If a bullet of mass 10 gm moving with velocity 150m/sec penetrates the wooden block upto 6 cm. Then the average force imposed by the bullet on the block is

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Tardigrade Admin · Accepted Answer

Using v 2= u 2+2 as 0 = (150)2 + 2 × a × 0.06 (-22500/2 × 0.06) = a ⇒ a = (-22500/0.12) (-ve sign indicates opposite direction) F = 102 × (22500/0.12) = 1875 N

Q. If a bullet of mass $10 g m$ moving with velocity $150 m / sec$ penetrates the wooden block upto $6 c m$ . Then the average force imposed by the bullet on the block is

1875 N

1900 N

2000 N

1500 N

Using $v^{2} = u^{2} + 2 a s$
$0 = (150)^{2} + 2 \times a \times 0.06$
$\frac{- 22500}{2 \times 0.06} = a$
$\Rightarrow a = \frac{- 22500}{0.12}$
(-ve sign indicates opposite direction)
$F = 1 0^{2} \times \frac{22500}{0.12} = 1875 N$

Q. If a bullet of mass 10gm moving with velocity 150m/sec penetrates the wooden block upto 6cm. Then the average force imposed by the bullet on the block is

1875 N

1900 N

2000 N

1500 N

Using v2=u2+2as 0=(150)2+2×a×0.06 2×0.06−22500​=a ⇒a=0.12−22500​ (-ve sign indicates opposite direction) F=102×0.1222500​=1875N

Q. If a bullet of mass $10 g m$ moving with velocity $150 m / sec$ penetrates the wooden block upto $6 c m$ . Then the average force imposed by the bullet on the block is

Using $v^{2} = u^{2} + 2 a s$
$0 = (150)^{2} + 2 \times a \times 0.06$
$\frac{- 22500}{2 \times 0.06} = a$
$\Rightarrow a = \frac{- 22500}{0.12}$
(-ve sign indicates opposite direction)
$F = 1 0^{2} \times \frac{22500}{0.12} = 1875 N$