Correct Answer - A
Given `m_(1) = 500 g = (1)/(2) kg, m_(2) = 250 g = (1)/(4)kg (b)`
`u_(1) = 10 m s^(-1), u_(2) = -2 m s^(-1) (d)`
Momentum before collision, `m_(1) u_(1) + m_(2) u_(2)`
`=(1)/(2) xx 10 + (1)/(4) xx (-2), 5 - (1)/(2) = (9)/(2) Kg m s^(-1)`
Momentum after collision, `(m_(1) + m_(2)) v = (3)/(4) v (c )`
From law of conservation of momentum,
`(9)/(2) = (3)/(4)v implies v = 6m s^(-1) (e)`