Question

A 0.400 ice puck, moving east with a speed of 6.80 , has a head-on collision with a 0.800 puck initially at rest.

Assuming a perfectly elastic collision, what will be the speed of the 0.400 object after the collision?

Answer 1

momentum conserved:
\[m _{1}V _{1i} = m _{1}V _{1f} + m _{2}V _{2f} \]
\[\frac{m _{1} }{m _{2}}= \frac{ V _{2f} }{ (V _{1i} -V _{1f}) }\]

elastic, so KE conserved:
\[ m _{1}V ^2_{1i} = m _{1}V^2 _{1f} + m _{2}V^2 _{2f} \]
\[\frac{m _{1} }{m _{2}}= \frac{ V^2 _{2f} }{ (V^2 _{1i} -V^2 _{1f}) }\]
\[\frac{ V^2 _{2f} }{ (V^2 _{1i} -V^2 _{1f}) }=\frac{ V _{2f} }{ (V _{1i} -V _{1f}) }\]
\[{ V^2 _{2f} (V _{1i} -V _{1f}) }={ V _{2f} (V^2 _{1i} -V^2 _{1f}) }\]
\[{ V^2 _{2f} (V _{1i} -V _{1f}) }= V _{2f} (V_{1i} -V_{1f}) (V_{1i} +V_{1f})\]
\[ V _{2f} = (V_{1i} +V_{1f})\]
sub.s back into original momentum equation and you get:
\[ V_{1i} \frac{ m _{1} -m _{2} }{m _{1} +m _{2}} = 2V _{1f}\]

Answer 2

It's not really necessary to sue all the steps... you can just use the result. I just went over the derivation it so you could see where it comes from.

Answer 3

Whoops, a '2' slipped into that last equation. Ignore it.
\[V _{1i}\frac{ m _{1} -m _{2}}{ m _{1} +m _{2}}= V_{1f}\]

Answer 4

Should I use the last equation. It seems so complicated

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Answer 5

It's not really... your book does the same derivation, I'm sure... but yes, use the result.