Question

A bullet of mass 10 gram travelling horizontally with a velocity of 150ms strikes a stationary wooden block and comes to rest in 0.03s. 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

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.

Answer 2

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