Why Two Equal Balls Fly Apart at Right Angles After a Glancing Hit

A tube-well motor and a hand pump can lift the same tank of water, but the motor does it far faster — that difference is power. And when a bat meets a ball or two vehicles collide, momentum and energy decide what happens next.

This lesson covers average and instantaneous power, elastic and perfectly inelastic collisions in one dimension, and oblique collisions in two dimensions.

**Power is the rate of doing work: average power is total work divided by total time, and instantaneous power is $P=\vec{F}\cdot \vec{v}$, measured in watts.**

$$P_{\text{av}}=\frac{W}{t}P=\frac{dW}{dt}=\vec{F}\cdot \vec{v}=Fv\cos \theta$$

- Unit: the watt, $1\text{ W}=1{\text{ J s}}^{-1}$; one horsepower is 746 W
- The kilowatt hour is a unit of energy, not power: $1\text{ kW h}=3.6\times 10^6$ J

Worked example 1 — a water pump. A motor lifts 500 kg of water into a tank 12 m high in 60 s, with $g=9.8$ m s${}^{-2}$:

$$P_{\text{av}}=\frac{mgh}{t}=\frac{500\times 9.8\times 12}{60}=980\text{ W}$$

If the motor is 70 per cent efficient, it must draw $\frac{980}{0.70}=1400$ W.

Worked example 2 — a car engine. An engine exerts a forward force of 3000 N at 25 m s${}^{-1}$:

$$P=Fv=3000\times 25=7.5\times 10^4\text{ W}=75\text{ kW}$$

An everyday example. A 1.5 hp tube-well motor on a Punjab farm delivers about $1.5\times 746=1119$ W, the rate at which it can lift water from underground.

The substance. At fixed power, force and speed trade off — since $P=Fv$, a truck climbing a steep hill must slow down to exert the larger force it needs.

Momentum is conserved in every collision; in an elastic collision kinetic energy is also conserved, while in a perfectly inelastic collision the bodies stick together and lose the most kinetic energy.

Elastic collision, with $m_2$ initially at rest:

$$m_1{u}_1=m_1{v}_1+m_2{v}_2\frac{1}{2}{m}_1{u}_1^2=\frac{1}{2}{m}_1{v}_1^2+\frac{1}{2}{m}_2{v}_2^2$$

$$v_1=\frac{(m_1-m_2)u_1}{m_1+m_2}{v}_2=\frac{2m_1{u}_1}{m_1+m_2}$$

- Equal masses exchange velocities: $v_1=0$ and $v_2=u_1$
- A light body hitting a heavy one bounces back with nearly its original speed

Worked example 1. A 2.0 kg ball at 6.0 m s${}^{-1}$ hits a stationary 1.0 kg ball elastically:

$$v_1=\frac{(2.0-1.0)\times 6.0}{3.0}=2.0{\text{ m s}}^{-1}{v}_2=\frac{2\times 2.0\times 6.0}{3.0}=8.0{\text{ m s}}^{-1}$$

Kinetic energy before is 36 J, and after it is $4.0+32=36$ J.

Perfectly inelastic collision:

$$v=\frac{m_1{u}_1+m_2{u}_2}{m_1+m_2}$$

Worked example 2 — a bullet in a block. A 20 g bullet at 300 m s${}^{-1}$ embeds in a 1.98 kg block at rest:

$$v=\frac{0.020\times 300}{2.00}=3.0{\text{ m s}}^{-1}\text{KE lost}=900-9=891\text{ J}$$

Coefficient of restitution. $e=\frac{\text{speed of separation}}{\text{speed of approach}}$, equal to 1 for elastic and 0 for perfectly inelastic collisions.

An everyday example. A carrom striker hitting a lighter coin head-on sends the coin shooting ahead faster than the striker was moving, while the heavier striker carries on slowly behind it.

The substance. Momentum is conserved in every collision, but kinetic energy is not — in the bullet example, almost all the kinetic energy becomes heat and deformation.

In a two-dimensional collision, momentum is conserved separately along two perpendicular directions, so you write one equation for the x-components and one for the y-components, adding energy conservation if the collision is elastic.

Equations. Ball 1 hits ball 2 at rest, and they move off at angles ${\theta }_1$ above and ${\theta }_2$ below the original line:

$$m_1{u}_1=m_1{v}_1\cos {\theta }_1+m_2{v}_2\cos {\theta }_{2}0=m_1{v}_1\sin {\theta }_1-m_2{v}_2\sin {\theta }_2$$

Worked example. A ball at 4.0 m s${}^{-1}$ strikes an identical ball at rest. Afterwards, the first moves at 30° above its original line and the second at 60° below it:

- y-direction: $v_1\sin 30^{\circ }=v_2\sin 60^{\circ }$, so $v_1=\sqrt{3}{v}_2$
- x-direction: $4.0=\sqrt{3}{v}_2\cos 30^{\circ }+v_2\cos 60^{\circ }=1.5v_2+0.5v_2=2v_2$

$$v_2=2.0{\text{ m s}}^{-1}{v}_1=3.46{\text{ m s}}^{-1}$$

Checking energy: $3.46^2+2.0^2=12.0+4.0=16.0=4.0^2$, so the collision is elastic, and the two balls separate at exactly 90°.

The right-angle rule. In an elastic glancing collision between equal masses with one at rest, the two always move off at 90° to each other.

An everyday example. Snooker players lining up a cut shot use this rule to predict that the cue ball will roll away at nearly a right angle to the ball it strikes.

The substance. Momentum is a vector, so its sideways parts must cancel — in a glancing collision, the two bodies' momenta perpendicular to the original line are always equal and opposite.