On a friction surface a block a mass `M` moving at speed `v` collides elastic with another block of same mass `M` which is initially at rest . After collision the first block moves at an angle `theta` to its initial direction and has a speed `(v)/(3)`. The second block's speed after the collision is

Options

A. `3 sqrt2v`
B. `sqrt3/2 v`
C. `2 sqrt 2/3 v`
D. `3/4 v`

Answer

`(3)`

Explanation

Let v. be speed of second block after the collision. As the collision is elastic, so kinetic energy is conserved. According to conservation of kinetic energy,
`1/2 (Mv)^(2) + 0 = 1/2 M(v/3)^(2) + 1/2 Mv^(2)`
`v_(2) = (v^(2))/9 + v.^(2)`
or ` v.^(2) = v^(2) - (v^(2))/9 = (9v^(2) - v^(2))/9 = 8/9 v^(2)`
`v. = sqrt((8/9)v^(2))= sqrt 8 /3 v = (2 sqrt 2)/3 v`