Linear Momentum, Impulse and Collisions

Lessons

Senior 3 · Physics

Linear Momentum, Impulse and Collisions

4 lessons

Lesson 4 of 4

Elastic and Inelastic Collisions

25 min

Two ways to collide

In a game of marbles, a marble rolls straight into another one that is at rest. With a sharp click, the first marble stops dead and the second rolls away at almost the same speed. Now think of a car that runs into the back of a stationary minibus: the two vehicles lock together, crumple, and slide on a short way as one. Momentum is conserved in both collisions, so what is different? The answer is the kinetic energy.

Experiment 4: bouncing and sticking trolleys

You need: two trolleys of equal mass (1.0 kg each), a pair of springy bumpers (two magnets facing each other, or a stretched rubber band across the front of one trolley), two Velcro pads or a lump of plasticine, a level bench, a metre rule and a phone to film the motion.

Fit the springy bumpers. Put trolley B at rest in the middle of the bench.
Push trolley A towards B and let go. Film both trolleys beside the metre rule and find their speeds before and after the collision.
Replace the bumpers with Velcro pads so that the trolleys stick together. Repeat.
For each run, calculate the total momentum and the total kinetic energy $\frac{1}{2}mv^2$ before and after.

Figure 4.1: Illustrative results for the two collisions. Momentum values are in $\mathrm{kg\,m\,s^{-1}}$ .

Collision	Total $p$ before	Total $p$ after	Total KE before	Total KE after
springy bumpers	0.60	0.59	0.18 J	0.17 J
Velcro pads (stick)	0.60	0.58	0.18 J	0.084 J

These are illustrative results. In both collisions the total momentum is the same before and after, within the experiment's error. The kinetic energy is almost unchanged when the trolleys bounce apart, but half of it disappears when they stick together. It is not destroyed: it becomes heat in the Velcro, sound and a little deformation.

Important

Safety. Push the trolleys gently; they should never leave the bench. Put soft stops at both ends. Keep strong magnets away from phones, bank cards and pacemakers.

Definition

A collision is a short, strong interaction between bodies. In a head-on (one-dimensional) collision, the bodies move along the same straight line before and after.

In an elastic collision, both the total momentum and the total kinetic energy are conserved. Example: two steel balls or two snooker balls colliding (almost elastic).

In an inelastic collision, the total momentum is conserved but some kinetic energy is changed into other forms (heat, sound, deformation). When the bodies stick together and move with a common velocity, the collision is perfectly (completely) inelastic, and the greatest possible share of kinetic energy is lost. Example: a car running into another car and locking to it.

How to analyse a collision

Draw a before diagram and an after diagram. Label every mass and velocity.
Choose a positive direction and give every velocity a sign.
Write conservation of momentum: $m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2}$ . For bodies that stick: $m_{1} u_{1} + m_{2} u_{2} = (m_{1} + m_{2}) V$ .
To decide whether the collision is elastic, compare the total kinetic energy $\sum \frac{1}{2}mv^2$ before and after.

	Elastic	Inelastic	Perfectly inelastic
Momentum	conserved	conserved	conserved
Kinetic energy	conserved	some lost	greatest possible loss
After the collision	bodies separate	bodies separate	bodies move together
Examples	steel balls, marbles, gas molecules (nearly)	a football kicked, a ball bouncing on the ground	a vehicle collision in which the vehicles lock together

Worked example 1: a vehicle collision

Figure 4.2: Before and after a perfectly inelastic vehicle collision.

A 1200 kg car moving at $15\ \mathrm{m\,s^{-1}}$ runs into the back of a stationary 2400 kg minibus, and the two lock together. Find (a) their common velocity just after the collision and (b) the kinetic energy lost.

(a) $1200 \times 15 + 2400 \times 0 = (1200 + 2400)\,V \Rightarrow V = \frac{18\,000}{3600} = 5.0\ \mathrm{m\,s^{-1}}$

(b) $\text{KE}_{\text{before}} = \frac{1}{2} \times 1200 \times 15^2 = 135\,000\ \text{J}$ ; $\text{KE}_{\text{after}} = \frac{1}{2} \times 3600 \times 5.0^2 = 45\,000\ \text{J}$ .

