Need help with collisions (M2 Edexcel)

This might sound a bit daft, but I don't understand how to arrange simultaneous equations relative to direction of movement.

If you have two particles of mass 2m and 3m approaching each other at the same speed u ms-1. Naming the speed after collision w and v, respectively, does this give:

2mu – 3mu = 2mw + 3mv OR 2mu – 3mu = 2mw – 3mv?

And for the coefficient of restitution would you write the speed of separation as v + w or v – w?

In the June '11 paper Q2 it seems they're doing 2mu – 3mu = 2mw – 3mv then v+w, but in Solomon A Q1 they've done 2mu – 3mu = 2mw + 3mv and v – w.

So I understand they have to be opposite since the equations are m1u1 + m2u = m1v1 + m2v2, and (v2 – v1) / (u1 – u2) = e, but how do I know which way around to have it for different questions?

In the June '11 paper it doesn't seem to make any difference, but in the Solomon paper I'm getting 1 and 3 when I should get -1 and 3. Perhaps I've done the equations wrong? Mark scheme has: m(5) – m(3) = mv1 + mv2 and (v2 – v1) / (5 + 3) = (1/2), while I wrote: m(5) – m(3) = mv2 – mv1 and (v2 + v1) / (5 + 3) = (1/2).

Question is: [Two identical particles are approaching each other along a straight horizontal track. Just before they collide, they are moving with speeds 5 ms-1 and 3 ms-1 respectively.The coefficient of restitution between the particles is 1/2 .

Find the speeds of the particles immediately after the impact.]

I must be missing something. :s-smilie:

Thank you for reading!

Reply 1

poorform

It doesn't matter.

The only important thing is that you are consistent with your directions.

So to begin draw a diagram and define a direction (+ve to the right say), you can indicate this by writing ------->+ve in the direction you wish to define as the positive direction.

Then whenever you are writing down equations by using conservation on linear momentum consider the direction of each particle/velocity. So say at the start you have 3m travelling to the right and you have defined right as the positive direction then 3m has a velocity of um/s and the 2m has a velocity of -um/s (since it is travelling in the opposite i.e. negative direction). The final result will always be the same regardless of which direction you choose to be + ve as long as you are consistent.

Reply 2

kiasas

Thank you!

No matter which masses and velocities I use I keep getting the same speed for both but not the same direction when trying both ways. Same as with the example in my original post. Am I not being consistent or does this method not work for velocities?

(edited 8 years ago)

Reply 3

poorform

Original post by kiasas

You should be getting the same result regardless of direction. Perhaps you have just made a mistake with the algebra or something.

Reply 4

kiasas

Oh, sorry, I phrased that poorly! I'm using the same direction, but I'm changing whether I use 2mu – 3mu = 2mw + 3mv or 2mu – 3mu = 2mw – 3mv, and then using the appropriate direction for the speed of separation. (w-v and w+v, respectively).

I then get two different equations for each of them, and for the 3 scenarios I've tried one way gives two positive velocities while the other gives one positive and one negative, but the speeds are the same for both ways.

Need help with collisions (M2 Edexcel)

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