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A block of mass m₁ lies on the top of fixed wedge as shown in Fig. (a) and another block of mass m₂ lies on top of wedge which is free to move as shown in Fig. (b). 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 T₁ and T₂ be the time taken by the blocks, respectively, to just reach the horizontal surface. Then

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$T_{1} < T_{2}$

$T_{1} = T_{2}$

$T_{1} > T_{2}$

Data insufficient

answer is A.

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Detailed Solution

In first case, acceleration of m₁ will be $a_{1} = g \sin θ$ down the inclined plane. In second case, acceleration of m₂ w.r.t. incline is

$\begin{array}{l} a_{2} = \frac{m_{2} g \sin θ + m_{2} a \cos θ}{m_{2}} \\ \Rightarrow a_{2} = g \sin θ + a \cos θ \end{array}$

Since $a_{2} > a_{1}, so T_{2} < T_{1}$

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