Work Energy And Power Ques 10
10. 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: 10.(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$
BETA
