Question

A ball of mass $20\text{kg}$ is suspended from a massless string of length $5\text{m}$ as shown in figure. A bullet of mass $5\text{kg}$ moving with velocity $v_0\text{m/s}$ collides with the ball and sticks to it. Find the minimum value of $v_0$ so that the combined mass completes the vertical circular motion. [Take $g=10{\text{m/s}}^2$]

• A. $100\text{m/s}$
• B. $15\text{m/s}$
• C. $79\text{m/s}$
• D. $47.4\text{m/s}$

Answer 1

The correct option is C $79\text{m/s}$

Given, mass of ball $\left(M\right)=20\text{kg}$
Mass of bullet $\left(m\right)=5\text{kg}$
Initial velocity of bullet $=v_0$

For completing the vertical circular motion, velocity at the bottommost point , $u\ge \sqrt{5gl}$

From PCLM,
$m{v}_0=(M+m)u$ where $u$ is the common velocity of ball + bullet after collsion.
$\Rightarrow u=\frac{m{v}_0}{M+m}$

$\therefore \frac{m{v}_0}{M+m}\ge \sqrt{5gl}$
$\Rightarrow v_0\ge \frac{M+m}{m}\times \sqrt{5gl}$
$\Rightarrow (v_0{)}_{min}=\frac{M+m}{m}\times \sqrt{5gl}$
$=\frac{25}{5}\sqrt{5\times 10\times 5}=79\text{m/s}$