Question

A ball with a mass of `15 kg` moving at `6 m/s` hits a still ball with a mass of `21 kg`. If the first ball stops moving, how fast is the second ball moving? How much kinetic energy was lost as heat in the collision?

Answer 1

`4.3ms^-1` and `75.9J`

Explanation:

Conservation of momentum states that before and after a physical reaction there must always be the same amount of momentum.

Momentum is the product of mass and velocity, or

`p = mv`

where `p` is momentum, `m` is mass and `v` is velocity.

Before the reaction, there is a total momentum of

`15kg * 6ms^-1 + 21kg * 0ms^-1 = 90kgms^-1`

because the question describes the `21kg` mass as stationary, therefore it has a velocity of `0` and a momentum of `0`.

After the reaction, we know that the `15kg` mass stops moving, so the velocity and momentum are `0`. The total momentum also must be `90kgms^-1`, according to the law of conservation of momentum.

Therefore, after the reaction,

`p = 15kg * 0ms^-1 + 21kg * v = 90kgms^-1`
`21kg * v = 90kgms^-1`
`v = 90/21 ms^-1 = 4.3ms^-1`

The answer is rounded to the first decimal place.

For the second part of the question, kinetic energy is given by the formula

`E = 1/2 mv^2`

where `E` is kinetic energy, `m` is mass, and `v` is velocity.

Before the reaction, total kinetic energy would therefore be

`(15 * 6^2)/2 + (21 * 0^2)/2 = 270kgm^2s^-2 = 270J`

After the reaction, we would have

`(15 * 0^2)/2 + (21 * 4.3^2)/2 = 194.1J`

The change in kinetic energy is

`270J - 194.1J = 75.9J` lost as heat.

Feel free to check my working, there's some room for error-carried-forward due to rounding. Hope it's somewhat clear.