If a bullet of mass 5 gm moving with velocity 100 m /sec, penetrates the wooden block upto 6 cm. Then the average force imposed by the bullet on the block is

8300 N

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417 N

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830 N

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Zero

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Solution

The correct option is B 417 N
u = 100 m/s, v = 0, s = 0.06 m
Retardation = $a = \frac{u^{2}}{2 s} = \frac{(100)^{2}}{2 \times 0.06} = \frac{1 \times 10^{6}}{12}$
$∴ F o r c e = m a = \frac{5 \times 10^{- 3} \times 1 \times 10^{6}}{12} = \frac{5000}{12} = 417 N$

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If a bullet of mass 5 gm moving with velocity 100 m /sec, penetrates the wooden block upto 6 cm. Then the average force imposed by the bullet on the block is

The correct option is B 417 Nu = 100 m/s, v = 0, s = 0.06 m Retardation = a=u22s=(100)22×0.06=1×10612 ∴Force=ma=5×10−3×1×10612=500012=417 N

The correct option is B 417 N
u = 100 m/s, v = 0, s = 0.06 m
Retardation = $a = \frac{u^{2}}{2 s} = \frac{(100)^{2}}{2 \times 0.06} = \frac{1 \times 10^{6}}{12}$
$∴ F o r c e = m a = \frac{5 \times 10^{- 3} \times 1 \times 10^{6}}{12} = \frac{5000}{12} = 417 N$