Question

A bullet of mass 10g 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 bullet into block. Also force exerted by wooden block on bullet.

Answer 1

Answer:Given, Mass of bullet, m = 10 g = 10/1000 kg = 0.01 kg
Initial velocity of bullet, u = 150 m/s
Since bullet comes to rest, thus final velocity, v =0
Time, t = 0.03 s
Distance of penetration, i.e. Distance, covered (s)=? 
Magnitude of force exerted by wooden block =?

We know that, v=u+atv=u+at
⇒0=150ms−1+a×0.03s⇒0=150ms-1+a×0.03s
⇒−150m/s=a×0.03s⇒-150m/s=a×0.03s
⇒a=−150m/s0.03s=−5000ms−2⇒a=-150m/s0.03s=-5000ms-2

We know that, s=ut+12at2s=ut+12at2
⇒s=150m/s×0.03s⇒s=150m/s×0.03s +12(−5000ms−2)×(0.03s)2+12(-5000ms-2)×(0.03s)2
⇒s=4.5m−2500ms−2×0.0009s2⇒s=4.5m-2500ms-2×0.0009s2
⇒s=4.5m−2.25m⇒s=4.5m-2.25m
⇒s=2.25m⇒s=2.25m

Magnitude of force exerted by wooden block

We know that, Force = mass x acceleration
Or, F=0.01kg×–5000ms−2=−50NF=0.01kg×–5000ms-2=-50N
Therefore, Penetration of bullet in wooden block = 2.25 m

Force exerted by wooden block on bullet = – 50 N. Here negative sign shows that force is exerted in the opposite direction of bullet


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Answer 2

hey this is question of hc verma I think.
and it seems you are in 9.

by the way your answer is here.

m¹=10gn,u¹=150m/s,v=0,t=0.3sec
v=u+at
0=150+a×3/10
-150=3a/10
150×10/3=a
a=-500m/s²

now s=?

v²=u²+2as
0²=150×150+2×-500s.
= 22500-1000s
1000s=22500
s= 22.5m