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10.5 Using Conservation of Momentum

GCSE Physics · AQA · Chapter 10: Force and Motion

A free GCSE Physics lesson for AQA (8463) from OBA School: (Higher tier, physics only) Solving collision and recoil problems in a straight line. Watch the video, read the explanation and worked examples, then try the revision questions. You can work through the full lesson, with every video, practice quizzes and printable worksheets with mark schemes, in our free online course.

What you will learn

  • (Higher tier) use conservation of momentum to find a velocity after a collision, including when objects stick together
  • (Higher tier) use + and − signs for objects moving in opposite directions
  • (Higher tier) explain why two objects pushed apart move off at different speeds if their masses are different
  • (Higher tier) calculate recoil velocities using mA vA + mB vB = 0

Key words

  • Conservation of momentum: In a closed system, total momentum before = total momentum after.
  • Closed system: Objects with no external resultant force acting on them.
  • Sign convention: Choosing one direction as positive so that velocities and momenta the other way are negative.
  • Coupled: Joined together after a collision so the objects move off with the same velocity.
  • Recoil: The backward motion of an object when it pushes or fires something forwards.

A method for collision problems

1. Sketch the objects before and after. Mark the positive direction.

2. Work out the total momentum before: add m v for each object, with signs.

3. Write the total momentum after, using a letter for the unknown velocity.

4. Put total before = total after and solve.

5. Use the sign of the answer to give the direction.

Worked example: railway wagons coupling

A 25 000 kg wagon moving at 3.0 m/s hits a stationary 15 000 kg wagon. They couple together. Find their velocity.
before: 25 000 × 3.0 + 15 000 × 0 = 75 000 kg m/s
after: (25 000 + 15 000) × v = 40 000 v
40 000 v = 75 000, so v = 1.875 m/s

Answer: about 1.9 m/s in the original direction

Worked example: snooker balls that do not stick

A 0.17 kg ball moving at 2.0 m/s hits a stationary 0.16 kg ball head-on and stops dead. Find the velocity of the second ball.
before: 0.17 × 2.0 = 0.34 kg m/s; after: 0.17 × 0 + 0.16 × v
0.16 v = 0.34, so v = 2.125 m/s

Answer: about 2.1 m/s forwards

Objects already moving: same and opposite directions

Worked example: a rear-end shunt (same direction)

A 1200 kg car at 15 m/s runs into the back of an 800 kg car moving at 5.0 m/s in the same direction. They move off together.
before: 1200 × 15 + 800 × 5.0 = 18 000 + 4000 = 22 000 kg m/s
after: 2000 v = 22 000, so v = 11 m/s

Answer: 11 m/s forwards

If the objects move towards each other, one velocity is negative:

Worked example: a head-on collision (opposite directions)

before: 150 × (+4.0) + 250 × (−2.0) = +600 − 500 = +100 kg m/s
after: (150 + 250) × v = 400 v; 400 v = 100, so v = +0.25 m/s

Answer: 0.25 m/s in the direction the 150 kg car was moving

Pushing apart: recoil

When two objects at rest push each other apart, the total momentum is zero before and after:

mA vA + mB vB = 0 so mA vA = −mB vB

The momenta are equal in size and opposite in direction. The object with the smaller mass moves off with the greater speed.

Worked example: recoil of a launcher

4.0 × v + 0.010 × 400 = 0
4.0 v = −4.0, so v = −1.0 m/s

Answer: the launcher recoils at 1.0 m/s backwards

Worked example: an astronaut in space

An astronaut (total mass 100 kg) at rest outside a space station throws a 2.0 kg tool forwards at 5.0 m/s.
100 v + 2.0 × 5.0 = 0, so v = −0.10 m/s

Answer: she drifts backwards at 0.10 m/s

Guns and artillery have heavy barrels or springs to make the recoil smaller and slow it down. A rifle is pressed into the shoulder so that the recoil is shared with the much larger mass of the person.

Exam tip: Check your answer: the lighter object must move faster, and the two momenta must be equal and opposite.

Common mistakes

  • Forgetting the minus sign for an object moving the other way, so the momenta add instead of partly cancelling.
  • Forgetting to add the masses when objects stick together.
  • Dividing by the wrong mass when finding a recoil velocity.
  • Thinking momentum is “used up” or “created” in a collision or explosion. In a closed system the total never changes.
  • Leaving out the direction when a question asks for a velocity.

Revision questions

Try these, then check your answers below each one.

Question 1. (Higher tier) A 2000 kg wagon moving at 4.0 m/s hits a stationary 3000 kg wagon and they couple together. Calculate their velocity just after the collision.

Answer: 1.6 m/s

Working: 2000 × 4.0 = 8000 kg m/s. v = 8000 ÷ (2000 + 3000) = 1.6 m/s.

Question 2. (Higher tier) A 2.5 kg rifle fires a 0.010 kg bullet at 500 m/s. Calculate the speed at which the rifle recoils.

Answer: 2 m/s

Working: Bullet: 0.010 × 500 = 5.0 kg m/s. Rifle speed = 5.0 ÷ 2.5 = 2.0 m/s backwards.

Question 3. (Higher tier) A 40 kg skater and a 60 kg skater push apart from rest. Which statement is correct? (Choose from: The 60 kg skater moves off 1.5 times as fast as the 40 kg skater, The 40 kg skater has more momentum, The 40 kg skater moves off 1.5 times as fast as the 60 kg skater, Both skaters move off at the same speed.)

Answer: The 40 kg skater moves off 1.5 times as fast as the 60 kg skater

Working: 40 × v₁ = 60 × v₂, so v₁ = 1.5 v₂. The lighter skater moves faster; their momenta are equal and opposite.

Related lessons

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