Question

A metal ball of mass $10\text{kg}$ is suspended from a massless string of length $2\text{m}$ as shown in figure. Another identical ball of mass $10\text{kg}$ is projected towards the suspended ball. Assuming the collision between the balls to be perfectly elastic, find the minimum value of velocity of second ball so that the first ball just completes the vertical circular motion. [Take $g=10{\text{m/s}}^2$]

• A. $5\text{m/s}$
• B. $10\text{m/s}$
• C. $15\text{m/s}$
• D. $20\text{m/s}$

Answer 1

The correct option is B $10\text{m/s}$

Given,
$m_1=m_2=10\text{kg}$
Length of the string, $\left(l\right)=2\text{m}$
Applying conservation of momentum,
$m_1{u}_1+m_2{u}_2=m_1{v}_1+m_2{v}_2$ $[\because u_2=0,v_1=0\text{and}{m}_1=m_2]$
$\Rightarrow u_1=v_2$
To just complete the vertical circular motion,
$v_2\ge \sqrt{5gl}$
$\Rightarrow u_1\ge \sqrt{5gl}$
${\left(u_1\right)}_{\text{min}}=\sqrt{5\times 10\times 2}$
$\Rightarrow v_0={\left(u_1\right)}_{\text{min}}=10\text{m/s}$

Hence, $\left(B\right)$ is the correct answer.