Question

A ball of mass $0.2\text{k}g$ rests on a vertical post of height $5\text{m}$. A bullet of mass $0.01\text{kg}$ travelling with a velocity $V\text{m/s}$ in a horizontal direction, hits the centre of the ball. After the collision, the ball and bullet travel independently. The ball hits the ground at a distance of $20\text{m}$ and the bullet at a distance of $100\text{m}$ from the foot of the post. The initial velocity $V$ of the bullet is,

• A. $250\text{m/s}$
• B. $250\sqrt{2}\text{m/s}$
• C. $400\text{m/s}$
• D. $500\text{m/s}$

Answer 1

The correct option is D $500\text{m/s}$

Since the height of the tower is $5\text{m}$ and just after collision, ball and bullet both travel in vertical direction. Thus,
time of flight,

$t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2\times 5}{10}}=1\text{s}$

After collision, let the velocity of ball be $V_{\text{ball}}$ and that of bullet be $V_{\text{bullet}}$. So,

$V_{\text{ball}}\times t=20$

$\Rightarrow V_{\text{ball}}=20\text{m/s}$

and $V_{\text{bullet}}=100\text{m/s}$

Applying the momentum conservation principle before and after the collision,

$m_{\text{bullet}}V=m_{\text{bullet}}{V}_{\text{bullet}}+m_{\text{ball}}{V}_{\text{ball}}$

$\Rightarrow 0.01\times V=0.01\times 100+0.2\times 20$

$\therefore V=500\text{m/s}$

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