A block A of mass m is moving on a frictionless horizontal plane towards a smooth wedge (B) of mass 2m, which is kept at rest on smooth horizontal plane at t = 0 . At t = t 1 , it is found that both

Question

A block A of mass m is moving on a frictionless horizontal plane towards a smooth wedge (B) of mass 2m, which is kept at rest on smooth horizontal plane at $t = 0$ . At $t = t_{1}$ , it is found that both block & wedge are moving with same velocity and at $t = t_{2}$ it is found that block & wedge are again moving on the horizontal frictionless plane. In List-I work done by different forces are given and in List-II values of work done by those forces are given. Match them.

List - I	List - II
(P) Work done by gravity on block A from $t = 0$ to $t = t_{1}$	(1) $- \frac{m v ^{2}}{9}$
(Q) Work done by normal force exerted by wedge on the block A from $t = 0$ to $t = t_{1}$	(2) $\frac{m v ^{2}}{3}$
(R) Work done by the normal force exerted by block A on the wedge for $t = t_{1}$ to $t = t_{2}$	(3) Zero
(S) Work done by the gravity on block A from $t = 0$ to $t = t_{2}$	(4) $- \frac{m v ^{2}}{3}$

TestHub · Accepted Answer

Momentum conservation: common velocity at $t_{1}$ is $\frac{v}{3}$ ; the elastic exchange returns the block at $t_{2}$ with $- \frac{v}{3}$

$m g h = \frac{1}{2} m v^{2} - \frac{1}{2} (3 m) \frac{v ^{2}}{9} = \frac{m v ^{2}}{3}$ , so $W_{g} (0 \to t_{1}) = - \frac{m v ^{2}}{3}$

Work-energy theorem: $W_{N} (A) = - \frac{m v ^{2}}{9}$ , $W_{N} (B, t_{1} \to t_{2}) = \frac{m v ^{2}}{3}$ , net gravity work over the full motion = 0

P→4, Q→1, R→2, S→3

Answer (D)

Physics - W.E.P. Question with Solution | TestHub

Options:

Doubts & Discussion