Conservation of Linear Momentum and Collision

Lessons

Senior 4 · Physics

Conservation of Linear Momentum and Collision

4 lessons

Lesson 3 of 4

Momentum, Impulse and the Conservation of Linear Momentum

25 min

Why a goalkeeper pulls the ball in

A good goalkeeper does not stop a hard shot with stiff arms; she draws the ball in towards her body as she catches it. A rider who falls from a moto is far safer in a helmet with a crushable lining. A modern car is built to crumple at the front. Each of these lengthens the time over which something is stopped. To see why that matters, you need a quantity that measures "how much motion" a body has, its momentum, and a law that says how forces change it.

Momentum and Newton's second law

The linear momentum of a body is the product of its mass and its velocity:

\vec{p} = m\vec{v} \qquad (\mathrm{kg\,m\,s^{-1}})

It is a vector, in the direction of the velocity. Newton actually stated his second law in terms of momentum: the resultant force on a body equals its rate of change of momentum,

\vec{F} = \frac{\Delta\vec{p}}{\Delta t}

For a body of constant mass, $\frac{\Delta(m\vec{v})}{\Delta t} = m\frac{\Delta\vec{v}}{\Delta t} = m\vec{a}$ , the familiar $F = m a$ . But the momentum form also works when mass flows, for example for a jet of water or the exhaust of a rocket.

Definition

The linear momentum of a body is $\vec{p} = m\vec{v}$ ; unit $\mathrm{kg\,m\,s^{-1}}$ . The impulse of a force is the product of the force and the time for which it acts, $\vec{J} = \vec{F}\Delta t$ ; unit N s (equal to $\mathrm{kg\,m\,s^{-1}}$ ). For a varying force, the impulse is the area under the force–time graph. The impulse–momentum theorem: impulse = change in momentum. Example: a 0.45 kg ball kicked from rest to $8.0\,\mathrm{m\,s^{-1}}$ receives an impulse of 3.6 N s.

Impulse

Rearranging $F = \frac{\Delta p}{\Delta t}$ gives

F\Delta t = \Delta p = mv - mu

The left side, force times time, is the impulse. If the force varies, the impulse is the area under the force–time graph. For a given change of momentum, the impulse is fixed; so the longer the collision lasts, the smaller the average force (Figure 3.1).

Figure 3.1: Two ways of stopping the same object. The areas (impulses) are equal; the forces are very different.

This explains many safety designs: crumple zones and airbags lengthen the time a passenger takes to stop, seat belts stretch slightly, helmet linings crush, gymnasts land on thick mats, and you bend your knees when you jump down. In sport the reverse is used: a long "follow-through" keeps the racket, bat or foot in contact with the ball for longer, giving a larger impulse and so a faster ball.

Worked example 1: a ball bouncing off a wall

A 0.15 kg ball hits a wall at $20\,\mathrm{m\,s^{-1}}$ and rebounds along the same line at $15\,\mathrm{m\,s^{-1}}$ . The contact lasts 0.010 s. Find the impulse and the average force on the ball.

Take the direction away from the wall as positive: $u = -20\,\mathrm{m\,s^{-1}}$ , $v = +15\,\mathrm{m\,s^{-1}}$ .

\Delta p = m(v - u) = 0.15 \times (15 - (-20)) = 5.25\,\mathrm{N\,s}, \qquad F = \frac{5.25}{0.010} = 525\,\mathrm{N} \approx 5.3 \times 10^{2}\,\mathrm{N}

The force is away from the wall. Forgetting the sign change would give only $0.15 \times 5 = 0.75\,\mathrm{N\,s}$ , a common mistake.

Worked example 2: why airbags save lives

In a crash, a 70 kg passenger travelling at $15\,\mathrm{m\,s^{-1}}$ (54 km/h) is brought to rest. Compare the average force if (a) her head hits the dashboard and stops in 0.010 s, (b) a seat belt and airbag stop her in 0.10 s.

\Delta p = 70 \times 15 = 1050\,\mathrm{N\,s}, \qquad \text{(a)}\;\; F = \frac{1050}{0.010} = 1.1 \times 10^{5}\,\mathrm{N}, \qquad \text{(b)}\;\; F = \frac{1050}{0.10} = 1.1 \times 10^{4}\,\mathrm{N}

Spreading the same impulse over ten times as long cuts the force tenfold, and the airbag also spreads it over a larger area of the body.

