SATHEE: Work Energy and Power - Result Question 15

Work Energy and Power - Result Question 15

17. A person holding a rifle (mass of person and rifle together is $100 k g$ ) 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 $800 m s^{- 1}$ . The final velocity acquired by the person and the average force exerted on the person are [NEET Kar. 2013]

(a) $- 1.6 m s^{- 1}; 8 N$

(b) $- 0.08 m s^{- 1}; 16 N$

(c) $- 0.8 m s^{- 1}; 8 N$

(d) $- 1.6 m s^{- 1}; 16 N$

Show Answer

Answer:

Correct Answer: 17. (c)

Solution:

(c) According to law of conservation of momentum

$M V + m n v = 0$

$\Rightarrow V = \frac{- m N v}{M} = \frac{- 0.01 k g \times 10 \times 800 m / s}{100}$

$\Rightarrow - 0.8 m / s$

According to work energy theorem,

Average work done $=$ Change in average

kinetic energy

i.e., $F_{a v} \times S_{a v} = \frac{1}{2} m V_{m} s^{2}$

$\Rightarrow \frac{F_{a v} V_{max} t}{2} = \frac{1}{2} m \frac{V_{rms}^{2}}{2}$

$\Rightarrow F_{a v} = 8 N$

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