Question

A bullet is fired at a velocity of 100ms-1 . It strikes a wooden block and stops after penetrating a certain distance in it. If the bullet takes 0.5s to stop, find its distance of penetration.​

Answer 1

Answer:

According to the first equation of motion,

v= u + at

Acceleration of the bullet (a)

0 = 150 + (a — 0.03 s)

a = -150 / 0.03

a = – 5000 m/s2

Here the negative sign indicates that the velocity of the bullet is decreasing.

According to the third equation of motion,

v2= u2+ 2as

0 = (150)2+ 2 (-5000)

s = 22500 / 10000

s = 2.25 m

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

From Newton’s second law of motion,

Force (F) = Mass × Acceleration

Mass of the bullet (m) = 10 g

m = 0.01 kg

Acceleration of the bullet (a) = 5000 m/s2

F = m × a

F = 0.01× -5000

F = -50 N

Explanation:

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