On a frictionless surface , a block of mass . M moving at speed v collides elastically with another block of same mass M which is intially at rest /After collision the first block moes 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. `(sqrt(3))/2v`
B. `(2sqrt(2))/3v`
C. `3/4` v
D. `3/(sqrt(2))v`

Answer

B

Explanation

In elastic collision energy of system remains constant .
` :. ` Kinetic energy before collision = kinetic energy after collision
` :. 1/2 mv^(2)+0=1/2m (v/3)^(2) +1/2m(v)^(2)`
After collision , v is the speed of second block .
` :. v^(2) = (v^(2))/9+(v)^(2)`
` :. (9v^(2)-v^(2))/9 =(v)^(2)`
` :. (8v^(2))/9 = (v)^(2)`
` :. v = (2sqrt(2))/3 v `