A small ball B of mass m is suspended with a light inelastic string of length L from a block A of same mass m which can move on a smooth horizontal surface as shown in the figure. The ball is displaced by the angle θ from equilibrium position then released. Maximum velocity of block during subsequent motion of the system after release of ball is

Question

A small ball B of mass m is suspended with a light inelastic string of length L from a block A of same mass m which can move on a smooth horizontal surface as shown in the figure. The ball is displaced by the angle θ from equilibrium position then released. <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-uwruosbveze1.png" /> Maximum velocity of block during subsequent motion of the system after release of ball is (A) ([ textitg L (1 - textcos θ )])1 / 2 (B) ([2 textitg L (1 - textcos θ )])1 / 2 (C) [ textitg L textcos θ ]1 / 2 (D) Insufficient information to decide

Tardigrade Admin · Accepted Answer

KE gain by system = mgL (1 - textcos θ ) half of that will be shared by block (1/2)( textitmv)2=(1/2) textitmg L(1 - textcos θ )⇒ textitv=√ textitgL (1 - textcos θ )

$([g L (1 - cos θ)])^{1/2}$

$([2 g L (1 - cos θ)])^{1/2}$

$[g L cos θ]^{1/2}$

Insufficient information to decide

KE gain by system = $m gL$ $(1 - cos θ)$
half of that will be shared by block
$\frac{1}{2} (mv)^{2} = \frac{1}{2} mg L (1 - cos θ) \Rightarrow v = gL (1 - cos θ)$

([gL(1−cosθ)])1/2

([2gL(1−cosθ)])1/2

[gLcosθ]1/2

Insufficient information to decide

KE gain by system = mgL (1−cosθ) half of that will be shared by block 21​(mv)2=21​mg L(1−cosθ)⇒v=gL(1−cosθ)​

$([g L (1 - cos θ)])^{1/2}$

$([2 g L (1 - cos θ)])^{1/2}$

$[g L cos θ]^{1/2}$

KE gain by system = $m gL$ $(1 - cos θ)$
half of that will be shared by block
$\frac{1}{2} (mv)^{2} = \frac{1}{2} mg L (1 - cos θ) \Rightarrow v = gL (1 - cos θ)$