JEE Mains · Physics · STD 11 - 4.1 newtons laws of motion
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 \(0.4\, {kg} \,{ms}^{-1}, 0.1\, {ms}^{-1}\)
- B \(0.2 \,{kg} \,{ms}^{-1}, 0.1\, {ms}^{-1}\)
- C \(0.2 \,{kg} \,{ms}^{-1}, 0.05\, {ms}^{-1}\)
- D \(0.4 \,{kg}\, {ms}^{-1}, 0.05 \,{ms}^{-1}\)
Answer & Solution
Correct Answer
(C) \(0.2 \,{kg} \,{ms}^{-1}, 0.05\, {ms}^{-1}\)
Step-by-step Solution
Detailed explanation
By momentum conservation \(4 \times 10^{-3}(50-v)-4 v=0\) \(v=\frac{4 \times 10^{-3} \times 50}{4+4 \times 10^{-3}} \approx 0.05\, m s^{-1}\) Impulse \(=J=m v=4 \times 0.05=0.2 \,{kgms}^{-1}\)
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