Question

A bullet of mass 10 g travelling horizontally with a velocity of 150 m/s 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:

Initial velocity, u= 150 m/s

Final velocity, v= 0 (since the bullet finally comes to rest)

Time taken to come to rest, t= 0.03 s

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 = -5000 m/s2

(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)

= 22500 / 10000

= 2.25 m

Answer 2

Answer:

Given parameters,

Initial velocity (u) = 150 m/s

Final velocity (v) = 0 (since the bullet finally comes to rest)

Time taken to come to rest (t) = 0.03 s

Find out

The distance of penetration of the bullet into the block. Also the magnitude of the force exerted by the wooden block on the bullet.

Solution

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

Hence, the magnitude of the force exerted by the wooden block on the bullet is 50 N.