Question

Class 9 Ch:8 motion


A bullet of mass 10 g travelling 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

Now, it is given that the bullet is travelling with a velocity of 150 m/s.

Thus, when the bullet enters the block, its velocity = 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/s^2

(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

s = -(150)^2/-2(5000) = 22500/1000

= 2.25m

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 = 0.01 kg

Acceleration of the bullet, a = 5000 m/s2

F = ma = 0.01 × 5000 = 50 N

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