The law of conservation of momentum is one of the most prominent laws in physics. The conservation of momentum law principle tells us that the total momentum of a system is always conserved for an isolated system. Let us learn more about the conservation of momentum along with derivation and solved problems.
| Table of Contents: |
Momentum Conservation Principle
Law of conservation of momentum states that
For two or more bodies in an isolated system acting upon each other, their total momentum remains constant unless an external force is applied. Therefore, momentum can neither be created nor destroyed.
The principle of conservation of momentum is a direct consequence of Newton’s third law of motion.
Read More: Conservation of Momentum
Derivation of Conservation of Momentum
Newton’s third law states that for a force applied by an object A on object B, object B exerts back an equal force in magnitude, but opposite in direction. This idea was used by Newton to derive the law of conservation of momentum.
Consider two colliding particles A and B whose masses are m1 and m2 with initial and final velocities as u1 and v1 of A and u2 and v2 of B. The time of contact between two particles is given as t.
|
\(\begin{array}{l}m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}\end{array} \) |
Therefore, above is the equation of law of conservation of momentum where
Related Articles:
Examples of Law of Conservation of Momentum
Following are the examples of law of conservation of momentum:
- Air-filled balloons
- System of gun and bullet
- Motion of rockets
Solved Problems on Law of Conservation of Momentum
Q1. There are cars with masses 4 kg and 10 kg respectively that are at rest. The car having the mass 10 kg moves towards the east with a velocity of 5 m.s-1. Find the velocity of the car with mass 4 kg with respect to ground.
Ans: Given,
m1 = 4 kg
m2 = 10 kg
v1 = ?
v2 = 5 m.s-1
Pinitial = 0, as the cars are at rest
Pfinal = p1 + p2
Pfinal = m1.v1 + m2.v2
= (4 kg).(v1) + (10 kg).(5 m.s-1)
We know from the law of conservation of momentum that,
Pinitial = Pfinal
0=4 kg.v1+50 kg.m.s-1
v1 = 12.5 m.s-1
Q2. Find the velocity of a bullet of mass 5 grams which is fired from a pistol of mass 1.5 kg. The recoil velocity of the pistol is 1.5 m.s-1.
Ans: Given,
Mass of bullet, m1 = 5 gram = 0.005 kg
Mass of pistol, m2 = 1.5 kg
The velocity of a bullet, v1 = ?
Recoil velocity of pistol, v2 = 1.5 m.s-1
Using law of conservation of momentum,
m1u1 + m2u2 = m1v1 + m2v2
Here, Initial velocity of the bullet, u1 = 0
Initial recoil velocity of a pistol, u2 = 0
∴ (0.005 kg)(0) + (1.5 kg)(0) = (0.005 kg)(v1) + (1.5 kg)(1.5 m.s-1)
0 = (0.005 kg)(v1)+(2.25 kg.m.s-1)
v1=-450 m.s-1
Hence, the recoil velocity of the pistol is 450 m.s-1.
Frequently Asked Questions – FAQs
What is momentum?
Give the formula for law of conservation of momentum
The formula for the law of conservation of momentum is:
List some examples of law of conservation of momentum.
Is momentum a scalar or a vector quantity?
State TRUE or FALSE: As friction increases, momentum decreases.
Put your understanding of this concept to test by answering a few MCQs. Click ‘Start Quiz’ to begin!
Select the correct answer and click on the “Finish” button
Check your score and answers at the end of the quiz
Visit BYJU’S for all Physics related queries and study materials
Your result is as below
Good
Thanks so much for the derivation
Thank you
Very Nice notes and explain
thank you
thank you very much
Very well explained and thanks for giving Questions. This makes it easier 2 understand the concept
Thanks for the Explanation
Excellent
Thank you so much for this derivation.
Extremely helpful
thank you so much i understood it really well