Law of conservation of Momentum:
From Newtons third law of motion we know that whenever a force is applied on a body there will be an equal and opposite reaction. Action and reaction forces result in change in velocities of both the bodies which in turn change the momentum of the bodies.
In an elastic collision the initial momentum of the bodies before collision is found to be equal to the final momentum of the bodies after collision.
Law of conservation of motion states that if a group of bodies are exerting force on each other. ie. Interacting with each other, their total momentum remains constant before and after the collision provided there is no external 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 [Fig.(a)]. And there are no other external unbalanced forces acting on them. Let uA > uB and the two balls collide with each other as shown in Fig.(b). During collision which lasts for a time t, the ball A exerts a force FAB on ball B and the ball B exerts a force FBA on ball A. Suppose vA and vB are the velocities of the two balls A and B after the collision, respectively [Fig.(c)].
the momenta (plural of momentum) of ball A before and after the collision are mAuA and mAvA, respectively. The rate of change of its momentum (or FAB) during collision will be
`"m"_"A"("v"_"A" - "u"_"A")/"t"`
Similarly, the rate of change of momentum of ball B=(FBA) during the collison will be
`"m"_"B"("v"_"B" - "u"_"B")/"t"`
According to the third law of motion, the force FAB exerted by ball A on ball B and the force FBA exerted by the ball B on ball A must be equal and opposite to each other. Therefore, FAB = – FBA or
`"m"_"A"("v"_"A" - "u"_"A")/"t"`= `"m"_"B"("v"_"B" - "u"_"B")/"t"`
mAuA + mBuB = mAvA + mBvB
Since (mAuA + mBuB) is the total momentum of the two balls A and B before the collision and (mAvA + mBvB) is their total momentum after the collision, we observe that the total momentum of the two balls remains unchanged or conserved provided no other external force acts. As a result of this ideal collision experiment, we 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. This is known as the law of conservation of momentum. This statement can alternatively be given as the total momentum of the two objects is unchanged or conserved by the collision.
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A hockey ball of mass 200 g travelling at 10 m s−1 is struck by a hockey stick so as to return it along its original path with a velocity at 5 m s−1. Calculate the change of momentum occurred in the motion of the hockey ball by the force applied by the hockey stick.
An object of mass 1 kg travelling in a straight line with a velocity of 10 m s−1 collides with, and sticks to, a stationary wooden block of mass 5 kg. Then they both move off together in the same straight line. Calculate the total momentum just before the impact and just after the impact. Also, calculate the velocity of the combined object.
Akhtar, Kiran and Rahul were riding in a motorocar that was moving with a high velocity on an expressway when an insect hit the windshield and got stuck on the windscreen. Akhtar and Kiran started pondering over the situation. Kiran suggested that the insect suffered a greater change in momentum as compared to the change in momentum of the motorcar (because the change in the velocity of the insect was much more than that of the motorcar). Akhtar said that since the motorcar was moving with a larger velocity, it exerted a larger force on the insect. And as a result the insect died. Rahul while putting an entirely new explanation said that both the motorcar and the insect experienced the same force and a change in their momentum. Comment on these suggestions.