A block of mass `m_(1)` lies on the top of fixed wedge as shown in fig. and another block of mass `m_(2)` lies on top of wedge which is free to move as shown in fig. At time t=0 both the blocks are released from rest from a vertical height h above the respective horizontal surface on which the wedge is placed as shown. There is no friction between block and wedge in both the figures. Let `T_(1)` and `T_(2)` be the time taken by the blocks respectively to just reach the horizontal surface. then

Options

A. `T_(1) gt T_(2)`
B. `T_(1) lt T_(2)`
C. `T_(1) = T_(2)`
D. Data insufficient

Answer

A

Explanation

In first case, acceleration of `m_(1)` willbe `a_(1)=g sin theta` down the inclined plane. In second case, acceleration of `m_(2)` w.r.t incline is
`a_(2) = (m_(2)g sin theta + m_(2)a cos theta)/(m_2)`
`implies a_(2)=g sin theta + a cos theta`
since `a_(2) gt a_(1). so T_(2) lt T_(1)`.