Question

Two particles of masses m₁ and m₂ in projectile motion have velocities Vector v₁ and Vector v₂, respectively, at time t = 0. They collide at t = t₀. Their velocities become Vector v₁' and Vector v₂' at t  2t₀ while still moving in air. The value of |m₁ Vector v₁' + m₂ Vector v₂') - (m₁ Vector v₁ + m₂ Vector v₂)| is

(a) zero

(b) (m₁ + m₂)gt₀

(c) 2(m₁ + m₂)gt₀

(d) 1/2 (m₁ + m₂)gt₀

Answer 1

Correct Answer is: (c) 2(m₁ + m₂)gt₀

The momentum of the two-particle system at t = 0 is given by Vector pi = m₁ Vector v₁ + m₂ Vector v₂.

A collision between the two does not affect the total momentum of the system. 

A constant external force (m₁ + m₂)g acts on the system. The impulse given by this force in time t = 0 to t = 2t₀ is (m₁ + m₂)g x 2t₀.

∴ the absolute change in momentum in this interval is

|(m₁ Vector v₁' + m₂ Vector v₂') - (m₁ Vector v₁ + m₂ Vector v₂)| = 2 (m₁ + m₂)gt₀.