staye and derive mathematical formula for conservation of momentum
Answers
Answer:
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.
A=m1(v1−u1) (change in momentum of particle A)
B=m2(v2−u2) (change in momentum of particle B)
FBA=−FAB (from third law of motion)
FBA=m2∗a2=m2(v2−u2)t FAB=m1∗a1=m1(v1−u1)t m2(v2−u2)t=−m1(v1−u1)t m1u1+m2u2=m1v1+m2v2
Therefore, above is the equation of law of conservation of momentum where m1u1+m2u2 is the representation of total momentum of particles A and B before the collision and m1v1+m2v2 is the representation of total momentum of particles A and B after the collision.
Explanation:
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Answer:
Explanation:
Law of conservation of momentum states that when two objects collide with each other , the sum of their linear momentum always remains same or we can say conserved and is not effected by any action, reaction only in case is no external unbalanced force is applied on the bodies.
Let,
mA = Mass of ball A
mB= Mass of ball B
uA= initial velocity of ball A
uB= initial velocity of ball B
vA= Velocity after the collision of ball A
vB= Velocity after the collision of ball B
Fab= Force exerted by A on B
Fba= Force exerted by B on A
Now,
Change in the momentum of A= momentum of A after the collision - the momentum of A before the collision
= mAvA−mAuA
Rate of change of momentum A= Change in momentum of A/ time taken
= mAvA−mAuA /t
Force exerted by B on A (Fba);
Fba=mAvA−mAuA/t........ [i]
In the same way,
Rate of change of momentum of B=
mbvB−mBuB/t
Force exerted by A on B (Fab)=
Fab=mBvB−mBuB/t.......... [ii]
Newton's third law of motion states that every action has an equal and opposite reaction, then,
Fab=−Fba
Using [i] and [ii] , we have
mBvB−mBuB/t = −mAvA−mAuA/t
mBvB−mBuB=−mAvA+mAuA
Finally we get,
mBvB+mAvA=mBuB+mAuA
This is the derivation of conservation of linear momentum.
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