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Q. A block of mass $m_{1}$ lies on the top of fixed wedge as shown in Fig. (a) and another block of mass $m_{2}$ 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 $T_{1}$ and $T_{2}$ be the time taken by the blocks, respectively, to just reach the horizontal surface. ThenPhysics Question Image

Laws of Motion

Solution:

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In first case, acceleration of $m_{1}$ will be $a_{1}=g \sin \theta$ down the inclined plane. In second case, acceleration of $m_{2}$ w.r.t. incline is
$a_{2}=\frac{m_{2} g \sin \theta+ m_{2} a \cos \theta}{m_{2}}$
$\Rightarrow a_{2}=g \sin \theta +a \cos \theta$
Since $a_{2} > a_{1}$, so $T_{2} < T_{1}$