Question

A smooth ball A of mass is attached to one end of a light inextensible string, and is suspended from fixed point O. Another identical ball B, is dropped from a height h, so that the string just touches the surface of the sphere.

$\text{Column I}$	$\text{Column II}$
(A) If collision between balls is completely elastic than speed of A just after collision is	(P) $\frac{3m}{5}\sqrt{2gh}$
(B) If collision between balls is completely elastic than impulsive tension provided by string is	(Q) $\frac{\sqrt{6gh}}{5}$
(C) If collision between balls is completely inelastic than speed of A just after collision is	(R) $\frac{6m}{5}\sqrt{2gh}$
(D) If collision between balls is completely inelastic than impulsive tension provided by string is	(S) $\frac{2\sqrt{6gh}}{5}$
	(T) None of these

• A. $\left(A\right)\to \left(S\right),\left(B\right)\to \left(R\right),\left(C\right)\to \left(Q\right),\left(D\right)\to \left(P\right)$
• B. $\left(A\right)\to \left(R\right),\left(B\right)\to \left(S\right),\left(C\right)\to \left(P\right),\left(D\right)\to \left(Q\right)$
• C. $\left(A\right)\to \left(P\right),\left(B\right)\to \left(Q\right),\left(C\right)\to \left(R\right),\left(D\right)\to \left(T\right)$
• D. $\left(A\right)\to \left(Q\right),\left(B\right)\to \left(S\right),\left(C\right)\to \left(P\right),\left(D\right)\to \left(S\right)$

Answer 1

The correct option is A $\left(A\right)\to \left(S\right),\left(B\right)\to \left(R\right),\left(C\right)\to \left(Q\right),\left(D\right)\to \left(P\right)$

In an elestic collision,
For (A),
$v_0=\sqrt{2gh},\sin \theta =\frac{R}{2R}=\frac{1}{2}$
$\cos \theta =\frac{\sqrt{3}R}{2R}=\frac{\sqrt{3}}{2}$

By definition of $e$,
$e=1=\frac{v_1\sin \theta +v_2}{v_0\cos \theta }$


Let impulse given by ball be $N$ then, by impulse momentum theorem $N=m(v_2+v_{0}cos\theta )$
and $Nsin\theta =m{v}_1$
$\Rightarrow v_1=\frac{2v_0\sin \theta \cos \theta }{1+{\sin }^2\theta }=\frac{\left(2\sqrt{2gh}\right)\left(\frac{1}{2}\right)\left(\frac{\sqrt{3}}{2}\right)}{1+{\left(\frac{1}{2}\right)}^2}=\frac{2\sqrt{6gh}}{5}$
Hence, $\left(A\right)\to \left(Q\right)$

For (B):
Impulsive tension = $N\cos \theta$
$N\cos \theta =\left(\frac{m{v}_1}{\sin \theta }\right)\cos \theta =\left(\frac{m\times \frac{2\sqrt{6gh}}{5}}{\frac{1}{2}}\right)\times \frac{\sqrt{3}}{2}=\frac{6m}{5}\sqrt{2gh}$
Hence, $\left(B\right)\to \left(R\right)$

For (C):
For completely inelastic collision e= 0
so, $v_1\sin \theta +v_2=0\to v_1=\frac{v_0\sin \theta \cos \theta }{1+{\sin }^2\theta }=\frac{\sqrt{6gh}}{5}$
Hence, $\left(C\right)\to \left(Q\right)$

For (D):
Impulsive tension for inelastic collision:
$N\cos \theta =\left(\frac{m{v}_1}{\sin \theta }\right)\cos \theta =m{v}_1\cot \theta =\frac{3m}{5}\sqrt{2gh}$
Hence, $\left(D\right)\to \left(P\right)$