Question

In the Figure, the ball A is released from rest when the spring is at its natural
length. For the block B, of mass M to leave contact with the ground at some
stage, the minimum mass of A must be:
(A) 2M
(B) M
(C) M/2
(D) A function of M and the force constant of the spring.​

Answer 1

Answer :

Option => C.

M/2.

Explanation :

Given that :

The ball A is released from rest when the spring is at its natural  length.

The block B, of mass M to leave contact with the ground at some stage.

To Find :

The minimum mass of A.

Solution :

$\bigstar\;\textbf{\underline{\underline{Consider as -}}}}$

Minimum mass of ball as - "m".

Mass A moves downwards by "x".

We know that :

Law of conservation of energy :

$\bigstar\;\boxed{\sf mgx = \dfrac{1}{2}kx^2}}\\ \\ \\\star\;\textbf{\underline{\underline{Values}}},\\ \sf \\ \implies x = \left(\dfrac{2mg}{k}\right)\\ \\\star\;\textbf{\underline{\underline{Mass M to leave contact with ground.}}}\\ \sf \\ \implies kx = Mg\\ \\\implies k\left(\dfrac{2mg}{k}\right) = Mg\\ \\ \\\implies M = \dfrac{M}{2}\\ \\ \\\bigstar\;\textbf{\underline{\underline{Hence,Option C is correct.}}}$

Answer 2

:Solution :

Answer :

C) M/2.

For showing full answer click on photo .