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Two objects with initial masses of m1 and m2 and initial velocity of u1 and u2 with final velocities after collision to be v1 and v2 after time t.
F1 = M2(V2-U2)/T
F2 = M1(V1-U1)/T
F1 = −F2
M2(V2-U2)/T=-( M1(V1-U1)/T)
M2V2 - M2U2 /T = - (M1V1 - M1U1 )/T
M2V2 - M2U2 = -M1V1 + M1U1
M2V2+M1V1 = M2U2 - M1U1
Therefore total momentum before and after collision is equal.
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Suppose 2 bodies A & B are moving on frictionless surface of masses m1 & m2 such that (m1>m2) with velocity u1 & u2 before collision. After their collision they move in their original velocities v1 & v2.

The change in momentum in A = m1v1 - m1u1

The force by A on B [F1] = m1v1 - m1u1 / t (change in momentum / time)........(i)

The change in momentum in B = m2v2 - m2u2

The force by B on A [F2] = m2v2 - m2u2 / t.............(ii)

BY NEWTON'S 3RD LAW F1 = - F2 ( F action = - F reaction)

m1v1 - m1u1 / t = - ( m2v2 - m2u2 / t)
m1v1 - m1u1 = - m2v2 + m2u2
m2u2 + m1u1 = m1v1 + m2v2 (INITIAL MOMENTUM = FINAL MOMENTUM)
So, momentum before collision = momentum after collision
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