Question

A $100 \,kg$ gun fires a ball of $1\, kg$ horizontally from a cliff of height $500 \,m. $ It falls on the ground at a distance of $400\, m $ from the bottom of the cliff. The recoil velocity of the gun is (Take $\,g=10 \,m\, s^{-2}$)

• A. $0.2\, m$ $s^{-1}$
• B. $0.4 \,m$ $s^{-1}$
• C. $0.6 \,m$ $s^{-1}$
• D. $0.8\, m$ $s^{-1}$

Answer 1

Answer: (B)

Here, Mass of the gun, $M = 100\, kg$
Mass of the ball, $m$ $= 1 \,kg$
Height of the cliff, $h$ $= 500\, m$
$g=10 \,m$ $s^{-2}$
Time taken by the ball to reach the ground is
$t=\sqrt{\frac{2 h}{g}}$ $=\sqrt{\frac{2\times500 \,m}{10 \, m \,s^{-2}}}=10 \, s$
Horizontal distance covered $=ut$
$\therefore \,$ $400=u \times10$
where $u$ is the velocity of the ball.
$u=40\, m$ $s^{-1}$
According to law of conservation of linear momentum, we get,
$0 = Mv + mu$
$v=-\frac{m \,u}{M}$ $=-\frac{\left(1 \, kg\right)\left(40\, m \,s^{-1}\right)}{100 \, kg}$ $=-0.4 \, m \, s^{-1}$
$-ve $ sign shows that the direction of recoil of the gun is opposite to that of the ball.