Kinetic energy lost $= 135\,000 - 45\,000 = 90\,000\ \text{J}$ , two thirds of the total. This energy goes into bending metal, breaking glass, heat and sound; the crumple zones are designed to absorb it slowly.

Worked example 2: is it elastic?

A 2.0 kg trolley moving at $3.0\ \mathrm{m\,s^{-1}}$ hits a 1.0 kg trolley at rest. Afterwards the 2.0 kg trolley moves at $1.0\ \mathrm{m\,s^{-1}}$ and the 1.0 kg trolley at $4.0\ \mathrm{m\,s^{-1}}$ , both in the original direction. Is momentum conserved? Is the collision elastic?

p_{\text{before}} = 2.0 \times 3.0 = 6.0\ \mathrm{kg\,m\,s^{-1}}

p_{\text{after}} = 2.0 \times 1.0 + 1.0 \times 4.0 = 6.0\ \mathrm{kg\,m\,s^{-1}}

\text{KE}_{\text{before}} = \frac{1}{2} \times 2.0 \times 3.0^2 = 9.0\ \text{J}

\text{KE}_{\text{after}} = \frac{1}{2} \times 2.0 \times 1.0^2 + \frac{1}{2} \times 1.0 \times 4.0^2 = 1.0 + 8.0 = 9.0\ \text{J}

Both are conserved, so the collision is elastic.

Worked example 3: a head-on collision from opposite directions

A 0.50 kg trolley moving at $4.0\ \mathrm{m\,s^{-1}}$ to the right meets a 1.0 kg trolley moving at $1.0\ \mathrm{m\,s^{-1}}$ to the left. They stick together. Find their common velocity.

Take right as positive: $u_2 = -1.0\ \mathrm{m\,s^{-1}}$ .

0.50 \times 4.0 + 1.0 \times (-1.0) = 1.5\,V \quad\Rightarrow\quad V = \frac{1.0}{1.5} = 0.67\ \mathrm{m\,s^{-1}}\ \text{to the right}

Kinetic energy falls from $4.0 + 0.5 = 4.5\ \text{J}$ to $\frac{1}{2} \times 1.5 \times 0.667^2 = 0.33\ \text{J}$ .

Bouncing balls

When a ball bounces on the ground, the ball and the Earth form a closed system, so momentum is conserved, but the Earth's recoil is far too small to see. The bounce is never perfectly elastic: the ball rises to a lower height each time. The height of the bounce tells you how much kinetic energy was kept, because at the top of each bounce all the kinetic energy has become gravitational potential energy $m g h$ .

For example, a ball dropped from 1.00 m that rebounds to 0.64 m hits the ground at $v = \sqrt{2gh} = \sqrt{2 \times 9.8 \times 1.00} = 4.4\ \mathrm{m\,s^{-1}}$ and leaves at $\sqrt{2 \times 9.8 \times 0.64} = 3.5\ \mathrm{m\,s^{-1}}$ . It keeps $\frac{0.64}{1.00} = 64\%$ of its kinetic energy in each bounce; the rest becomes heat and sound. A well-pumped football bounces higher than a soft one because less of its energy goes into squashing it.

Warning

Momentum is conserved in EVERY collision; kinetic energy is not. Do not use $\frac{1}{2}mv^2$ to find an unknown velocity unless you are told the collision is elastic. Start with momentum, and use the kinetic energy only to test whether the collision was elastic or to find the energy lost.

A desk Newton's cradle: identical steel balls hang in a row from pairs of threads. When one end ball is pulled aside and released, it stops when it strikes the row and the ball at the far end swings out, because the collisions between steel balls are nearly elastic. Photo: Nysam1 at English Wikipedia, public domain, via Wikimedia Commons

Tip

Stuck together? One velocity. If the bodies stick, there is only one unknown, $V$ , and momentum alone gives it. For two EQUAL masses in a head-on elastic collision with one at rest, they simply swap velocities: the moving one stops and the other moves off at the same speed, as in the marbles and the Newton's cradle.

Try it: in PhET: Collision Lab (version 1.1.20), set the elasticity slider to 100% and then to 0% (inelastic, "stick"), and compare the kinetic energy before and after.

Collisions: Crash Course Physics #10 (CrashCourse). Watch for the difference between elastic and inelastic collisions and how momentum is used in both.

