Question

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

Hey Dear,

◆ Answer -

s = 2.25 m

F = -50 N

● Explanation -

# Given -

m = 10 g

u = 150 m/s

t = 0.03 s

# Solution -

Acceleration of the bullet is calculated by -

a = (v-u) / t

a = (0-150) / 0.03

a = -5000 m/s^2

Let s be the distance of penetration of bullet in the block.

s = ut + 1/2 at^2

s = 150×0.03 + 1/2 ×(-5000)×0.03^2

s = 2.25 m

Now, magnitude of the force exerted by block on bullet is -

F = m.a

F = 10×10^-3 × (-5000)

F = -50 N

Thanks dear...