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}

]

Aicharya · Accepted Answer

Let us assume that just after the collision, the speed of the particle plus $1 kg$ block is $u$

$2 u = 1 \times 10 \Rightarrow u = 5 m s^{- 1}$

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

$2 \times 5 = 5 v$

$\Rightarrow v = 2 m s^{- 1}$

A light inelastic thread passes over a small frictionless pulley. Two blocks of masses m = 1 k g and M = 3 k…

Solution