Q. A shell of mass 200 g is ejected from a gun of mass 4 kg by an explosion that generates 1.05 kJ of energy. The initial velocity of the shell is
A) 120 m s⁻¹
B) 100 m s⁻¹
C) 80 m s⁻¹
D) 40 m s⁻¹
Explanation
[Solution from source] Momentum conservation: 0.2v = 4V, so the gun recoils with V = v/20. Energy: (1/2) × 0.2 × v² + (1/2) × 4 × (v/20)² = 0.1v² + 0.005v² = 0.105v² = 1050, so v² = 10⁴ and v = 100 m s⁻¹. Ignoring the gun's recoil energy would give about 102 m s⁻¹, so the recoil share must be included. Answer (2).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.