Linear Momentum, Impulse and Collisions

Lessons

Senior 3 · Physics

Linear Momentum, Impulse and Collisions

4 lessons

Lesson 3 of 4

Conservation of Linear Momentum

25 min

Kick-back

When a rifle is fired, it kicks back hard against the shooter's shoulder. A rocket far out in space, with no air or ground to push against, still speeds up when its engine fires. A pupil who jumps forward out of a small boat sends the boat drifting backwards. These look like three different events, but one law explains them all, and it follows directly from Newton's third law.

Experiment 3: an "explosion" between two trolleys

You need: two trolleys (or one trolley with a spring-loaded plunger and a second trolley), extra masses, strong thread, scissors, two wooden blocks as stoppers, and a metre rule. Use a smooth, level bench.

Make both trolleys the same mass. Push them together so that the spring between them is compressed, and tie them with the thread. They are at rest.
Place a stopper 60 cm from each trolley. Cut the thread and listen: do the trolleys hit their stoppers at the same moment (one "click")?
Now double the mass of trolley B. Move B's stopper until both trolleys hit their stoppers at the same moment. Record the distances.
Repeat with B three times as heavy as A.

Figure 3.1: The trolleys travel for the same time, so the distance each covers is proportional to its speed.

Mass of A	Mass of B	Distance for A (cm)	Distance for B (cm)	Mass × distance: A	Mass × distance: B
$m$	$m$	60	60	$60 m$	$60 m$
$m$	$2 m$	60	30	$60 m$	$60 m$
$m$	$3 m$	60	20	$60 m$	$60 m$

These are illustrative results; real ones differ by a few centimetres because of friction. Because both trolleys move for the same time, distance is proportional to speed, so the results show that $m_{A} v_{A} = m_{B} v_{B}$ : the trolleys move apart with equal and opposite momenta. Before the thread was cut, the total momentum was zero; afterwards, $(+ m_{A} v_{A}) + (- m_{B} v_{B})$ is still zero.

Important

Safety. Keep fingers away from the spring while it is compressed, and cut the thread away from your body. Put a soft bag behind each stopper so that nothing falls off the bench.

Explanation: why momentum is conserved

While the spring pushes, it pushes trolley A with a force $F$ and, by Newton's third law, pushes trolley B with a force $- F$ , equal and opposite. Both forces act for the same time $\Delta t$ . So the impulses are equal and opposite:

\Delta p_A = F\,\Delta t

\Delta p_B = -F\,\Delta t \qquad\Rightarrow\qquad \Delta p_A + \Delta p_B = 0

Whatever momentum A gains, B loses (or gains in the opposite direction). The total momentum of the two trolleys does not change. This is true for any interaction, a push, a collision or an explosion, as long as no resultant force from outside acts on the bodies.

Definition

A closed (isolated) system, for momentum, is a group of bodies on which no resultant external force acts: the bodies push and pull only on each other. (You met isolated systems in S2 when studying energy.)

Principle of conservation of linear momentum: in a closed system, the total linear momentum stays constant: the total momentum before an interaction equals the total momentum after it.

m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2}

Example: in Experiment 3 the total momentum before is 0, so afterwards $m_{A} v_{A} + m_{B} v_{B} = 0$ .

Applications: recoil, rockets, boats and running

Figure 3.2: In both cases the total momentum was zero before, so the two parts move off in opposite directions with equal momenta.

Gun recoil. The bullet gains forward momentum, so the rifle gains an equal backward momentum. Pressing the rifle firmly into the shoulder makes the rifle and the shooter's body move back together: the mass is much larger, so the recoil speed is much smaller.
Rocket propulsion. The engine pushes hot gas out of the nozzle at very high speed. The gas carries momentum backwards, so the rocket gains the same momentum forwards. The rocket does not need air to push on, which is why rockets work in space.
Boats and swimmers. A paddle or an oar pushes water backwards; the boat moves forwards. A swimmer pushes water backwards with her hands and feet and moves forwards. A person who jumps forward out of a boat sends the boat backwards.
Jumping and running. When you run or jump, your feet push the Earth backwards (or down) and the Earth pushes you forwards (or up). The Earth gains an equal and opposite momentum, but its mass is so huge that its change in velocity is far too small to notice.

