Ncert Exercises & Solutions

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. Find the recoil velocity of the gun.
(acceleration due to gravity = 10 ${ms}^{- 2}$ )

Hint: Apply momentum conservation.

Step 1: Find the time of flight of the ball

Given, the mass of the gun (

m_{1}

) = 100 kg

Mass of the ball (m) = 1 kg
Height of the cliff (h) = 500 m
Horizontal distance traveled by the ball (x) = 400 m

$\begin{array}{l} From h = \frac{1}{2} {gt}^{2} (∵ Initial velocity in downward direction is zero) \\ 500 = \frac{1}{2} \times 10 t^{2} \\ t = \sqrt{100} = 10 s \\ Step 2 : Find horizontal speed of ball \\ From x = ut, u = \frac{x}{t} = \frac{400}{10} = 40 m / s . \end{array}$

Step 2: Find the recoil velocity of the gun by applying momentum conservation.

If v is the recoil velocity of the gun, then according to the principle of conservation of linear momentum,

$m_{1} v = m_{2} u$
$v = \frac{m_{2} u}{m_{1}} = \frac{1}{100} \times 40 = 0 .4 m / s$

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