Question

expand law of conservation of momentum in mathematically​

Answer 1

Let us consider a situation wherein: a truck of mass m1, velocity u1 and its momentum = m1u1 and a car of mass m2, velocity u2 and its momentum m2u2; are moving in the same direction but with different speeds. Therefore, total momentum=m1u1 + m2u2.

Now suppose the car and truck collide for a short time t, their velocities will change. So now the velocity of the truck and car become v1 and v2 respectively. However, their mass remains the same. Hence, now the total momentum = m1v1 + m2v2

Acceleration of car (a) = (v2–u2)/t

Also, F = ma

F1 = Force exerted by truck on the car

F1 = m2(v2–u2)/t

Acceleration of truck =(v1–u1)/t

F2 = m1(v1–u1)/t and F1 = –F2

m2(v2– u2)/t = –m1(v1– u1)/t

m1u1+m2u2 = m2v2+m1v1

Hope it helps you....

Answer 2

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 collision and m1v1+m2v2 is the representation of total momentum of particles A and B after collision.