Elastic and Inelastic Collisions – Conservation of Linear Momentum and Collision (Senior 4 Physics) | Ganzaa

Conservation of Linear Momentum and Collision

Lessons

Senior 4 · Physics

Conservation of Linear Momentum and Collision

4 lessons

Lesson 4 of 4

Elastic and Inelastic Collisions

25 min

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Lift one steel ball at the end of a Newton's cradle and let it go: it stops dead, and the ball at the far end flies out at almost the same speed. A cue ball can stop completely when it hits another ball head-on. But when a lorry runs into a car, the two crumple and move off together. Momentum is conserved in all three; what differs is what happens to the kinetic energy. This lesson sorts collisions by that difference and shows how to solve each kind.

Types of collision

In every collision between free bodies, momentum is conserved (lesson 3). Kinetic energy may or may not be:

Elastic collision: total kinetic energy is the same before and after. Collisions between gas molecules, and very nearly between steel or billiard balls, are elastic.
Inelastic collision: some kinetic energy becomes internal energy, sound or permanent deformation; the bodies separate.
Perfectly (completely) inelastic collision: the bodies stick together and move with a common velocity. This loses the largest amount of kinetic energy that momentum conservation allows.

Figure 4.1: Momentum is the same before and after in all three; kinetic energy is conserved only in the elastic collision.

Definition

An elastic collision is one in which total kinetic energy is conserved; in an inelastic collision it is not. In a perfectly inelastic collision the bodies move off together. The coefficient of restitution is $e = \frac{\text{relative speed of separation}}{\text{relative speed of approach}} = \frac{v_2 - v_1}{u_1 - u_2}$ , with $e = 1$ for elastic and $e = 0$ for perfectly inelastic collisions. Example: a ball dropped from 1.80 m that bounces to 1.25 m has $e = \sqrt{\frac{1.25}{1.80}} = 0.83$ .

Elastic collisions in one dimension

Body 1 (mass $m_{1}$ , velocity $u_{1}$ ) hits body 2 (mass $m_{2}$ , velocity $u_{2}$ ) head-on; afterwards their velocities are $v_{1}$ and $v_{2}$ . Write the two conservation laws with each body's terms grouped:

m_1(u_1 - v_1) = m_2(v_2 - u_2) \qquad (\text{momentum})

m_1(u_1^2 - v_1^2) = m_2(v_2^2 - u_2^2) \qquad (\text{kinetic energy, after cancelling } \tfrac{1}{2})

Factorise the second as $m_{1} (u_{1} - v_{1}) (u_{1} + v_{1}) = m_{2} (v_{2} - u_{2}) (v_{2} + u_{2})$ and divide by the first:

u_1 + v_1 = u_2 + v_2 \quad\Rightarrow\quad u_1 - u_2 = v_2 - v_1

In an elastic collision the relative speed of approach equals the relative speed of separation. Solving with the momentum equation gives

v_1 = \frac{(m_1 - m_2)u_1 + 2m_2u_2}{m_1 + m_2}, \qquad v_2 = \frac{(m_2 - m_1)u_2 + 2m_1u_1}{m_1 + m_2}

Three special cases with body 2 at rest ( $u_{2} = 0$ ):

Masses	Result	Example
$m_{1} = m_{2}$	$v_{1} = 0$ , $v_{2} = u_{1}$ : the velocities are exchanged	Newton's cradle; a head-on billiard shot
$m_1 \gg m_2$	$v_1 \approx u_1$ , $v_2 \approx 2u_1$	a bat hitting a ball
$m_1 \ll m_2$	$v_1 \approx -u_1$ , $v_2 \approx 0$	a ball bouncing off a wall

Worked example 1: an elastic collision of trolleys

A 2.0 kg trolley moving at $3.0\,\mathrm{m\,s^{-1}}$ collides elastically head-on with a 1.0 kg trolley at rest. Find their velocities afterwards and check that kinetic energy is conserved.

v_1 = \frac{(2.0 - 1.0) \times 3.0}{3.0} = 1.0\,\mathrm{m\,s^{-1}}, \qquad v_2 = \frac{2 \times 2.0 \times 3.0}{3.0} = 4.0\,\mathrm{m\,s^{-1}}

Check: $E_{k}$ before $= \tfrac{1}{2} \times 2.0 \times 3.0^2 = 9.0\,\mathrm{J}$ ; after $= \tfrac{1}{2} \times 2.0 \times 1.0^2 + \tfrac{1}{2} \times 1.0 \times 4.0^2 = 1.0 + 8.0 = 9.0\,\mathrm{J}$ . Relative speeds: approach 3.0, separation $4.0 - 1.0 = 3.0\,\mathrm{m\,s^{-1}}$ .

