Question

A small steel ball $A$ is suspended by an inextensible thread of length $l=1.5\text{m}$ from $O$. Another identical ball 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 completes vertical circle after collision, calculate the velocity (in $\text{cm/s}$) of the falling ball just before collision $(g=10{\text{ms}}^{-2})$.

• A. $12.5\text{cm/s}$
• B. $1250\text{cm/s}$
• C. $12.5\text{m/s}$
• D. $1250\text{m/s}$

Answer 1

The correct option is B $1250\text{cm/s}$


Given that,
Length of string, $l=1.5\text{m}$
Let the radius of each ball be $R$ and mass be $m$.

When the ball $B$ hits $A$, it transfers an impulse $J$ and $J\sin \theta$ provides the velocity $v_2$ for completing the vertical circle.

From the above diagram:

$\sin \theta =\frac{R}{2R}$

$\Rightarrow \sin \theta =\frac{1}{2}$

$\Rightarrow \theta ={30}^{\circ }$

The suspended ball $A$ will just complete the vertical circle if the horizontal velocity($v_2)$ provided by the collision is equal to

$v_2=\sqrt{5gl}$

$v_2=\sqrt{5\times 10\times 1.5}=5\sqrt{3}$

Therefore $J\sin \theta =\frac{J}{2}=5m\sqrt{3}$

$J=m10\sqrt{3}{\text{kg ms}}^{-1}$

For ball $A:$

i.e. $J=m(v_1+u\cos \theta )$

$\Rightarrow m10\sqrt{3}=m(v_1+u\cos {30}^{\circ })$

$\Rightarrow 20\sqrt{3}=2v_1+u\sqrt{3}\dots \dots \left(1\right)$

Now coefficient of restitution of the system is

$e=\frac{v_2\sin \theta +v_1}{u\cos \theta }=1$

$\Rightarrow u\sqrt{3}=5\sqrt{3}+2v_1\dots \dots \left(2\right)$

Subtracting equation ($2$) from ($1$)

$\Rightarrow 20\sqrt{3}-u\sqrt{3}=u\sqrt{3}-5\sqrt{3}$

$\Rightarrow 2u=25$

$\Rightarrow u=12.5\text{m/s}\text{or}1250\text{cm/s}$

Hence, option (b) is the correct answer.