A light inelastic thread passes over a small frictionless pulley. Two blocks of masses m = 1 k g and M = 3 k g , respectively, are attached with the thread and heavy block rests on a surface. A particle P of mass 1 kg moving upward with a velocity of 10 m s -1 collides with the lighter block and sticks to it. The speed of the bigger block just after the string is taut will be [ g = 10 m s - 2 ]

Question

A light inelastic thread passes over a small frictionless pulley. Two blocks of masses m = 1 k g and M = 3 k g , respectively, are attached with the thread and heavy block rests on a surface. A particle P of mass 1 kg moving upward with a velocity of 10 m s -1 collides with the lighter block and sticks to it. The speed of the bigger block just after the string is taut will be [ g = 10 m s - 2 ], 2.5 m s - 1, 4 m s - 1, 5 m s - 1, 2 m s - 1

MARKS App · Accepted Answer

Let us assume that just after the collision, the speed of the particle plus 1 kg block is u 2 u = 1 × 10 ⇒ u = 5 m s - 1 After the collision and until the string is taut again, the particle block system is in free fall. So just before the string is taut, the particle plus 1 kg block will be moving downward with a velocity of 5 m s - 1 and just after the string is taut, let us assume that the speed of both the blocks and the particle is v , then 2 × 5 = 5 v ⇒ v = 2 m s - 1

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