Two particles of mases `m_(1)` and `m_(2)` in projectile motion have velocitie `vecv_(1)` and `vecv_(2)`, respectively at time `t=0`. they collide at time `t_(0)`. Their velocities become `vecv_(1)^(')` and `vecv_(2)^(')` at time `2t_(0)` while still moving in air. The value of
`|(m_(1)vecv_(1)^(')+m_(2)vecv_(2)^('))|-|(m_(1)vecv_(1)+m_(2)vecv_(2))|` is

Answer

C

Explanation

By applying impulse-momentum theorem
`=|(m_(1)vec(v)_(1) + m_(2)vec(v)_(2)) - (m_(1)vec(v)_(1) + m_(2)vec(v)_(2))|`
`= |(m_(1) + m_(2)) vec(g) (2L_(0))| - 2(m_(1) + m_(2)) g t_(0)`