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

Given :-

● A bullet of mass 10g moving with a velocity of 150m/s

● A bullet comes to rest in 0.03 s

Solution :-

Here,

Initial velocity of bullet ( u ) = 150 m/s

Final velocity of bullet ( v ) = 0

( it comes to at rest position)

Time taken by the bullet to come to at rest position ( t) = 0.03 s

Now,

By using first equation of motion,

v = u + at

Put the required values,

0 = 150 + a * 0.03

0 = 150 + 0.03a

-0.03a = 150

a = 150/-0.03

a = -5000m/s^2

( Negative sign indicates that the velocity of bullet is decreasing)

Thus, The acceleration of bullet is -5000m/s^2

Now,

( We have to find distance of penetration of the bullet into block)

Therefore,

By using third equation of motion

2as = v^2 - u^2

2( -5000) *s = 0 - ( 150)^2

-10000s = -22500

s = -22500/-10000

s = 2.25m

Thus, The distance of penetration of the bullet into the block is 2.25m

Now,

( We have to calculate the magnitude)

As we know that,

F = mass * acceleration

Here,

Mass = 10g = 0.01kg

( SI unit of mass is kg)

Acceleration = 5000

( Acceleration be taken in positive because we have to find the magnitude)

Therefore,

Force = 0.01 * 5000

Force = 50N

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

Three equations of motion are :-

● v = u + at

● S = ut + 1/2at^2

● 2as = v^2 - u^2