Worked example 3: a water jet

A hose sends 2.0 kg of water each second horizontally at $15\,\mathrm{m\,s^{-1}}$ against a wall. The water stops and runs down the wall. Find the force on the wall.

F = \frac{\Delta p}{\Delta t} = \frac{\Delta m}{\Delta t}\,v = 2.0 \times 15 = 30\,\mathrm{N}

Here the velocity of each bit of water changes, and new water keeps arriving: $F = m a$ is awkward, but $F = \frac{\Delta p}{\Delta t}$ gives the answer in one line.

Conservation of linear momentum

When two bodies A and B interact (collide, push apart, pull on each other), Newton's third law says that at every instant the force of A on B is equal and opposite to the force of B on A: $\vec{F}_{AB} = -\vec{F}_{BA}$ . The forces act for the same time $\Delta t$ , so the impulses are equal and opposite, and so are the changes of momentum:

\Delta\vec{p}_A = -\Delta\vec{p}_B \quad\Rightarrow\quad \Delta(\vec{p}_A + \vec{p}_B) = 0

The principle of conservation of linear momentum: in a closed system on which no external resultant force acts, the total momentum stays constant:

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

The principle holds for every kind of collision and explosion, however complicated the forces inside, as long as external forces (friction, for example) are negligible during the short interaction. In two dimensions it holds separately for the $x$ and $y$ components.

Explosions and recoil

If a system starts at rest, its total momentum is zero, and after an "explosion" the pieces move off with momenta that add to zero. A gun recoils when it fires; a rocket moves forward because it throws exhaust gas backwards; a person stepping off a canoe pushes it away from the shore.

Experiment 3.1: an explosion of two trolleys

Apparatus: two trolleys of different masses, one with a spring-loaded plunger, a level track (or smooth floor), two wooden stops, a metre rule.

Figure 3.2: Two trolleys pushed apart by a spring. The stops are placed so that both trolleys hit them at the same moment.

Place the trolleys together in the middle with the spring compressed. Release it and watch which stop is hit first.
Move the stops until both trolleys hit them at the same moment (you hear a single click). Measure the distances $d_{1}$ and $d_{2}$ travelled.
Both trolleys move for the same time, so their speeds are proportional to the distances. Compare $m_{1} d_{1}$ with $m_{2} d_{2}$ .

With 1.0 kg and 2.0 kg trolleys you find about 60 cm and 30 cm: $1.0 \times 60 = 2.0 \times 30$ . The momenta are equal and opposite, adding to zero, as they were before release.

Definition

The principle of conservation of linear momentum: the total linear momentum of a system of bodies remains constant provided no resultant external force acts on the system. Example: a 4.0 kg rifle that fires a 10 g bullet at $400\,\mathrm{m\,s^{-1}}$ recoils at $1.0\,\mathrm{m\,s^{-1}}$ .

Worked example 4: recoil of a rifle

A 4.0 kg rifle fires a 10 g bullet at $400\,\mathrm{m\,s^{-1}}$ . Find the recoil velocity of the rifle, and compare the kinetic energies of the bullet and the rifle.

0 = 0.010 \times 400 + 4.0\,v \;\Rightarrow\; v = -1.0\,\mathrm{m\,s^{-1}}

The rifle moves back at $1.0\,\mathrm{m\,s^{-1}}$ . Kinetic energies: bullet $\tfrac{1}{2} \times 0.010 \times 400^2 = 800\,\mathrm{J}$ ; rifle $\tfrac{1}{2} \times 4.0 \times 1.0^2 = 2.0\,\mathrm{J}$ . Momentum is shared equally (and oppositely), but almost all the kinetic energy goes to the lighter body.

Brazilian goalkeeper Gilmar catching the ball during the 1958 World Cup final. Drawing the ball in with the arms lengthens the time of the catch, so the same change of momentum needs a smaller force on the hands. Photo: Scanpix, public domain, via Wikimedia Commons.

Warning

Common mistakes. (1) Treating momentum as a scalar: a rebound reverses the sign of the velocity, so $\Delta p = m(v - u)$ with signs. (2) Applying conservation of momentum when an external force acts for a long time (for example friction during a long slide); it applies to the short collision itself. (3) Thinking kinetic energy is always conserved when momentum is: it is not (lesson 4). (4) Writing the unit of impulse as N s⁻¹: it is N s.

