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. `sqrt(3)upsilon//2`
B. `2sqrt(2)upsilon//3`
C. `3 upsilon//4`
D. `3 upsilon//sqrt(2)`

Answer

B

Explanation

Refer to figure. As the collision is perfectly elastic hence the laws of conservation of linear momentum and kinetic energy are obeyed.
Accordin to law of conservation of kinetic energy
Total KE before collision `=` total KE after collision
`i.e., (1)/(2)Mupsilon^(2)+0=(1)/(2)M(upsilon//3)^(2)+(1)/(2)Mupsilon_(2)^(2)`
or `upsilon^(2)=(upsilon^(2))/(9)+upsilon_(2)^(2)` or `upsilon_(2)^(2)=upsilon^(2)-(upsilon^(2))/(9)=(8upsilon^(2))/(9)`
or `upsilon_(2)=(2sqrt(2)upsilon)/(3)`