1 + m2u2 = (m1 + m2) v.
Example: a 3.0 kg trolley moving at 4.0 m/s hits a stationary 1.0 kg trolley and they stick together. 3.0 × 4.0 = (3.0 + 1.0) × v, so v = 3.0 m/s.
For objects moving towards each other, give one direction + and the other −.
Explosions and recoil
Before an explosion everything is at rest, so the total momentum is zero.
Afterwards the total momentum must still be zero, so the parts move in opposite directions with momentum of equal size.
Example: when a gun fires a bullet forwards, the gun recoils backwards. The gun moves much more slowly because its mass is much larger.
Example: a 60 kg skater and a 40 kg skater push apart from rest. If the 40 kg skater moves at 3.0 m/s, the 60 kg skater moves at 40 × 3.0 ÷ 60 = 2.0 m/s in the opposite direction.
Cheatsheet
Closed system: total momentum before = total momentum after
p = m v for each object; add them, using signs for direction
Objects stick together: m1u1 + m2u2 = (m1 + m2) v
Explosion from rest: total momentum = 0 before and after
Recoil: parts move in opposite directions with equal-sized momentum
The heavier part moves more slowly
How to answer each type of question
State the law of conservation of momentum
2 marks6
Say 'in a closed system'.
Say 'total momentum before = total momentum after' (the event).
Example. State the law of conservation of momentum.
Show the model answer
In a closed system (1) the total momentum before an event is equal to the total momentum after the event (1).
Collision in which the objects join
4 marks7
Calculate the total momentum before.
State that the momentum after is the same.
Divide by the combined mass to find the velocity.
Example. A railway wagon of mass 8000 kg moves at 3.0 m/s and collides with a stationary wagon of mass 4000 kg. The wagons couple together. Calculate their velocity just after the collision.
Show the model answer
momentum before = 8000 × 3.0 = 24 000 kg m/s (1) momentum after = 24 000 kg m/s (momentum is conserved) (1) 24 000 = (8000 + 4000) × v (1) v = 2.0 m/s (1)
Explosion or recoil calculation
4 marks7
State that the total momentum before is zero.
Calculate the momentum of the part you know.
The other part has equal momentum in the opposite direction: divide by its mass.
Example. A cannon of mass 900 kg fires a cannonball of mass 6.0 kg forwards at 150 m/s. Both are at rest before firing. Calculate the recoil velocity of the cannon.
Show the model answer
total momentum before = 0 (1) momentum of cannonball = 6.0 × 150 = 900 kg m/s forwards (1) cannon's momentum = 900 kg m/s backwards, so v = 900 ÷ 900 (1) v = 1.0 m/s backwards (1)
Explain an event using momentum
3 marks7
Give the total momentum before the event.
Say that momentum is conserved.
Explain what this means for the objects afterwards.
Example. An astronaut is floating at rest in space. She throws a tool away from her. Explain why she moves in the opposite direction.
Show the model answer
The total momentum before the throw is zero (1). Momentum is conserved, so the total momentum after the throw is also zero (1). The tool has momentum in one direction, so the astronaut must have momentum of equal size in the opposite direction (1).
Collision with objects moving in opposite directions
4 marks8
Choose a positive direction and give each velocity a sign.
Add the momenta to find the total.
Divide by the combined mass; the sign of the answer gives the direction.
Example. Trolley A (mass 2.0 kg) moves to the right at 3.0 m/s. Trolley B (mass 1.0 kg) moves to the left at 4.0 m/s. They collide and stick together. Calculate their velocity after the collision.
Show the model answer
taking right as positive: p of A = +6.0 kg m/s; p of B = 1.0 × (−4.0) = −4.0 kg m/s (1) total momentum = +2.0 kg m/s (1) 2.0 = (2.0 + 1.0) × v (1) v = +0.67 m/s, i.e. 0.67 m/s to the right (1)
Shortcuts and memory tricks
Before = After. Writing 'total momentum before = total momentum after' first often earns a mark on its own.
Stuck together? Add the masses on the 'after' side.
