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Q4.19: NCERT Physics - Part 1 LAWS OF MOTION 4.19 Solution: A shell of mass 0.020 kg is fired by a gun of mass 100 kg. If…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often forget that the gun‑shell system is initially at rest, so the total momentum is zero; they either omit the negative sign or incorrectly add the velocities, leading to a wrong recoil direction or magnitude. Always set $m_{shell}v_{shell}+m_{gun}V=0$ and solve $V=-(m_{shell}v_{shell})/m_{gun}$, taking the magnitude for speed.

A shell of mass 0.020 kg is fired by a gun of mass 100 kg. If the muzzle speed of the shell is 80 m s__, what is the recoil speed of the gun?

Step-by-Step Verified Solution

**Answer:** V = 0.016 m/s **Step-by-Step Explanation:** By conservation of momentum:
0=mshellvshell+mgunV0 = m_{shell}v_{shell} + m_{gun}V
V=−mshellvshellmgun=−0.020×80100=−0.016 m/sV = -\frac{m_{shell}v_{shell}}{m_{gun}} = -\frac{0.020 \times 80}{100} = -0.016 \text{ m/s}
Recoil speed = 0.016 m/s [citation:10].

All Questions in this Chapter

Q4.1 Q4.2 Q4.3 Q4.4 Q4.5 Q4.6 Q4.7 Q4.8 Q4.9 Q4.10 Q4.11 Q4.12 Q4.13 Q4.14 Q4.15 Q4.16 Q4.17 Q4.18 Q4.19 Q4.20 Q4.21 Q4.22 Q4.23

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