Question

A trolley of mass 200 kg moves with a 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 m s–1 relative to the trolley in a direction opposite to the its 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 ?

Answer 1

Given that the mass of trolley is 200 kg, speed of trolley is 36  km/h , mass of boy is 20 kg, distance covered by the child is 10 m and speed of child is 4 m/s .

Speed of trolley v is

v=36  km/h =36× 5 18 =10 m/s

Let p i be the initial momentum of the trolley and child system then,

p i =( m+M )⋅v

Here, m is the mass of the child and M is the mass of the trolley and v is the speed of the trolley, then,

p i =( 200+20 )×10 =2200  kg⋅m/s

Let the final speed of the trolley in reference to ground be v f , then the final speed of the child in reference to ground v b is given as,

v b =v−V

Here, V is the speed of child on trolley.

Substitute the values in the above expression.

v b =( v−4 ) m/s

Let p f be the final momentum of trolley and child, then

p f =Mv+m( v b ) =Mv+m( v−V ) =v( m+M )−mV

Substitute the values in the above expression.

p f =v( 200+20 )−20×4 =220v−80

From law of conservation of momentum,

p i = p f

Substitute the values in the above expression.

2200=220v−80 220v=2280 v=10.36 m/s

Hence, the final speed of the trolley is 10.36 m/s .

Let t be the time taken by the child to run on the trolley, then

t= d v b

Substitute the values in the above expression.

t= 10 4 =2.5 s

Let D be the distance covered by the trolley, then

D=t×v

Substitute the values in the above expression.

D=10.36×2.5 =25.9 m

Hence, the trolley moved 25.9 m from the time the child begins to run.