Figure 3.3: Before-and-after diagrams make momentum problems clear. Draw both, label every mass and velocity, and choose a positive direction.

Worked example 1: the recoil of a rifle

A 4.0 kg rifle fires a 0.010 kg bullet at $400\ \mathrm{m\,s^{-1}}$ . Find the recoil velocity of the rifle (a) if it is free to move; (b) if it is pressed firmly against a 70 kg shooter.

Before firing, the total momentum is zero. Take the bullet's direction as positive.

\text{(a)}\quad 0 = 0.010 \times 400 + 4.0\,V \quad\Rightarrow\quad V = -\frac{4.0}{4.0} = -1.0\ \mathrm{m\,s^{-1}}

\text{(b)}\quad 0 = 0.010 \times 400 + (4.0 + 70)\,V \quad\Rightarrow\quad V = -\frac{4.0}{74} = -0.054\ \mathrm{m\,s^{-1}}

The minus sign shows that the rifle moves backwards. Holding it firmly makes the recoil speed about 18 times smaller.

Worked example 2: a spacecraft fires its engine

A spacecraft of total mass 1000 kg is at rest in deep space. Its engine pushes out 10 kg of gas at $2000\ \mathrm{m\,s^{-1}}$ backwards. Find the spacecraft's speed afterwards.

0 = 10 \times (-2000) + (1000 - 10)\,V \quad\Rightarrow\quad V = \frac{20\,000}{990} = 20\ \mathrm{m\,s^{-1}}\ \text{forwards}

A real rocket pushes gas out continuously, so it keeps gaining speed for as long as the engine runs.

Worked example 3: jumping out of a boat

A 50 kg pupil stands at the front of a 100 kg boat at rest on a lake (Figure 3.3). She jumps horizontally forwards onto a jetty at $2.0\ \mathrm{m\,s^{-1}}$ . Find the velocity of the boat just afterwards (ignore the water's resistance).

0 = 50 \times 2.0 + 100\,v \quad\Rightarrow\quad v = -\frac{100}{100} = -1.0\ \mathrm{m\,s^{-1}}

The boat moves backwards at $1.0\ \mathrm{m\,s^{-1}}$ . If she does not allow for this, she may fall short of the jetty.

Worked example 4: does a runner move the Earth?

A 60 kg sprinter starts from rest and reaches $3.0\ \mathrm{m\,s^{-1}}$ in her first stride. The Earth's mass is $6.0 \times 10^{24}\ \text{kg}$ . Find the Earth's change in velocity.

60 \times 3.0 + (6.0 \times 10^{24})\,V = 0 \quad\Rightarrow\quad V = -\frac{180}{6.0 \times 10^{24}} = -3.0 \times 10^{-23}\ \mathrm{m\,s^{-1}}

The Earth does recoil, but far too little to ever be measured.

Warning

Signs, and the whole system. Momentum is conserved for the whole system, not for each body alone, and only if no outside resultant force acts. Always include every body that takes part, choose a positive direction, and give each velocity its sign. A common error is to write $0.010 \times 400 = 4.0 \times V$ and report the recoil as $+1.0\ \mathrm{m\,s^{-1}}$ , forwards.

Tip

Before = after. Write the total momentum before on the left and after on the right, then solve. If everything starts at rest, the left-hand side is zero and the parts must fly apart in opposite directions.

Try it: in PhET: Collision Lab (version 1.1.20), open the Explore 1D screen, show the momentum values, and check that the total stays the same in every collision.

GCSE Physics: Momentum, part 1, conservation of momentum (Cognito). Watch the before-and-after method and how the signs of the velocities are handled.

