Question

A light inelastic thread passes over a small frictionless
pulley. Two blocks of masses m 1 kg and M
3 kg,
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' 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 ms?)
-1
m
Р
M​

Answer 1

The speed of the bigger block after the string is taut is 2 m/s

Explanation:

Given:

Two blocks of masses m = 1 kg  and M = 3 kg

A particle P of mass 1 kg is moving with a velocity of 10 m/s.

Let us assume  that after the collision the speed of the particle plus 1 kg block is v.

$p_{1} =$ 1 × 10

⇒ 10 kg m/s

2v = 10 kg m/s

v = 10/2 kg m/s

v = 5 m/s

After the collision and the string is taut again, the particle block system is in free fall.

Before the string is taut, the particle plus 1 kg block will be moving downward with a velocity of 5 m/s.

After the string is taut, let we take that the speed of both the blocks and the particle is u.

Then,

5v = 2× 5

5v = 10

v = 2 m/s

Final answer:

The speed of the bigger block after the string is taut is 2 m/s

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