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Practice

Define an elastic collision and a perfectly inelastic collision. Which quantity is conserved in both?
A 0.50 kg trolley moving at $2.0\ \mathrm{m\,s^{-1}}$ hits and sticks to a 1.5 kg trolley at rest. Find their common velocity and the kinetic energy lost.
A 3.0 kg ball moving at $4.0\ \mathrm{m\,s^{-1}}$ hits a 1.0 kg ball at rest. Afterwards the 3.0 kg ball moves on at $2.0\ \mathrm{m\,s^{-1}}$ . Find the velocity of the 1.0 kg ball, and show whether the collision is elastic.
A 1000 kg car moving east at $20\ \mathrm{m\,s^{-1}}$ collides head-on with a 1500 kg car moving west at $10\ \mathrm{m\,s^{-1}}$ . The cars lock together. Find their velocity just after the crash.
A ball is dropped from 2.0 m and bounces up to 1.5 m. Find its speed just before and just after the bounce, and the fraction of its kinetic energy kept ( $g = 9.8\ \mathrm{m\,s^{-2}}$ ).
A 0.010 kg bullet moving at $300\ \mathrm{m\,s^{-1}}$ embeds itself in a 2.0 kg wooden block at rest on ice. Find the velocity of the block afterwards and the percentage of kinetic energy kept.
A pupil says: "When a ball bounces on the floor, momentum is not conserved, because the ball changes direction." Explain her mistake.
Two identical marbles collide head-on; one was at rest. Describe what happens if the collision is (a) elastic; (b) perfectly inelastic.

Show answersHide answers

Elastic: total momentum and total kinetic energy both conserved. Perfectly inelastic: the bodies stick together; momentum conserved, the greatest possible loss of kinetic energy. Momentum is conserved in both.
$0.50 \times 2.0 = 2.0\,V \Rightarrow V = 0.50\ \mathrm{m\,s^{-1}}$ . KE before $= 1.0\ \text{J}$ , after $= \frac{1}{2} \times 2.0 \times 0.50^2 = 0.25\ \text{J}$ ; lost $0.75\ \text{J}$ .
$3.0 \times 4.0 = 3.0 \times 2.0 + 1.0\,v \Rightarrow v = 6.0\ \mathrm{m\,s^{-1}}$ . KE before $= 24\ \text{J}$ ; after $= 6.0 + 18 = 24\ \text{J}$ , so it is elastic.
East positive: $1000 \times 20 + 1500 \times (-10) = 2500\,V \Rightarrow V = \frac{5000}{2500} = 2.0\ \mathrm{m\,s^{-1}}$ east.
Before: $\sqrt{2 \times 9.8 \times 2.0} = 6.3\ \mathrm{m\,s^{-1}}$ ; after: $\sqrt{2 \times 9.8 \times 1.5} = 5.4\ \mathrm{m\,s^{-1}}$ ; fraction kept $= \frac{1.5}{2.0} = 0.75$ (75%).
$0.010 \times 300 = 2.01\,V \Rightarrow V = 1.5\ \mathrm{m\,s^{-1}}$ . KE before $= 450\ \text{J}$ , after $= \frac{1}{2} \times 2.01 \times 1.49^2 = 2.2\ \text{J}$ : only about 0.5% is kept.
The ball alone is not a closed system: the floor pushes on it. The ball's momentum changes, but the Earth gains an equal and opposite momentum, so the total momentum of ball + Earth is conserved.
(a) The moving marble stops and the other moves off with the first marble's velocity. (b) They move off together at half the first marble's speed, and half the kinetic energy is lost.

Key points

Momentum is conserved in all collisions in a closed system; kinetic energy is conserved only in elastic ones.
Perfectly inelastic: the bodies stick, $m_{1} u_{1} + m_{2} u_{2} = (m_{1} + m_{2}) V$ , and the most kinetic energy is lost.
Test for an elastic collision by comparing the total $\frac{1}{2}mv^2$ before and after.
Lost kinetic energy becomes heat, sound and deformation; in vehicles, crumple zones absorb it.

1Two ways to collide

2Experiment 4: bouncing and sticking trolleys

3How to analyse a collision

4Bouncing balls

Two ways to collide

Experiment 4: bouncing and sticking trolleys

How to analyse a collision

Bouncing balls