A ball with a mass of #2 kg# moving at #3 m/s# hits a still ball with a mass of #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 #0.6ms^-1#.

#7.2J# or #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

#(mv)_1 = (mv)_2#

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

#(mv)_1 = 2kg xx 3ms^-1 = 6Ns#

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

#(mv)_2 = 10kg xx vms^-1#

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

#10 xx v = 6 -> v = 0.6ms^-1#

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

#E = 1/2mv^2#

which, before the reaction, is

#E = 1/2 xx 2 xx 3^2 = 9J#

after the reaction, it will be

#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.