Question

A bullet of mass 10 g travelling horizontally with a velocity of 150m/s strikes a stationary wooden block and come to rest in 0.03s.Calculate the distance of penetration of the bullet into the block. Also, calculate the magnitude of force exerted by the wooden block of the bullet ?​

Answer 1

Answer:

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Explanation:

mass =0.01kg

initial velocity=150m/s

t=0.03s

so,

a=(v-u)/t

=(0-150)/0.03

=-5000m/s²

so by using 2nd equation of motion,i.e.,

s=ut+1/2at²

s=150*0.03+1/2*-5000*0.03*0.03

s=22.5-4.5m

s=18m

force=mass * acceleration

=0.01*5000N

=50N

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Answer 2

Answer:

Initial velocity of the bullet, u=150 ms

−1

Final velocity of the bullet, v=0

Mass of the bullet, m=10 g=0.01 kg

Time taken by the bullet to come to rest, t=0.03 s

Let distance of penetration be s

v=u+at

0=150+a(0.03)

⟹a=−5000 ms

−2

(- sign shows retardation)

v

2

=u

2

+2as

0=150

2

+2×(−5000)×s

s=2.25 m

Magnitude of the force exerted by the wooden block, ∣F∣=m∣a∣

∣F∣=0.01×5000=50 N