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    Stuck on this question:
    Three supermarket trolleys, each of mass 20 kg, are placed on a straight line. As these trolleys move they are not subject to any resistance to motion. The first one is set in motion so that it moves at a speed of 4.5 m s–1 towards the second trolley. It then collides with the second trolley. After this collision the two trolleys continue to move together along the straight line at constant speed until they collide with the third trolley. After this collision all three trolleys move together along the straight line.
    (a) Show that the speed of the two moving trolleys after the first collision is 2.25 m s–1.
    (b) Find the speed of the trolleys after the second collision.
    (c) After the second collision the combined trolleys are subject to a single, horizontal resistance force of magnitude 30 N.
    (i) Calculate the acceleration of the trolleys while this force acts.
    (ii) Find the distance, that the trolleys move after the second collision, before they come to rest.

    Have hit a brick wall at part c(i), (got 1.5m/s for b) For c is it just F=MA? If so is F just the 30N or does the 1.5m/s have some part in it?
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    Yes, I think it is just F=ma using -30N as the resultant force
    (taking the trolleys' direction of motion as the positive direction)
 
 
 
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