zatynlpv

2023-03-08

A bullet of mass 10 g travelling horizontally with a velocity of $150m{s}^{-1}$ strikes a stationary wooden block and comes to rest in 0.03 s. Determine the bullet's penetration depth into the block. Calculate the force exerted by the wooden block on the bullet as well.

energiasim4c

Beginner2023-03-09Added 2 answers

Given, initial velocity, $u=150m{s}^{-1}$, final velocity, $v=0$ (because the bullet has finally come to rest), mass of the bullet, $m=10g=0.01kg$ and time taken to come to rest, t=0.03 s.

Let the acceleration of a bullet be a.

We employ the first equation of motion, v=u+at,

$\Rightarrow 0=150+a\times 0.03$

$\Rightarrow a=\frac{-150}{0.03}=-5000m{s}^{-2}$

(A negative sign indicates that the bullet's velocity is decreasing.)

Now we use third equation of motion:

${v}^{2}={u}^{2}+2$as,

$\Rightarrow 0={150}^{2}+2\times (-5000)s$

$\Rightarrow s=\frac{22500}{10000}=2.25m$

Thus, the distance of penetration of the bullet into the block is 2.25 m.

Newton's second law of motion states:

Force, F=m×a=0.01×5000=50 N

Therefore, the magnitude of force exerted by the wooden block on the bullet is 50 N.

Let the acceleration of a bullet be a.

We employ the first equation of motion, v=u+at,

$\Rightarrow 0=150+a\times 0.03$

$\Rightarrow a=\frac{-150}{0.03}=-5000m{s}^{-2}$

(A negative sign indicates that the bullet's velocity is decreasing.)

Now we use third equation of motion:

${v}^{2}={u}^{2}+2$as,

$\Rightarrow 0={150}^{2}+2\times (-5000)s$

$\Rightarrow s=\frac{22500}{10000}=2.25m$

Thus, the distance of penetration of the bullet into the block is 2.25 m.

Newton's second law of motion states:

Force, F=m×a=0.01×5000=50 N

Therefore, the magnitude of force exerted by the wooden block on the bullet is 50 N.

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