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