A block of mass 'm' moving on a frictionless surface at speed 'v' collides elastically with a block of same mass, initially at rest. Now the first block moves at an angle `theta` with its initial direction and has speed `v_(1)` . The speed of the second block after collision is

Options

A. `sqrt(v_(1)^(2) - v^(2))`
B. `sqrt(v^(2) - v_(1)^(2))`
C. `sqrt(v^(2) + v_(1)^(2))`
D. `sqrt(v-v_(1))`

Answer

B

Explanation

The situation can be shown as

Applying law of conservation of kinetic energy , KE (before collison) = KE (after collison)
`(1)/(2)mv^(2) + (1)/(2)m(0)^(2) = (1)/(2)mv_(1)^(2) +(1)/(2)mv_(2)^(2)`
`rArr v^(2)= v_(1)^(2) + v_(2)^(2) rArr v_(2) = sqrt(v^(2) - v_(1)^(2))`
Thus, the velocity of second block after collision is `sqrt(v^(2) -v_(1)^(2))`