Tip

Before any momentum calculation, draw an arrow for the positive direction and write each velocity with its sign. Most wrong answers in this topic are sign errors.

Impulse and change in momentum (vt.physics). Watch how F Δt = Δp explains seat belts and crumple zones, and how the impulse is found from the area under a force–time graph.

✦

Practice

Find the momentum of a 1500 kg car moving at $20\,\mathrm{m\,s^{-1}}$ and of a 0.057 kg tennis ball served at $50\,\mathrm{m\,s^{-1}}$ .
A force of 200 N acts on a 5.0 kg body at rest for 0.30 s. Find the impulse and the final speed.
A 60 g egg falls from 1.0 m onto a tiled floor and stops in 2.0 ms. Find the average force. What is the force if it lands on a cushion and stops in 0.10 s?
A force on a 0.45 kg football rises steadily from 0 to 900 N in 4.0 ms and falls steadily to 0 in the next 4.0 ms. Find the impulse and the ball's speed if it started at rest.
A 50 kg skater at rest pushes a 70 kg skater, also at rest, who moves off at $1.5\,\mathrm{m\,s^{-1}}$ . Find the velocity of the first skater.
A fire hose delivers 8.0 kg of water per second at $25\,\mathrm{m\,s^{-1}}$ . Find the force the firefighter must exert to hold it.
A student says: "When I jump off a wall and land with stiff legs, the ground's impulse on me is larger than when I bend my knees." Evaluate this statement.

Show answersHide answers

$3.0 \times 10^{4}\,\mathrm{kg\,m\,s^{-1}}$ ; $2.9\,\mathrm{kg\,m\,s^{-1}}$ .
$J = 200 \times 0.30 = 60\,\mathrm{N\,s}$ ; $v = \frac{60}{5.0} = 12\,\mathrm{m\,s^{-1}}$ .
$v = \sqrt{2 \times 9.8 \times 1.0} = 4.43\,\mathrm{m\,s^{-1}}$ , $\Delta p = 0.060 \times 4.43 = 0.266\,\mathrm{N\,s}$ . Tile: $F = \frac{0.266}{0.0020} = 1.3 \times 10^{2}\,\mathrm{N}$ ; cushion: $F = \frac{0.266}{0.10} = 2.7\,\mathrm{N}$ .
Area of the triangle $= \tfrac{1}{2} \times 900 \times 0.0080 = 3.6\,\mathrm{N\,s}$ ; $v = \frac{3.6}{0.45} = 8.0\,\mathrm{m\,s^{-1}}$ .
$0 = 50v + 70 \times 1.5$ , so $v = -2.1\,\mathrm{m\,s^{-1}}$ : 2.1 m s⁻¹ in the opposite direction.
$F = 8.0 \times 25 = 200\,\mathrm{N}$ (backwards along the hose, which the firefighter must resist).
Wrong. In both landings the change of momentum is the same (from $m v$ to zero), so the impulse is the same. With stiff legs the stop takes less time, so the force is much larger, which is what causes injury.

Quick check

Why does an airbag reduce injury in a crash?

Key points

Momentum $\vec{p} = m\vec{v}$ is a vector; $\vec{F} = \frac{\Delta\vec{p}}{\Delta t}$ .
Impulse $F\Delta t = \Delta p$ = area under the force–time graph; a longer stopping time means a smaller force.
With no external resultant force, total momentum is conserved: $m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2}$ . It follows from Newton's third law.
In an explosion from rest the momenta add to zero; the lighter body gets most of the kinetic energy.

Sources and credits

8.1: Linear Momentum and Force, 8.2: Impulse and 8.3: Conservation of Momentum, from College Physics (1st edition) by Paul Peter Urone and Roger Hinrichs, OpenStax, hosted by LibreTexts, licensed under CC BY 4.0. Adapted by Ganzaa.
Photo: Crash-test-with-airbag-and-safty-belt.jpg by Transport for NSW, Wikimedia Commons, licensed under CC BY-SA 4.0.
Photo: Gilmar catching a ball.jpg by Scanpix, Wikimedia Commons, public domain.
Video: Impulse and Change in Momentum - GCSE & A Level Physics, vt.physics, embedded from YouTube.

The derivations, the two diagrams, the experiment, the worked examples and the practice exercises are written by Ganzaa.

1Why a goalkeeper pulls the ball in

2Momentum and Newton's second law

3Impulse

4Conservation of linear momentum