Momentum - Impulse collision to find initial speeds and directions

[dahoom102]

Homework Statement

Two particles P and Q of masses 3 kg and 2 kg respectively are moving along the same straight line on a smooth horizontal surface. The particles collide. After the collision both the particles are moving in the same direction, the speed of P is 1 m/s and the speed of Q is 1.5 m/s. The magnitude of the impulse of P on Q is 9N s. Find:
a) Speed and direction of P before the collision
b) Speed and direction of Q before the collision

Relevant Equations

I=m(v-u)
Pf=Pi

Hi
I've tried solving this question but it seems that I flipped the direction of the impulse, what did I interpret wrong? the question didn't give any clue on their direction before so I couldn't infer the direction of the impulse. It also just gave me the magnitude without the direction. I would appreciate if you could help me know why is my diagram wrong.

 

[haruspex]

why is my diagram wrong.

If after two particles collide they are moving in the same direction, which is moving faster, the one in the lead or the one behind?

 

[dahoom102]

If after two particles collide they are moving in the same direction, which is moving faster, the one in the lead or the one behind?

Oh, i just realized! how couldn't I be logical in the first place ;(. Thanks a ton haruspex!

 

[Delta2]

Your equations would yield the same results as those of book (with the roles of u,v swapped) , if you change the sign of impulse in each one, that is if you had the equations $$-9=3(1-v)$$$$9=2(1.5-u)$$.

However I am confused too why we should consider the impulse as negative in the first equation with v , and as positive in the second equation with u.

 

[Steve4Physics]

Can I add a few words as I can see some sources of confusion.

In the official answer (shown in Post #1), it looks like the author has mixed sign conventions:
- for P, they have taken the positive direction to be same as the direction of the impulse acting on P (to the left);
- for Q, they have taken the positive direction to be same as the direction of the impulse acting on Q (to the right).

Very silly and/or a bit of a bodge (IMO)!

(Also note, in the official answer, ‘u’ is P’s initial velocity and ‘v’ is Q’s initial velocity. That seems logical - based on alphabetical ordering.)

A more sensible approach to the sign convention would be to use the known common direction of the final velocities as the reference. Take it to be the +x direction (to the right).

The final velocities of P and Q are then both positive (+1m/s and +1.5m/s).

It is not hard to deduce that the impulse of Q on P must have been negative (-9Ns) and the impulse of P on Q must have been positive (+9Ns).

We then get:
For P: -9 = 3(1-u) ⇒ u = 4m/s
For Q: +9 = 2(1.5 – v) ⇒ v = -3m/s

 

[bob012345]

@Steve4Physics, are you suggesting the author wasn't minding their P's and Q's?

 

[PeroK]

Following the book's bad example and ignoring units, I would say.

Q has momentum $+3$ after the collision and was given an impulse of $+9$ so must have had a momentum of $-6$ before the collision.

P has momentum $3$ after an impulse of $-9$ so must have had $12$.

It seems logical to me use momentum rather than velocity. And then get the velocity from the momentum.