Question

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

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

Answer 1

Answer: (B)

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