A bullet of mass $10\ g$ traveling horizontally with a velocity of $150\ ms^{-1}$ strikes a stationary wooden block and comes to rest in $0.03\ s$. Calculate the distance of penetration of the bullet into the block. Also, calculate the magnitude of the force exerted by the wooden block on the bullet.

Given: A bullet of mass $10\ g$ traveling horizontally with a velocity of $150\ ms^{-1}$ strikes a stationary wooden block and comes to rest in $0.03\ s$.

To do: To calculate the distance of penetration of the bullet into the block. Also, to calculate the magnitude of the force exerted by the wooden block on the bullet.

Solution:

Mass of the bullet $m=10\ g=\frac{10}{1000}\ kg.=\frac{1}{100}\ kg.$

Initial velocity $u=150\ ms^{-1}$

Final velocity $v=0$

Time $t=0.03\ s$

Therefore, acceleration $a=\frac{v-u}{t}$

$=\frac{0-150}{0.03}$

$=-500\ ms^{-2}$

Let $s$ be the distance of penetration of the bullet into the block.

On using $v^2=u^2+2as$

$0^2=150^2+2\times(-150)s$

Or $0=22500-300s$

Or $300s=22500$

Or $s=\frac{22500}{300}$

Or $s=75\ m$

And the force exerted by the wooden block on the bullet $F=ma$

$=\frac{1}{100}\ kg.\times -500\ ms^{-2}$

$=-5\ N$     [$-ve$ sign indicates the opposite direction]

Thus, the distance of penetration of the bullet into the block is $75\ m$, and the force exerted by the wooden block on the bullet is $-5\ N$.

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Updated on: 10-Oct-2022

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