Mathematical Example: Newton’s 3rd Law

Example 1: A force of 2000 N acts on a body of mass 20 kg for a time of 0.1 s. What is the change of momentum of the body?

Solution:

Here,

applied force, F= 2000 N

time duration, t = 0.1 s

change of momentum, mv -mu =?

We know,

change of momentum = force x time,

mv — mu = Ft

=2000 N x 0.1 s

= 200 kg ms^-2 s =200 kg ms^-1

Ans: change of momentum = 200 kg ms^-1

Mathematical Example 2: A bullet of mass 10 g was shot from a gun with a velocity of 500 ms^-1. If the mass of the gun is 2 kg, find the backward velocity of the gun.

Solution:

Let the direction of the bullet’s velocity i.e. the forward direction be positive. From the conservation of momentum,

Here,

mass of the bullet, m₁ = 10 g = 10 x 10^-3 kg = 10^-2 kg

mass of the gun, m₂ = 2 kg

initial velocity of the bullet, u₁ = 0 ms^-1

initial velocity of the gun, u₂ = 0 ms^-1

final velocity of the bullet, v₁ = 500 ms^-1

final velocity of the gun, v₂ = ?

We get,

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

 m₁ x 0 ms^-1 + m2 kg x 0 ms^-1 = 10^-2 kg x 500 ms^-1 + 2 kg x v₂

v₂ =  kg ms^-1/ 2kg = -2.5 ms^-1

Here, the velocity of the gun is negative, i.e. the gun will move backward.

Ans: backward velocity = 2.5 ms^-1