Inelastic collisions

For a perfectly inelastic collision, there is one final velocity $v$ :

m_1u_1 + m_2u_2 = (m_1 + m_2)v \quad\Rightarrow\quad v = \frac{m_1u_1 + m_2u_2}{m_1 + m_2}

For a collision that is neither elastic nor perfectly inelastic, use momentum conservation together with the coefficient of restitution, $v_{2} - v_{1} = e (u_{1} - u_{2})$ . For a ball dropped from height $h_{1}$ onto a fixed floor that rebounds to $h_{2}$ , the speeds are $\sqrt{2gh_1}$ and $\sqrt{2gh_2}$ , so $e = \sqrt{\frac{h_2}{h_1}}$ .

Worked example 2: a car crash

A 1200 kg car travelling at $15\,\mathrm{m\,s^{-1}}$ runs into the back of a stationary 800 kg car, and the two lock together. Find their common velocity and the kinetic energy lost.

v = \frac{1200 \times 15}{1200 + 800} = 9.0\,\mathrm{m\,s^{-1}}

E_k\ \text{before} = \tfrac{1}{2} \times 1200 \times 15^2 = 135\,\mathrm{kJ}, \qquad E_k\ \text{after} = \tfrac{1}{2} \times 2000 \times 9.0^2 = 81\,\mathrm{kJ}

54 kJ (40%) of the kinetic energy goes into crushing metal, heat and sound. Crumple zones are designed to absorb this energy so that the passengers do not.

The ballistic pendulum

Figure 4.2: A ballistic pendulum measures the speed of a fast projectile.

The speed of a bullet is too high to time directly. Fire it into a heavy hanging block (Figure 4.2). The impact is perfectly inelastic and very short, so momentum is conserved during it: $m u = (m + M) V$ . Afterwards the block and bullet swing up, and mechanical energy is conserved during the swing: $\tfrac{1}{2}(m + M)V^2 = (m + M)gh$ . Combining the two stages:

V = \sqrt{2gh}, \qquad u = \frac{m + M}{m}\sqrt{2gh}

You must not use energy conservation across the impact itself: most of the bullet's kinetic energy becomes internal energy in the wood.

Worked example 3: measuring a bullet's speed

A 10 g bullet is fired into a 2.0 kg block, which rises 5.0 cm. Find the bullet's speed and the fraction of its kinetic energy lost in the impact.

V = \sqrt{2 \times 9.8 \times 0.050} = 0.99\,\mathrm{m\,s^{-1}}, \qquad u = \frac{2.01}{0.010} \times 0.99 = 2.0 \times 10^{2}\,\mathrm{m\,s^{-1}}

Kinetic energy after the impact divided by before is $\frac{\frac{1}{2}(m + M)V^2}{\frac{1}{2}mu^2} = \frac{m}{m + M} = \frac{0.010}{2.01} = 0.005$ , so 99.5% of the bullet's kinetic energy is lost in the impact.

Collisions in two dimensions

When bodies collide off-centre they move off at angles. Momentum is a vector, so it is conserved separately along any two perpendicular directions (Figure 4.3).

Figure 4.3: A glancing collision between identical balls.

A useful result: in an elastic glancing collision between equal masses, one initially at rest, the two move off at right angles. (Momentum: $\vec{u} = \vec{v}_1 + \vec{v}_2$ ; energy: $u^{2} = v_{1}^{2} + v_{2}^{2}$ . Both can be true only if $\vec{v}_1$ and $\vec{v}_2$ are perpendicular, by Pythagoras.) Snooker players use this every time they "cut" a ball into a pocket.

Worked example 4: a glancing collision

In Figure 4.3, ball 1 (moving at $2.0\,\mathrm{m\,s^{-1}}$ ) hits an identical ball at rest. Afterwards ball 1 moves at 30° to its original direction and ball 2 at 60° on the other side. Find their speeds.

Along the original direction: $2.0 = v_1\cos 30^\circ + v_2\cos 60^\circ$ . At right angles: $0 = v_1\sin 30^\circ - v_2\sin 60^\circ$ , so $v_1 = \sqrt{3}\,v_2$ .

2.0 = \sqrt{3}v_2 \times 0.866 + 0.500\,v_2 = 2.00\,v_2 \;\Rightarrow\; v_2 = 1.00\,\mathrm{m\,s^{-1}},\quad v_1 = 1.73\,\mathrm{m\,s^{-1}}

Check: $v_{1}^{2} + v_{2}^{2} = 3.0 + 1.0 = 4.0 = u^{2}$ , so kinetic energy is conserved: the collision is elastic.

Experiment 4.1: how bouncy is a ball?

Apparatus: balls of different kinds (tennis, table-tennis, rubber, clay), a metre rule fixed upright, a phone to film the bounce.

Drop each ball from a measured height $h_1 = 1.00\,\mathrm{m}$ onto a hard floor, filming it beside the rule. From the video, read the height $h_{2}$ of the first bounce.
Repeat three times and average. Calculate $e = \sqrt{\frac{h_2}{h_1}}$ and the fraction of kinetic energy kept, $\frac{h_2}{h_1}$ .
Rank the balls from most to least elastic. A clay ball that does not bounce at all has $e = 0$ .

A Newton's cradle: five steel balls hanging in a row. When one end ball swings in and strikes the row, the ball at the other end swings out with almost the same speed, because nearly elastic collisions between equal masses exchange their velocities. Photo: AntHolnes, CC BY-SA 3.0, via Wikimedia Commons.

