Question

A small steel ball $A$ is suspended by an inextensible thread of length $l=1.5\text{m}$ from $O$ as shown in the figure. Another identical ball $B$ is thrown vertically downwards such that its surface remains just in contact with thread during downward motion and collides elastically with the suspended ball. If the suspended ball just complete vertical circle after collision, calculate the Impulse on the steel ball $A$ due to the ball $B$. (Take $g=10{\text{m/s}}^2$ and mass of ball $m=5\text{kg}$)

• A. $50\sqrt{2}\text{Ns}$
• B. $25\sqrt{3}\text{Ns}$
• C. $50\sqrt{3}\text{Ns}$
• D. $25\text{Ns}$

Answer 1

The correct option is C $50\sqrt{3}\text{Ns}$

Velocity of ball $A$ just after collision should be $\sqrt{5gl}$ in order to complete the vertical circle.
Let radius of each ball be $r$ and the line joining centers of the two balls make an angle $\theta$ with the vertical at the instant of collision, then


$\sin \theta =\frac{r}{2r}=\frac{1}{2}\Rightarrow \theta ={30}^{\circ }$
Now, we have F.B.D of ball $A$ as



Here, $J$ is Impact due to the ball $B$ on $A$.
$v_2=$ velocity of the ball $A$ after impact.
Now, horizontal velocity $v_2$ of ball $A$ is given by
$J\sin {30}^{\circ }=m{v}_2$
$\Rightarrow J\times \frac{1}{2}=5\times \sqrt{5gl}$
$\Rightarrow J\times \frac{1}{2}=5\times \sqrt{5\times 10\times 1.5}$
$\Rightarrow J=50\sqrt{3}\text{Ns}$