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:

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)_