Answer to Question #90832 in Mechanics | Relativity for Gordon Huy

Question

A mass m moves with velocity v and collides with another mass 2m initially at rest. After collision the first moves with velocity v/√2 in the direction perpendicular to initial direction of motion. Find the speed of 2nd mass after collision.

Answer 1

Consider a plane in which the movement takes place before and after the collision. Consider in this plane an orthonormal coordinate system, the first axis of which is directed along the velocity of the first body before the collision, and the second - along the velocity of the first body after the collision. Then, in this coordinate system, the law of conservation of momentum has the form

$$m(v,0)+2m(0,0)=m\left(0,\frac{v}{\sqrt{2}}\right)+2m(x,y),$$

where (x,y) is the velocity coordinates of the second body after the collision. We write this equation coordinatewise:

$$mv=2mx,\; 0=\frac{mv}{\sqrt{2}}+2my.$$

From here

$$x=\frac{v}{2},\; y=-\frac{v}{2 \sqrt{2}}.$$

The length of the velocity vector is

$$\sqrt{x^2+y^2}=\frac{1}{2}\sqrt{\frac{3}{2}}v.$$

Answer:

$$\frac{1}{2}\sqrt{\frac{3}{2}}v.$$