Warning

Common mistakes. (1) Using conservation of kinetic energy for an inelastic collision. (2) Using energy conservation across the impact of a ballistic pendulum: use momentum for the impact and energy only for the swing. (3) Forgetting directions: a body that rebounds has a negative velocity. (4) Thinking "perfectly inelastic" means all the kinetic energy is lost: the joined bodies still move unless the total momentum was zero.

Tip

For a head-on elastic collision, use two linear equations instead of the energy equation: momentum conservation and "speed of approach = speed of separation", $u_{1} - u_{2} = v_{2} - v_{1}$ . No squares, fewer mistakes.

Collisions demo: two carts (Physics Demos). Watch two carts collide elastically (bouncing apart) and perfectly inelastically (sticking together), and compare their speeds before and after.

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Practice

State which quantities are conserved in (a) an elastic collision, (b) a perfectly inelastic collision.
A 0.50 kg ball moving at $4.0\,\mathrm{m\,s^{-1}}$ collides elastically head-on with a 1.5 kg ball at rest. Find both velocities afterwards.
A 3000 kg truck moving east at $10\,\mathrm{m\,s^{-1}}$ collides head-on with a 1000 kg car moving west at $20\,\mathrm{m\,s^{-1}}$ . They lock together. Find their common velocity and the kinetic energy lost.
A 1.0 kg trolley at $6.0\,\mathrm{m\,s^{-1}}$ hits a 2.0 kg trolley at rest; the coefficient of restitution is 0.50. Find both velocities afterwards.
A ball dropped from 2.0 m rebounds to 1.2 m. Find the coefficient of restitution and the percentage of kinetic energy lost in the bounce.
A 20 g bullet hits a 3.0 kg ballistic pendulum, which rises 8.0 cm. Find the bullet's speed.
An investigator finds two cars locked together after a crash and claims she can find the speed of one car before the crash from the skid marks after it. Evaluate this method: which conservation laws can she use, and for which stages?

Show answersHide answers

(a) momentum and kinetic energy (and total energy); (b) momentum (and total energy, but not kinetic energy).
$v_1 = \frac{(0.50 - 1.5) \times 4.0}{2.0} = -2.0\,\mathrm{m\,s^{-1}}$ (it bounces back); $v_2 = \frac{2 \times 0.50 \times 4.0}{2.0} = 2.0\,\mathrm{m\,s^{-1}}$ .
$v = \frac{3000 \times 10 - 1000 \times 20}{4000} = 2.5\,\mathrm{m\,s^{-1}}$ east. $E_{k}$ before $= 1.5 \times 10^{5} + 2.0 \times 10^{5} = 3.5 \times 10^{5}\,\mathrm{J}$ ; after $= \tfrac{1}{2} \times 4000 \times 2.5^2 = 1.25 \times 10^{4}\,\mathrm{J}$ ; lost $3.4 \times 10^{5}\,\mathrm{J}$ .
Momentum: $6.0 = v_{1} + 2.0 v_{2}$ ; restitution: $v_2 - v_1 = 0.50 \times 6.0 = 3.0$ . So $3.0 v_{2} = 9.0$ : $v_2 = 3.0\,\mathrm{m\,s^{-1}}$ , $v_{1} = 0$ .
$e = \sqrt{\frac{1.2}{2.0}} = 0.77$ ; kinetic energy kept $= \frac{1.2}{2.0} = 60\%$ , so 40% lost.
$V = \sqrt{2 \times 9.8 \times 0.080} = 1.25\,\mathrm{m\,s^{-1}}$ ; $u = \frac{3.02}{0.020} \times 1.25 = 1.9 \times 10^{2}\,\mathrm{m\,s^{-1}}$ .
The method is sound if done in two stages. After the crash, the work done by friction during the skid equals the kinetic energy of the joined cars (work–energy theorem), giving their common speed from the skid length and the friction force. For the crash itself, momentum is conserved (not kinetic energy, since the collision is perfectly inelastic), so knowing both masses and the other car's initial velocity gives the first car's speed. She needs the friction coefficient of the road and the other car's speed; any error there carries into the result.

Quick check

Two identical balls collide head-on elastically; ball 2 was at rest. What happens?

Key points

Momentum is conserved in all collisions; kinetic energy only in elastic ones. Perfectly inelastic: bodies stick, $v = \frac{m_1u_1 + m_2u_2}{m_1 + m_2}$ .
Elastic, 1D: speed of approach = speed of separation; equal masses exchange velocities.
Coefficient of restitution $e = \frac{v_2 - v_1}{u_1 - u_2}$ ; for a bounce $e = \sqrt{\frac{h_2}{h_1}}$ .
Ballistic pendulum: momentum for the impact, energy for the swing; in 2D, conserve each momentum component.

1Click, click, click

2Types of collision

3Elastic collisions in one dimension

4Inelastic collisions

5The ballistic pendulum

6Collisions in two dimensions

7Experiment 4.1: how bouncy is a ball?