✦

Practice

State the principle of conservation of linear momentum. Under what condition does it hold?
Use Newton's third law to explain why two bodies that push each other apart gain equal and opposite momenta.
A 1.5 kg toy cannon at rest fires a 0.025 kg ball at $12\ \mathrm{m\,s^{-1}}$ . Find the recoil velocity of the cannon.
A 70 kg person standing still on a smooth, slippery floor throws a 5.0 kg bag forwards at $4.0\ \mathrm{m\,s^{-1}}$ . How fast does the person slide backwards?
A 40 kg girl jumps from a 60 kg canoe at rest, with a horizontal velocity of $3.0\ \mathrm{m\,s^{-1}}$ . Find the velocity of the canoe.
A 5.0 kg object at rest explodes into two pieces. The 2.0 kg piece moves at $6.0\ \mathrm{m\,s^{-1}}$ to the east. Find the velocity of the other piece.
Explain why a rocket can speed up in space, where there is no air to push against.
When you jump up from the ground, the Earth moves the other way. Why can nobody ever notice this?

Show answersHide answers

In a closed system (no resultant external force), the total momentum before an interaction equals the total momentum after it.
Each body pushes the other with an equal and opposite force, for the same time, so they receive equal and opposite impulses and therefore equal and opposite changes in momentum.
$0 = 0.025 \times 12 + 1.5\,V$ , so $V = -0.20\ \mathrm{m\,s^{-1}}$ , that is $0.20\ \mathrm{m\,s^{-1}}$ backwards.
$0 = 5.0 \times 4.0 + 70\,V$ , so $V = -0.29\ \mathrm{m\,s^{-1}}$ : $0.29\ \mathrm{m\,s^{-1}}$ backwards.
$0 = 40 \times 3.0 + 60\,v$ , so $v = -2.0\ \mathrm{m\,s^{-1}}$ : $2.0\ \mathrm{m\,s^{-1}}$ in the opposite direction.
$0 = 2.0 \times 6.0 + 3.0\,v$ , so $v = -4.0\ \mathrm{m\,s^{-1}}$ : $4.0\ \mathrm{m\,s^{-1}}$ to the west.
The rocket pushes exhaust gas backwards, giving the gas backward momentum; the total momentum is conserved, so the rocket gains an equal forward momentum. It pushes on its own exhaust, not on the air.
The Earth receives the same momentum as you, but its mass is about $10^{23}$ times larger, so its change in velocity is about $10^{23}$ times smaller.

Key points

In a closed system the total momentum is constant: $m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2}$ .
It follows from Newton's third law: interacting bodies receive equal and opposite impulses.
Bodies that start at rest and push apart move off with equal and opposite momenta: gun recoil, rockets, jumping from a boat.
Always draw before-and-after diagrams and use signs for directions.

Sources and credits

8.3: Conservation of Momentum and 8.7: Introduction to Rocket Propulsion, from College Physics (1st edition) by Paul Peter Urone and Roger Hinrichs, OpenStax, hosted by LibreTexts, licensed under CC BY 4.0. Consulted for the physics; no text was copied.
Photo: Fisherman riding a traditional boat.jpg by Isma250, Wikimedia Commons, licensed under CC BY-SA 4.0.
Collision Lab (version 1.1.20). Simulation by PhET Interactive Simulations, University of Colorado Boulder, licensed under CC BY 4.0 (https://phet.colorado.edu)
Video: GCSE Physics - Momentum (Part 1 of 2) | Conservation of Momentum, Cognito, embedded from YouTube.

Written by Ganzaa: the lesson text, the trolley experiment and its illustrative results, the diagrams, worked examples and practice exercises.

1Kick-back

2Experiment 3: an "explosion" between two trolleys

3Explanation: why momentum is conserved

4Applications: recoil, rockets, boats and running

Kick-back

Experiment 3: an "explosion" between two trolleys

Explanation: why momentum is conserved

Applications: recoil, rockets, boats and running