# Impulse and momentum of two particles

## Homework Statement

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.

## Homework Equations

impulse = change in momentum (mv-mu)

## The Attempt at a Solution

(c)

I get

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??

alphysicist
Homework Helper
Hi Pagey,

They asked for the magnitude of the impulse so they just need the absolute value.

O so the maths is right. Ok it makes sense now, cheers!

What is a negaitve impulse then, what does the - 0.216 mean? why is the answer negative (ingorning the fact it asks for magnitude)?

What is a negaitve impulse then, what does the - 0.216 mean? why is the answer negative (ingorning the fact it asks for magnitude)?

B consumed that momentum.

alphysicist
Homework Helper
The negative sign in the impulse indicates direction. The direction of the impulse is the same direction as the average net force during the collision.

So you called the velocities of particle A positive (they are both in the same direction in this problem). That set your coordinate system so that postive direction is in the direction of the velocities of A. But if A is slowing down, which direction is the average force on it? It is in the opposite direction of the velocities, and so since you already called the velocites positive, the average force and therefore the momentum must be negative.

(If particle A was moving to the right, then it needs a force to the left to slow it down.)