# Prove how can LAW of CONSERVATION of MOMEMTUM IS SAME BEFORE AND AFTER COLLISION?

Principle of conservation of linear momentum states that if the net external force on a system of particles is zero, the linear momentum of the system remains constant. Or we can say that the sum of momenta of the two objects before collision is equal to the sum of momenta after the collision provided there is no external unbalanced force acting on them.

Suppose two objects (two balls A and B, say) of masses mA and mB are travelling in the same direction along a straight line at different velocities uA and uB, respectively and there are no other external unbalanced forces acting on them. Let vA and vB are the velocities of the two balls A and B after the collision, respectively.

$\\ \text{The momenta (plural of momentum) of ball A before and after the collision are}\ m_{A} u_{A} and m_{A} v_{A}, respectively.\\ The rate of change of its momentum (or F_{AB}, action) during the collision will be m_{A} \frac{\left(v_{A}-u_{A}\right)}{t} \\ Similarly, the rate of change of momentum of ball \mathrm{B}\left(=F_{BA}, \text { or reaction }\right) during the collision will be m_{B} \frac{\left(v_{B}-u_{B}\right)}{t}$

$\\ \text{According to the third law of motion, the force} \ F_{A B} \ \text{exerted by ball A on ball B} (action)\\ and\ the\ force\ F_{BA} \ \text{exerted by the ball B on ball A }\ (reaction) \text{must be equal and opposite to each other. Therefore}$

$\\ \begin{array}{l} F_{A B}=-F_{B A} \\ m_{A} \frac{\left(v_{A}-u_{A}\right)}{t}=-m_{B} \frac{\left(v_{B}-u_{B}\right)}{t} \end{array} \\ This \ gives\\ m_{A} u_{A}+m_{B} u_{B}=m_{A} v_{A}+m_{B} v_{B}$

### Preparation Products

##### JEE Main Rank Booster 2021

This course will help student to be better prepared and study in the right direction for JEE Main..

₹ 13999/- ₹ 9999/-
##### Rank Booster NEET 2021

This course will help student to be better prepared and study in the right direction for NEET..

₹ 13999/- ₹ 9999/-
##### Knockout JEE Main April 2021 (Easy Installments)

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 4999/-
##### Knockout NEET May 2021

An exhaustive E-learning program for the complete preparation of NEET..

₹ 22999/- ₹ 14999/-