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. `(2sqrt2)/(3)`
B. `(3)/(4)v`
C. `(3)/(sqrt2)v`
D. `(sqrt3)/(2)v`

Answer

a

Explanation

(a)
According to law of conservation of kinetic energy, we have
`(1)/(2)Mv^(2)+0=(1)/(2)M((v)/(3))^(2)+(1)/(2)Mv_(2)^(2)`
`impliesv^(2)=(v^(2))/(9)+v_(2)^(2)`
`impliesv^(2)-(v^(2))/(9)=v_(2)^(2)implies(8v^(2))/(9)`
Velocity of second block after collision `v^(2)=(2sqrt2)/(3)v`