JEE Main 2021 · 22 July · evening · Single correct · +4 / −1
A bullet of '4 g' mass is fired from a gun of mass 4 kg. If the bullet moves with the muzzle speed of 50 ms⁻¹, the impulse imparted to the gun and velocity of recoil of gun are :
- (a) 0.2 kg ms⁻¹, 0.1 ms⁻¹
- (b) 0.4 kg ms⁻¹, 0.05 ms⁻¹
- (c) 0.2 kg ms⁻¹, 0.05 ms⁻¹
- (d) 0.4 kg ms⁻¹, 0.1 ms⁻¹
Answer and solution
Answer (c) 0.2 kg ms⁻¹, 0.05 ms⁻¹
Solution
Physics sir · 4 steps
Whatever momentum the bullet gets forward, the gun gets backward. Find that first, then divide by 4 kg.
- Momentum is conserved. Gun and bullet start at rest, so their total momentum is zero before and after firing. The gun must carry momentum equal and opposite to the bullet's.
- Bullet's momentum. Convert the mass first: 4 g = 0.004 kg. Then p = 0.004 × 50 = 0.2 kg m s⁻¹.
- Impulse on gun. Impulse is the change in momentum. The gun goes from 0 to 0.2 kg m s⁻¹ backward, so the impulse on it is 0.2 kg m s⁻¹.
V_(gun) = (0.2 kg m s⁻¹)/(4 kg) = 0.05 m s⁻¹
- So. Impulse is 0.2 kg m s⁻¹ and the recoil speed is 0.05 m s⁻¹, which is option (c).
The idea to keep. In firing and explosion problems, the impulse on one part equals the momentum the other part gains. Divide that impulse by a body's mass to get its velocity.
Answer from the NTA final answer key. Solution by PrepWiser.