Question

A ball hanging from a support by an inextensible string of length $l$ is strucked perfectly inelastically by an identical ball of mass $m$ moving with horizontal velocity of $u=\sqrt{2gl}$. Assuming the balls execute circular motion just after collision, find the tension in the string just after collision.

• A. $mg$
• B. $3mg$
• C. $2mg$
• D. $0$

Answer 1

The correct option is B $3mg$


Given:
Collision is perfectly inelastic, therefore both ball will stick to each other and share common velocity $\left(v\right)$ after collision.

As there are no external force in horizontal direction.
$\therefore$ Using conservation of linear momentum of the system of bodies just before and after collision,
$\Rightarrow mu=(m+m)v\Rightarrow v=\frac{u}{2}...\left(i\right)$

After the collision body starts circular motion of radius $l$



Tension in the string is $T$,
$T-2mg=2m{a}_c=2m\left(\frac{v^2}{l}\right)...\left(ii\right)$
Using equation $\left(i\right)$ and $\left(ii\right)$
$T=\frac{2m{u}^2}{4l}+2mg$
Given:
$u=\sqrt{2gl}$
$\therefore T=mg+2mg=3mg$