Question

A bullet if mass 1kg travelling horizontally with the vellocity of 5 m/s strikes a stationary wooden block and and comes back to rest in 0.2 second.calculate the acceleration produced.​

Answer 1

Answer :

Here, we are given the mass of the bullet, initial velocity and the time in which the bullet comes to rest. We need to calculate the acceleration produced by the bullet. To find the acceleration we will use the first equation of motion.

Solution :

Given :

• Mass of the bullet = 1 kg

• Initial velocity of the bullet = 5 m/s

• Final velocity of the bullet = 0 m/s (because it comes to rest.)

• Time taken by the bullet to come to rest = 0.2 seconds

To find :

• Acceleration produced by the bullet

Required Solution :

First equation of motion $:$

• v = u + at

where,

• v denotes the final velocity
• u denotes the initial velocity
• a denotes the acceleration
• t denotes the time

we have,

• v = 0 m/s
• u = 5 m/s
• t = 0.2 seconds

Substituting the given values :

⇒ 0 = 5 + (a)(0.2)

⇒ - 5 = a × 0.2

⇒ - 5 = a × 2/10

⇒ - 5 × 10/2 = a

⇒ - 5 × 5 = a

⇒ - 25 = a

Acceleration produced by the bullet = - 25 m/s²