Question

A bullet of mass 10 g travelling at a speed of 300m/s penetrates
deeply into a fixed target and comes to rest after travelling 2 m.
Calculate the average force exerted on the bulletq​

Answer 1

Required Answer :

The average force exerted on the bullet = - 225 N

Given :

• Mass of the bullet = 10 g

• Initial velocity of the bullet = 300 m/s

• Final velocity of the bullet = 0 m/s [Final velocity of the bullet is zero because it comes to rest.]

• Distance travelled by the bullet = 2 m

To find :

• Average force exerted on the bullet

Solution :

To calculate the average force exerted on the bullet, firstly we will calculate the acceleration of the bullet by using the third equation of motion. And then by using the formula of force we will get our required answer.

Calculating the acceleration :-

Third equation of motion :-

• v² - u² = 2as

where,

• v denotes the final velocity
• u denotes the initial velocity
• a denotes the acceleration
• s denotes the distance/displacement

we have,

• Final velocity (v) = 0 m/s
• Initial velocity (u) = 300 m/s
• Distance (s) = 2 m

Substituting the given values :-

→ (0)² - (300)² = 2(a)(2)

→ - (300)² = 4 × a

→ - 90000 = 4 × a

→ - 90000 ÷ 4 = a

→ - 22,500 = a

Therefore, the acceleration of the bullet = - 22,500 m/s²

Now, calculating the force :-

Using formula,

• Force = ma

where,

• m denotes the mass
• a denotes the acceleration

we have,

• Mass (m) = 10 g
• Acceleration (a) = - 22,500 m/s²

Firstly, we will convert the mass of the bullet from grams into kg.

As we know,

• 1 kg = 1000 g

→ Mass = 10/1000

→ Mass = 1/100

→ Mass = 0.01 kg

Now, substituting the values ::

→ Force = 0.01 × - 22,500

→ Force = - 225

Therefore, the average force exerted on the bullet = - 225 N