Conservation of momentum and recoil
If no external force acts, the total momentum stays the same before and after a collision or explosion. A gun's recoil follows from this.
The law
When bodies push each other, momentum is not lost. It is only passed from one body to another. The law of conservation of momentum says: when bodies collide or push apart, their total momentum stays the same. This holds if no external force acts on them. A group of bodies with no outside force on it is called an isolated system.
Why momentum is conserved
- 1From the second law, force = rate of change of momentum: F = (p₂ − p₁)/t.
- 2If no external force acts on the system, F = 0.
- 3Then (p₂ − p₁)/t = 0, so p₂ − p₁ = 0.
- 4In a collision, the third law makes the forces on the two bodies equal and opposite, for the same time. So the momentum one body gains, the other loses.
- 5★So, p₁ = p₂: the total momentum after equals the total momentum before.
★ key step
Activity: marbles in a channel
You need a straight plastic channel (or a ruler with a groove) about 40 cm long and seven identical marbles, on a flat table.
- 1Put seven marbles of the same size in a row in the channel, touching each other.
- 2Take one marble away and place it about 15 cm from the row.
- 3Flick this single marble gently so that it hits the row.
- 4★Observe: the striking marble stops, and the last marble at the far end moves off at about the same speed.
- 5Repeat with two marbles striking the row. Observe: two marbles move off at the far end.
- 6Conclude: the momentum before the collision equals the momentum after it.
★ key step
Worked example: recoil of a rifle
A rifle of mass 4 kg fires a bullet of mass 20 g at 300 m s⁻¹. Find the recoil velocity of the rifle.
- Given: rifle m₁ = 4 kg; bullet m₂ = 20 g = 0.02 kg; both at rest at first, so u₁ = u₂ = 0; bullet's final velocity v₂ = 300 m s⁻¹.
- Conservation of momentum: m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂.
- 0 = 4 kg × v₁ + 0.02 kg × 300 m s⁻¹, so 0 = 4v₁ + 6.
- v₁ = −6 ÷ 4 = −1.5 m s⁻¹.
The rifle recoils at 1.5 m s⁻¹. The minus sign shows it moves opposite to the bullet
Worked example: two trolleys stick together
A 3 kg trolley moving at 4 m s⁻¹ hits a 1 kg trolley at rest. They stick together and move off as one. Find their common velocity.
- Total momentum before = 3 kg × 4 m s⁻¹ + 1 kg × 0 = 12 kg m s⁻¹.
- After the collision the joined mass is 3 kg + 1 kg = 4 kg, moving with velocity v.
- Momentum after = 4 kg × v.
- Conservation: 4v = 12, so v = 3 m s⁻¹.
They move together at 3 m s⁻¹, in the direction the 3 kg trolley was moving. The total momentum is 12 kg m s⁻¹ before and after
Where you see it
- A gun recoils (kicks back) when it fires a bullet.
- A boat moves back when a person jumps from it onto the bank.
- A person standing on a skateboard who throws a heavy bag forward rolls backward.
- A rocket gains forward momentum while its exhaust gases gain equal backward momentum.
- The recoil speed of a gun is small because its mass is much larger than the bullet's.
Police and army trainees learn to hold a rifle butt firmly against the shoulder. The shoulder then takes the recoil smoothly instead of being struck by it.
2 marks: “State the law of conservation of momentum.” When two or more bodies interact, their total momentum stays constant, provided no external force acts on them. Total momentum before = total momentum after.
3 marks: “A gun of mass 6 kg fires a 30 g bullet at 400 m s⁻¹. Find the recoil velocity of the gun.” Method: 0 = MV + mv, so V = −mv/M. mv = 0.03 × 400 = 12 kg m s⁻¹. V = −12 ÷ 6 = −2 m s⁻¹. Answer: the gun recoils at 2 m s⁻¹, opposite to the bullet.
1 marks: “Why does a gun move backward when it fires a bullet?” The total momentum is zero before firing. The bullet gains forward momentum, so the gun must gain equal backward momentum.
Before equals after: momentum is shared, never lost, when no outside force acts.