First slide

Third law

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 ${ms}^{- 2}$ )

Difficult

A

0.2 ${ms}^{- 1}$
B

0.4 ${ms}^{- 1}$
C

0.6 ${ms}^{- 1}$
D

0.8 ${ms}^{- 1}$

Solution

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

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App