Question

A bullet of mass 10 g travlling horizontally with a velocity of 150 m s^-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

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 = 0.01 kg x – 5000 m s–2 = – 50 N

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.