Work Energy and Power - Result Question 15
17. A person holding a rifle (mass of person and rifle together is $100 kg$ ) stands on a smooth surface and fires 10 shots horizontally, in $5 s$. Each bullet has a mass of $10 g$ with a muzzle velocity of $800 ms^{-1}$. The final velocity acquired by the person and the average force exerted on the person are [NEET Kar. 2013]
(a) $-1.6 ms^{-1} ; 8 N$
(b) $-0.08 ms^{-1} ; 16 N$
(c) $-0.8 ms^{-1} ; 8 N$
(d) $-1.6 ms^{-1} ; 16 N$
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Answer:
Correct Answer: 17. (c)
Solution:
- (c) According to law of conservation of momentum
$MV+mnv=0$
$\Rightarrow V=\frac{-m N v}{M}=\frac{-0.01 kg \times 10 \times 800 m / s}{100}$
$\Rightarrow-0.8 m / s$
According to work energy theorem,
Average work done $=$ Change in average
kinetic energy
i.e., $F _{av} \times S _{av}=\frac{1}{2} m V_ms^{2}$
$\Rightarrow \frac{F _{av} V _{\max } t}{2}=\frac{1}{2} m \frac{V _{\text{rms }}^{2}}{2}$
$\Rightarrow F _{av}=8 N$
