- Books Name
- Science Made Easy Science Book

- Publication
- Science Made Easy

- Course
- CBSE Class 9

- Subject
- Science

**Law of Conservation of Momentum**

→ When two (or more) bodies act upon one another, their total momentum remains constant (or conserved) provided no external forces are acting.

**• Initial momentum **= Final momentum

Suppose, two objects A and B each of mass m_{1} and mass m_{2} are moving initially with velocities u_{1} and u_{2}, strike each other after time t and start moving with velocities v_{1} and v_{2} respectively.

Now, Initial momentum of object A = m_{1}u_{1}

Initial momentum of object B = m_{2}u_{2}

Final momentum of object A = m_{1}v_{1}

Final momentum of object B = m_{2}v_{2}

So, Rate of change of momentum in A,

F_{1} = (m_{1}v_{1 - m1}u_{1})t = m_{1}(v_{1} - u_{1})/t ....(i)

Rate of change of momentum in B,

F_{2} = (m_{2}v_{2} – m_{2}u_{2})t = m_{2}(v_{2} - u_{2})/t ....(ii)

We know from 3rd law of motion,

F1 = −F2

So, m_{1}(v_{1} - u_{1})/t = - m_{2}(v_{2} - u_{2})/t

⇒ m_{1}v_{1 – }m_{1}u_{1 } = -m_{2}v_{2 }+ m_{2}u_{2}

⇒ m_{1}u_{1 + }m_{2}u_{2 }= m_{1}v_{1 }+ m_{2}v_{2}

Thus, Initial momentum = Final momentum

Example 1: A bullet of mass 20 g is fired horizontally with a velocity of 150 m/s from a pistol of mass 2 kg. Find the recoil velocity of the pistol.

Solution

Given, Mass (m1) of bullet = 20 g = 0.02 kg

Mass (m2) of pistol = 2 kg

Initially bullet is inside the gun and it is not moving.

∵ Mass = m1+m2 = (0.02 + 2) kg = 2.02 kg

and u1 = 0

∴ Initial momentum = 2.02 × 0 = 0 ....(i)

Let the velocity of pistol be v2 and v1 for bullet = 150

∴ Final momentum = m1v1 + m2v2 = 0.02×150 + 2v2 ...(ii)

We know that,

Initial momentum = Final momentum

∴ (0.02×150)/100 + 2v2 = 0 [From equations (i) and (ii)]

⇒ 3 + 2v2 = 0

⇒ 2v2 = −3

⇒ v2 = −1.5 m/s

(−)ve sign indicates that gun recoils in direction opposite to that of the bullet.

Example 2: Two hockey players viz A of mass 50 kg is moving with a velocity of 4 m/s and another one B belonging to opposite team with mass 60 kg is moving with 3 m/s, get entangled while chasing and fall down. Find the velocity with which they fall down and in which direction?

Solution

Given,

m_{A} = 50 kg, u_{A} = 4 m/s

m_{B} = 60 kg, u_{B} = 3 m/s

Initial momentum A = m_{A}u_{A} = 50 × 4 = 200 kgm/s

Initial momentum B = M_{B}u_{B} = 60 × 3 = 180 kgm/s

∴ Total initial momentum = 200 + 180 = 380 kgm/s ....(i)

Final momentum = (m_{A} + m_{B})v = (50 + 60)v = 110v ....(ii)

According to the law of conservation of momentum,

380 = 110v

⇒ v = 380/110 = 3.454 m/s

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###### Science Made Easy

Course : CBSE Class 9

Start Date : 26.09.2022

End Date : 31.12.2022

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###### Science Made Easy

Course : CBSE Class 9

Start Date : 27.09.2022

End Date : 31.12.2022

Types of Batch : Live Online Class