Physics · Laws Of Motion · NEET
Before firing, the gun and bullet are at rest, so total momentum is zero. Firing is an internal force (the explosion) and internal forces cannot change total momentum. So after firing the momentum must still add up to zero. The bullet moves forward with momentum m x v, so the gun must move backward with equal momentum M x V to cancel it. That backward motion of the gun is the recoil.
Use conservation of momentum. Initial momentum = 0. Final momentum = m x v (bullet, forward) + M x V (gun, backward). Set them equal: 0 = m x v + M x V. So V = -(m x v)/M. The minus sign just means the gun moves opposite to the bullet. Recoil speed = m x v / M.
Because the gun is much heavier than the bullet. Both carry the same size of momentum (m x v = M x V), but the gun has a large mass M, so its velocity V = (m/M) x v is small. A 4 kg gun firing a 0.02 kg bullet at 500 m/s recoils at only (0.02/4) x 500 = 2.5 m/s. Small mass fast, large mass slow, same momentum.
They are the same physics said two ways. The rocket pushes hot gas backward (action) and the gas pushes the rocket forward (reaction) - that is the third law. At the same time, the total momentum of rocket plus ejected gas is conserved, so as gas leaves backward the rocket gains forward momentum. In NEET, both statements are correct; pick whichever the question asks for.
A rocket does not push against air. It pushes against its own ejected gas. By conservation of momentum, throwing mass (gas) out the back at high speed gives the rocket forward momentum. This works even better in empty space because there is no air drag. The idea that a rocket needs air to push on is a common wrong belief.
Thrust is the forward force from ejecting gas. Thrust F = v(gas) x (dm/dt), where v(gas) is the speed of the exhaust relative to the rocket and dm/dt is the rate at which mass is thrown out per second. For NEET numericals, thrust force = exhaust speed x burn rate (kg/s).
Try the real previous-year questions from this chapter — each with the answer and a full solution.
V = (m x v) / M, where m and v are the mass and velocity of the bullet and M is the mass of the gun. The gun moves opposite to the bullet.
Yes. The explosion is an internal force to the gun+bullet system. Internal forces come in action-reaction pairs and cancel, so they cannot change the total momentum. Total momentum stays zero.
A lighter gun recoils faster because V = (m/M) x v. Higher recoil speed means more recoil kinetic energy delivered to your shoulder, so it kicks harder.
No. Momentum is conserved but kinetic energy is not. The chemical energy of the gunpowder becomes the kinetic energy of the bullet and gun plus heat and sound. The system starts with zero KE and ends with a lot, so KE increases.
A gun ejects mass once, so it is a single-shot momentum transfer. A rocket ejects gas continuously, so its mass keeps changing and it keeps accelerating. Both rely on the same conservation of momentum idea.
