`N` identical balls are placed on a smooth horizontal surface. An another ball of same mass collides elastically with velocity `u` with first ball of `N` balls. A process of collision is thus started in which first ball collides with second ball and the with the third ball and so on. The coefficient of restitution for each collision is `e`. Find the speed of `Nth` ball.

Options

A. `(1 + e)^(N) u`
B. `u(1 + e)^(N-1)`
C. `(u(1 + e)^(N-1))/(2^(N-1))`
D. `u^(N)(1 + e)^(N)`

Answer

C

Explanation

After collision, first ball start moving with velocity `u`
Now,`e = (V_(2)-V_(1))/(u)`
or `V_(1) = u - V_(2)`
so, `e = (V_(2) - u +V_(2))/(u)`
or `V_(2) = (u(1 + e))/(2)`
Incoming velocity of second ball is `V_(2)` and after collision velocity of third ball is `V_(3)` and velocity of second ball is `V'_(2)`
`e = (V_(3) - V'_(2))/(V_(2))`
and `mV_(2) - mV_(3) + 3 mV'_(2)`
or `V'_(2) = V_(2) - V_(3)`
`= (u(1 +e))/(2) - V_(3) = (u(1 +e))/(2) - V_(3)`
So, `e = V_(3) - ([(u + ue - 2V_(3))/(2)])/([(u(1 + e))/(2)])`
On solving `V_(3) = (u(1 + e)^(2))/(2^(2))`
so, velocity of Nth ball `= (u(1 + e)^(N-1))/(2^(N-1))`