**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, m_{1 }= 10 g = 10 x 10^{-3} kg = 10^{-2} kg

mass of the gun, m_{2} = 2 kg

initial velocity of the bullet, u_{1} = 0 ms^{-1}

initial velocity of the gun, u_{2} = 0 ms^{-1}

final velocity of the bullet, v_{1} = 500 ms^{-1}

final velocity of the gun, v_{2} = ?

We get,

**m _{1}u_{1} + m_{2}u_{2} = m_{1}v_{1} + m_{2}v_{2}**

** m _{1} x 0 ms^{-1} + m2 kg x 0 ms^{-1 }= 10^{-2} kg x 500 ms^{-1} + 2 kg x v_{2} **

**v _{2} = 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}**

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