Question

A 100 kg gun fires a ball of 1 kg horizontally from a cliff of height 500m. 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 = $10m{s}^{-2}$)

• 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)

0.4 $m{s}^{-1}$

Here,

Mass of the gun, M = 100 kg

Mass of the ball, m= 1 kg

Height of the chiff, h= 500 m

g = 10$m{s}^{-2}$

Time taken by the ball to reach the ground is

$t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2\times 500m}{10m{s}^{-1}}}$

Horizontal distance covered = ut

$\therefore 400=u\times 10$

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

$v=-\frac{mu}{M}=-\frac{\left(1kg\right)\left(40m{s}^{-1}\right)}{100kg}=-0.4m{s}^{-1}$

-ve sign shows that the direction of recoil of the gun is opposite to that of the ball.