write the law of conservation of momentum?find the velocity of bullet of mass 5gm which is fired from a pistol of mass 1.5kg. the recoil velocity of pistol is 1.5 m/s
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Derivation of Conservation of Momentum
Newton’s third law states that for a force applied by an object A on object B, object B exerts back an equal force in magnitude, but opposite in direction. This idea was used by Newton to derive the law of conservation of momentum.
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.