Question

A 0.500 ice puck, moving east with a speed of 6.40 , has a head-on collision with a 0.800puck initially at rest.

Assuming a perfectly elastic collision, what will be the speed of the 0.500 object after the collision?

What will be the speed of the 0.800 object after the collision?

Please provide a detailed solution, Im am confused

Answer 1

using Law of COnservation of Momentum
\[m_{1} v_{1} + m_{2} v_{2} = m_{1} v_{1}^{'} + m_{2} v_{2}^{'}\]
based on the question, becomes
\[m_{i} v_{i} + m_{p} v_{p} = m_{i} v_{i}^{'} + m_{p} v_{p}^{'}\]
where i is ice, and p is puck.,
look at the question, it's say "...a 0.800puck initially at rest." so the initial speed of puck is zero., then initial momentum of puck is zero too.,

Now., for perfectly elastic, e = 1 (where e is coefficient of restution)
\[e = -\frac{ (v_{i}^{'} - v_{p}^{'}) }{ v_{i} - v_{p} }\]
\[1 = -\frac{ (v_{i}^{'} - v_{p}^{'}) }{ v_{i} - v_{p} }\]
\[v_{i} -v_{p} = -v_{i}^{'} + v_{p}^{'}\]
\[v_{i} -0 = -v_{i}^{'} + v_{p}^{'}\]
\[v_{p}^{'} = v_{i} + v_{i}^{'}\]
substitution of equation into :
\[m_{i} v_{i} + m_{p} v_{p} = m_{i} v_{i}^{'} + m_{p} v_{p}^{'}\]
So You will get a speed of the 0.500 object after the collision (\(v_{i}^{'} \)) and the speed of the 0.800 object after the collision (\(v_{p}^{'}\))

Answer 2

Remember:
(\(v_{i}\)) is the 0.500 object before the collision
(\(v_{p}\)) is the 0.800 object before the collision