A particle of mass `m` is projected up from the bottom of an inclined plane with initial velocity `v_(0)` at angle `45^(@)` with an inclined plane of inclination `30^(@)` as shown in figure. At the same time a small block of same mass `m` is released from rest at a height `h`, the particle hits the block at some point on the inclined plane. Neglect friction at inclined.
Q. The value of height h is

Options

A. `(v_(0))/(sqrt(2))`
B. `(v_(0))/2`
C. `(v_(0))/(2sqrt(2))(1-4/(sqrt(3)))`
D. `(v_(0))/(2sqrt(2))(1+4/(sqrt(3)))`

Answer

C

Explanation

From conservation of momentum along the inclined plane
`mv_(x)+mv_(x)^(')=2nv_(t)`
`v_(t)=(v_(x)-v_(x)^('))/2=1/2[(v_(0))/(sqrt(2))-g/2 (2sqrt(2)v_(0))/(sqrt(3)g)-g/2 (2sqrt(2)v_(0))/(sqrt(3)g)]=(v_(0))/(2sqrt(2))(1-4/(sqrt(3)))`