A ball with a mass of $2 k g$ $2 kg$ moving at $3 \frac{m}{s}$ $3 m/s$ hits a still ball with a mass of $10 k g$ $10 kg$ . If the first ball stops moving, how fast is the second ball moving? How much kinetic energy was lost as heat in the collision?

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Answer 1 · 2017-03-05T12:52:06

The second ball moves at $0.6 m s^{- 1}$ $0.6ms^-1$ .

$7.2 J$ $7.2J$ or $80 %$ $80%$ of the original kinetic energy is lost.

Conservation of momentum says that momentum before a reaction and momentum after a reaction must be equal.

Momentum is the product of mass and velocity, so

${(m v)}_{1} = {(m v)}_{2}$ $(mv)_1 = (mv)_2$

Only moving objects have momentum, so all the momentum beforehand is in the moving ball, so

${(m v)}_{1} = 2 k g \times 3 m s^{- 1} = 6 N s$ $(mv)_1 = 2kg xx 3ms^-1 = 6Ns$

This ball stops moving, and now only the $10 k g$ $10kg$ ball moves, so momentum after the reaction is

${(m v)}_{2} = 10 k g \times v m s^{- 1}$ $(mv)_2 = 10kg xx vms^-1$

We know that this is equal to the momentum before, so

$10 \times v = 6 \to v = 0.6 m s^{- 1}$ $10 xx v = 6 -> v = 0.6ms^-1$

Now for the second part. Calculating kinetic energy is done by the equation

$E = \frac{1}{2} m v^{2}$ $E = 1/2mv^2$

which, before the reaction, is

$E = \frac{1}{2} \times 2 \times 3^{2} = 9 J$ $E = 1/2 xx 2 xx 3^2 = 9J$

after the reaction, it will be

$E = \frac{1}{2} \times 10 \times {0.6}^{2} = 1.8 J$ $E = 1/2 xx 10 xx 0.6^2 = 1.8J$

Overall, then, the kinetic energy lost is

$DeltaE = 9 - 1.8 = 7.2J$

or, as a percentage,

$(9-1.8)/9 xx 100% = 80%$ of the kinetic energy is lost.

A ball with a mass of 2 kg2kg2 kg moving at 3 m/s3ms3 m/s hits a still ball with a mass of 10 kg10kg10 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?