assign each mass to m1 and m2, u is the initial velocities. this particular example is a completely inelastic collision so there will be only 1 mass (m1 + m2) with a common velocity after the collision. Hope this helps

since the masses stick together the second half of that equation i wrote would be (m1 + m2) x common velocity of the two. there is no v1 and v2 because they have stuck together and will both be travelling at the same velocity. the final velocity wont be zero though. after they have stuck together they will move off together with a common velocity. does that help?

(m1 X u1) + (m2 X u2) = (m1 X v1) + (m2 X v2)
(3*3) + (2*2) = (3*v1)+(2*v2)
13 = (m1+m3) * v (is that the right letter to use???)
13 = 5 * v
v = 13/5
v = 2.6

Is that right, and if not where do you mind explaining where I went wrong please

You've used 2 kg as your mass for one of the trolleys, though above you said it was 24g. You made a typo by putting x on the RHS of your top line, when you meant +, but apart from that it looks ok. Yes, v is the right letter for velocity (were you not taught the usual symbols for things like velocity then?). Also, you don't need to bother putting v1 or v2, just a v in the 2nd line, as you know they're going to move off with the same velocity.

The letter you use is up to you really, as long as you dont confuse yourself. Make it clear that one side is the initial velocity and the other is the final velocity. i just used v because its what I use for final velocity. In this case I would have noted it as Vc (small c at bottom) simply to remind myself it was a common velocity as the two masses had stuck together and moved off with the same velocity. Does that make it clearer?

Sorry I meant 2kg, a typo there. I also mistyped the x on the RHS top line, they should have been +.

Thanks for all your help, I do not have my exercise book so couldn;t remember the equation for it. In case your wondering I'm an english pupil in year 10 which makes me 15 yrs old, so hopefully this was the right forum to posst in.

A field gun of mass 1000kg, which is free to move, fires a shell of mass 10 kg at a speed of 200 m/s.

(b) What is the momentum if the gun just after firing? - Do we use momentum = mass X velocity, and if so what is the velocity it moving back and if so how is that worked out.
(c) Calculate the recoil velocity of the gun. - Is that an impulse, i.e.g. Ft= mv-mu

The conservation of momentum applies to this one again, m1u1 + m2u2 = m1v1 + m2v2

Initially everything is at rest so the left hand side of that equation will equal 0. (1000*0) + (200*0). Whats the final velocities? Try it from there.

(Others feel free to correct me if I am leading him astray :) )