A gun of mass 4 kg fires a 0.02 kg bullet at 200 m per second. By conservation of momentum, the recoil velocity of the gun is:
A4 m per second
B1 m per second
C2 m per second
D0.5 m per second
Answer & Solution
Correct answer: B. 1 m per second
1. Before firing, total momentum of (gun + bullet) is zero.
2. After firing, momentum must still be zero (conservation of momentum).
3. Let V be the gun's recoil velocity. Then 4·V + 0.02 × 200 = 0.
4. 4·V + 4 = 0.
5. V = −1 m per second (negative sign means opposite direction to the bullet).
6. The magnitude of the recoil velocity is 1 m per second.
_Source: Selina Concise Physics ICSE Class 9, Ch 3 'Laws of Motion' (aplustopper.com extract)_
Related questions
A constant force of 6 N acts for 4 s on a 2 kg body initially at rest. The kinetic energy A stone of mass 0.5 kg is whirled in a horizontal circle of radius 1 m at constant angularA car of mass 1000 kg accelerates from rest to 20 m s⁻¹ in 5 s on a level road. The net hoA man weighing 50 kg jumps off a stationary boat of mass 250 kg into water with a horizontNewton's third law states that 'action and reaction are equal and opposite'. The most accuTwo identical billiard balls strike a smooth rigid wall with equal speeds u. Ball A strikeA force F acts on a particle making an angle θ with its instantaneous velocity. The componA cricketer draws his hands back while catching a fast ball. The physical reason is that: