Question

a block of mass $m_1$ lies on the top of fixed wedge as shown in $Fig.\left(a\right)$ and another block of mass $m_2$ lies on top of wedge which is free to move as shown in $Fig\left(b\right)$. 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 $T_1$ be the time taken by the blocks, respectively, to just reach the horizontal surface. Then

• A. $T_1>T_2$
• B. $T_1<T_2$
• C. $T_1=T_2$
• D. Data insufficient

Answer 1

The correct option is A $T_1>T_2$

In the $\left(a\right)$ case acceleration of the "$m_1$" along the incline is $a_1=g\sin \theta$

Whereas in the $\left(b\right)$ case the acceleration of "$m_1$" along the incline can be found by applying psuedo force on "$m_1$ as shown.

So, the acceleration $a_2=\frac{m_{1}g\sin \theta +m_{1}A\cos \theta }{m_1}$

hence $a_1<a_2$ so $T_1>T_2$

Therefore of option A is correct