Question

A bullet of mass 10g is fired with a rifle. The bullet taken 0.003 second to move through the barrel and leaves it with velocity of 300m/s. what is the force exerted on the bullet by the rifle ​

Answer 1

Given that, a bullet of mass 10g is fired with a rifle. The bullet taken 0.003 second to move through the barrel and leaves it with velocity of 300m/s.

Here, mass (m) of the bullet is 10 g and time (t) is 0.003 sec.

Also, the final velocity (v) with which the bullet leaves from the barrel is 300 m/s and Initial Velocity (u) of the bullet is 0 m/s.

We have to find the force exerted on the bullet by the rifle.

Firstly convert mass of the bullet in kg. To convert g into kg, divide the given value by 1000.

So, 10 g = 0.01 kg

Now, using the First Equation Of Motion,

v = u + at

300 = 0 + a(0.003)

300 = 0.003a

300/0.003 = a

10000 = a

Therefore, the acceleration of the bullet is 10000 m/s².

Now,

Force is defined as the product of mass and acceleration.

F = ma

F = 0.01 × 10000

F = 100 N

Therefore, the force exerted on the bullet by the rifle is 100N.

Answer 2

GIVEN:

• Mass( m ) of bullet = 10g = 0.01 kg
• Time( t ) taken = 0.003 s
• Final velocity ( v ) = 300 m/s
• Initial velocity ( u ) = 0

TO FIND:

• Force exerted on bullet

SOLUTION:

Using the first Kinematic equation of motion

v = u + at

Where

• v = final velocity = 300 m/s
• u = initial velocity = 0 m/s
• t = time = 0.003 s
• a = acceleration

Substituting the values we get

→ 300 = 0 + a(0.003)

→ a = 300/0.003

→ a = 10000 m/s²

Hence, acceleration = 1000 m/s²

We know that

Force ( f ) = mass ( m ) × acceleration ( a )

→ F = 0.01 × 10000

→ F = 100N

Hence , force exerted on bullet = 100N