# Conservation of Momentum

We saw that one unit of force (one Newton) is that quantity of force which can give an acceleration of 1 m/s^{2}to a body of mass 1 kg. Now, suppose, you have two balls, a red one A and a blue one B. Let their masses be m

_{A}and m

_{B}. You place them on a smooth and levelled floor and slowly roll them in the same direction with different velocities u

_{A}and u

_{B}. Let u

_{A}be greater than u

_{B}. Suppose there are no other external unbalanced forces acting on them. Let the two balls collide. Let the collision last for 't' seconds. During the collision, the red ball A exerts a force F

_{A}on the blue ball B and the blue ball exerts an equal force F

_{B}on the red ball.

Suppose the velocities change to v_{A} and v_{B} after the collision. The momenta of the ball A before and after the collision will be m_{A} x u_{A} and m_{A} x v_{A}. The rate of change of momentum for F_{A} (action) during the collision = m_{A} x (v_{A} â€“ u_{A}) / t.

Similarly, the rate of change of momentum for F_{B} (reaction) during the collision will be

m_{B} x (v_{B} â€“ u_{B}) / t.

By Newtonâ€™s third law we know action and reaction are equal and opposite.

Therefore, F_{A} = - F_{B }or m_{A} x (v_{A} â€“ u_{A}) / t = - m_{B} x (v_{B} â€“ u_{B}) / t.

m_{A} x v_{A} â€“ m_{A} x u_{A} = - m_{B} x v_{B }+ m_{B} x u_{B}

m_{A} x v_{A} + m _{B}x v_{B} = m_{B} x u_{B} + m_{A} x u_{A}

From the above equation, we see that the total momentum before and after the collision remains the same when there is external unbalanced force acting on them. In other words momentum is conserved; it remains unchanged after the collision.

You can also see that the quantity of force applied by the red ball on the blue will not depend only on the mass of the red ball but also on its velocity.

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