Two particles have mass 0.12kg and 0.08kg respectively. They are intially at rest on a smooth horizontal table. Particle A is then given an impulse in the direction AB so that it moves with speed 3m/s directly towards B.
Immediately after the particles collide the speed of A is 1.2m/s, its direction of motion being unchanged.
(c) find the magnitude of the impulse exerted on A in the collision.
impulse = change in momentum (mv-mu)
The Attempt at a Solution
impulse = mv-mu
= (0.12 x 1.2) - (0.12 x 3)
= - 0.216 Ns
But the mark scheme said the answer is + 0.216 Ns ... how can this be if the velocity after is less than the velocity before, i don't understand why??