A bullet of mass 10 g strikes a wooden block with a velocity of 300 ${m s}^{- 1}$ . After penetrating 20 cm into it, its velocity drops to 200 ${m s}^{- 1}$ . Then average resistance offered by the block is :

125 N

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250 N

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1250 N

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2500 N

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Solution

The correct option is B 1250 N
Given,

$u = 300 m / s$

$v = 200 m / s$

$s = 20 c m = 0.20 m$

$m = 10 g = 0.01 k g$

3rd equation of motion,

$2 a s = v^{2} - u^{2}$

retardation, $a = - 125000 m / s^{2}$

The resistance force, $F = m a$

$F = 0.01 \times 125000$

$F = 1250 N$

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A bullet of mass 10 g strikes a wooden block with a velocity of 300 ms−1. After penetrating 20 cm into it, its velocity drops to 200 ms−1. Then average resistance offered by the block is :

The correct option is B 1250 NGiven,u=300m/sv=200m/ss=20cm=0.20mm=10g=0.01kg3rd equation of motion,2as=v2−u2retardation, a=−125000m/s2The resistance force, F=maF=0.01×125000F=1250N

A bullet of mass 10 g strikes a wooden block with a velocity of 300 ${m s}^{- 1}$ . After penetrating 20 cm into it, its velocity drops to 200 ${m s}^{- 1}$ . Then average resistance offered by the block is :

The correct option is B 1250 N
Given,

$u = 300 m / s$

$v = 200 m / s$

$s = 20 c m = 0.20 m$

$m = 10 g = 0.01 k g$

3rd equation of motion,

$2 a s = v^{2} - u^{2}$

retardation, $a = - 125000 m / s^{2}$

The resistance force, $F = m a$

$F = 0.01 \times 125000$

$F = 1250 N$