Question

A bullet of mass 12g travelling horizontally with a

velocity of 155m/s strikes a stationary wooden block and

comes to rest in 0.05second. Calculate the acceleration

produced and magnitude of the force exerted by the

wooden block on the bullet.

​

Answer 1

Given :

• Mass of the bullet = 12g = 0.012 kg
• Initial velocity = 155 m/s
• Final velocity = 0 m/s [rest]
• Time taken = 0.05 s

Concept :

• First, we need to find the acceleration of the bullet. As the bullet is moving uniformly we can use the first equation of motion in order to find acceleration.

• Now that we have the acceleration of the bullet we can easily find the force by using Newton's second law.

Solution :

As per the first equation of motion,

$\implies\sf{v = u + at }$

On rearranging,

$\implies\sf{a = \dfrac{v-u}{t} }$

Now, let's substitute the values,

$\implies\sf{a = \dfrac{0 -155}{0.05} }$

$\implies\boxed{\sf{a = -3100 m/s^{2} }}$

The acceleration of the bullet is -3,100 m/s².

____________________

Note: here the negative sign denotes retardation.

As per Newton's Second Law,

$\implies\sf{F = ma }$

$\implies\sf{F = 0.012 \times -3100 }$

$\implies\boxed{\sf{F = - 37.2 N }}$

The magnitude of the force exerted by the wooden block on the bullet is 37.2N.