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. If the particlee stick to the block after collision the velocity of block parallel tot he inclined plane just after collision will be (take `g=10(m)/(s^(2))`

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)=2mv_(f)`
`v_(f)=(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)))`