Question

A body of mass m moves with a velocity v and collides with another mass 2m at rest.After collision the first mass moves with a velocity v+underoot 3 in a direction prependicular to the initial direction of motion.Find the speed of 2nd mass after collision.

Answer 1

dear student

Initial momentum of the first body =mv ( along the x direction)
Initial momentum of the second body =0 ( along the x direction)
momentum of the first body before collision in the y direction =0
momentum of the second body before collision in y direction =0

Momentum of the first body after collision in the x direction =0
Momentum of the first body after collision in the y direction=$m\left(v+\sqrt{3}\right)$
Momentum of the second body after collision in the x-direction is =2mv_x
Momentum of the second body after collision in the x-direction is =2mv_​y
Thus by conservation of momentum in x and y direction

$$\begin{gathered} momentuminxdirectionbeforecollision=momentuminxdirectionaftercollision \\ mv+0=0+2m{v}_x \\ {v}_x=\frac{v}{2} \\ momentuminydirectionbeforecollision=momentuminydirectionaftercollision \\ 0+0=m\left(v+\sqrt{3}\right)+2m{v}_y \\ {v}_y=-\frac{\left(v+\sqrt{3}\right)}{2} \\ thusspeedofsecondbodyisgivenby \\ \sqrt{{v_x}^2+{v_y}^2} \\ \sqrt{\frac{v^2}{4}+\frac{{\left(v+\sqrt{3}\right)}^2}{4}} \\ \sqrt{\frac{v^2}{4}+\frac{v^2+3+2v\sqrt{3}}{4}} \\ \frac{1}{2}\sqrt{2v^2+2v\sqrt{3}+3} \end{gathered}$$