Started at rest? The total momentum is zero, so afterwards the momenta are equal and opposite.
Sense check: if a moving object hits and sticks to a stationary one, the pair must move more slowly than the moving object did.
Where marks are lost
Forgetting to add the masses when the objects stick together.
Ignoring direction for objects moving towards each other (not using a negative sign).
Using kinetic energy instead of momentum to find the velocities after a collision.
Leaving the direction out of the final answer.
Exam technique
Draw a quick before-and-after sketch with the masses, the velocities and your positive direction.
Give the final velocity with a direction, or say what the sign means.
In 'explain' questions, mention the closed system, 'total momentum before = total momentum after', and equal and opposite momentum where relevant.
Quick recall
Cover the answers and test yourself. The app has these as flashcards that come back just before you'd forget them.
Two ice skaters stand still, facing each other. They push each other away. Skater A has a mass of 60 kg and skater B has a mass of 45 kg. Friction is negligible. In which direction does skater B move?
In the opposite direction to skater A.
A trolley of mass 1.5 kg moves at 2.0 m/s. Calculate its momentum. Use the equation: momentum = mass × velocity
3.0 kg m/s
Sample questions
Written for this site in the style of AQA exam questions. They are not taken from real past papers.
Question 1Easy5 marks
(a) Complete the sentence. In a closed system, the total momentum before an event is ............ the total momentum after the event. Tick (✓) one box.[1]
greater than
equal to
less than
(b) A trolley of mass 2.0 kg moving at 3.0 m/s collides with a stationary trolley of mass 1.0 kg. Calculate the total momentum before the collision. Use the equation: momentum = mass × velocity[2]
(c) The trolleys stick together. Calculate their velocity after the collision.[2]
Show the answer and mark scheme
(a)Answer: equal to
(b)Answer: 6.0 kg m/s
p = 2.0 × 3.0 (+ 0)
6.0 (kg m/s)
(c)Answer: 2.0 m/s
6.0 = (2.0 + 1.0) × v
v = 2.0 (m/s)
Question 2Medium6 marks
Two ice skaters stand still, facing each other. They push each other away. Skater A has a mass of 60 kg and skater B has a mass of 45 kg. Friction is negligible.
(a) What is the total momentum of the skaters before they push? Give a reason for your answer.[2]
(b) After the push, skater A moves at 1.5 m/s. Calculate the velocity of skater B.[3]
(c) In which direction does skater B move?[1]
Show the answer and mark scheme
(a)Answer: Zero, because neither skater is moving.
zero
neither skater is moving / both velocities are zero
(b)Answer: 2.0 m/s (in the opposite direction to A) m/s
total momentum after = 0, so 60 × 1.5 = 45 × v
v = 90 ÷ 45
2.0 (m/s)
(c)Answer: In the opposite direction to skater A.
in the opposite direction to skater A
Question 3Hard9 marks
An astronaut is floating at rest outside a space station. The total mass of the astronaut and spacesuit is 110 kg. The astronaut throws a bag of tools, of mass 2.5 kg, directly away from the space station at 6.0 m/s.
(a) Calculate the velocity of the astronaut after throwing the bag.[3]
(b) Calculate the total kinetic energy of the astronaut and the bag after the throw.[3]
(c) Before the throw the total kinetic energy was zero. Where did the kinetic energy come from?[1]
(d) The astronaut is 12 m from the space station. Calculate the time taken for the astronaut to reach the space station.[2]
Show the answer and mark scheme
(a)Answer: 0.14 m/s towards the space station m/s
total momentum before = 0, so 110 × v = 2.5 × 6.0
v = 15 ÷ 110
0.14 (m/s) towards the space station
(b)Answer: 46 J
kinetic energy of the bag = 0.5 × 2.5 × 6.02 = 45 (J)
kinetic energy of the astronaut = 0.5 × 110 × 0.1362 = 1.0 (J)
total = 46 (J)
(c)Answer: The chemical energy store of the astronaut’s muscles.
the chemical energy store (of the astronaut’s muscles)