[slop8]

Check out the attached question. How would you answer the second part 'State the total momentum...' ? and would you answer the first part explaining a set up consisting of light gates and use of cards?

I would appreciate any help. Thank you very much.

[SinghFello]

I think the Total Momentum is 0 - because if you find v (-29.47ms^1) and then times it by 0.95, it equals
-28kgms^1. So -28kgms^1 + 28kgms^1(momentum from the other trolley) = 0.
The experiment is just about using 2 light gates and 2 interpreters on each trolley. Find the time the light gates are interupted. Divide the displacement of the interpretters by the time taken to go through the light gates to get the velocity of each trolley. You probably have to say something about having trolleys of known mass, and using a frictionless airtrack to get the 3rd mark.

I might be wrong. Its been a years since I did U1.

[Quantophrenia]

I agree with SinghFello- the total momentum is 0, because the initial momentum is 0 (due to trolleys being at rest), and also so due to the conservation of momentum.

[slop8]

Thank you very much guys!!! Your answers have been very useful.