Question

. A trolley of mass 200 kg moves witha uniform speed of 36 km/ h on a frictionless track. A child of mass 20 kg runs on the trolley from one end to the other (10 m away) with a speed of 4 ms relative to the trolley in a direction opposite to the trolley's motion, and jumps out of the trolley. What is the final speed of the trolley? How much has the trolley moved from the time the child begins to run ? e Ans The child gives an imn



show the diagram
unnecessary ans I will report your last 10 answer​

Answer 1

Mass of the trolley, M=200kg

Speed of the trolley, v=36km/h=10m/s

Mass of the boy, m=20kg

Initial momentum of the system of the boy and the trolley

     =(M+m)v 

     =(200+20)×10

     =2200kg m/s

Let v′ be the final velocity of the trolley with respect to the ground.

Final velocity of the boy with respect to the ground  =v′−4

Final momentum =Mv′+m(v′−4)

                             =200v′+20v′−80

From the conservation of linear momentum:

Initial momentum= Final momentum

                  2200=220v′−80

                    ∴ v′=2280/220=10.36m/s

Length of the trolley, l=10m

Speed of the boy, v′′=4m/s 

Time taken by the boy to run, t=10/4=2.5s

∴ Distance moved by the trolley =v′′×t= 10.36×2.5=25.9m

Answer 2

Explanation:

∴ Final speed of trolley is 10.36 ms-1. The child take 2.5 s to run on the trolley. Therefore, the trolley moves a distance = 2.5 x 10.36 m = 25.9 m.

my answer is correct please don't report my questions