What is the derivation for law …
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Posted by Naman Kaur 5 years, 9 months ago
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Yogita Ingle 5 years, 9 months ago
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 v2of 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 collision and m1v1+m2v2 is the representation of total momentum of particles A and B after collision.
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