Question

A trolley of mass 200 kg moves with a uniform speed of 10 m/s on a frictionless track. A child of mass 20 kg runs on the trolley from one end to the other with a speed of 4 m/s relative to the trolley in a direction opposite to its motion, and jumps out of the trolley. The final speed of the trolley is .

Answer 1

$\sf\star\red{\bold{\underline{{QUESTION:-}}}}$

✓✓A trolley of mass 200 kg moves with a uniform speed of 10 m/s on a frictionless track. A child of mass 20 kg runs on the trolley from one end to the other (10m away ) with a speed of 4 m/s relative to the trolley in a direction opposite to 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?

$\sf\star\bold\green{\underline{{GIVEN:-}}}$

• $\sf\: Mass\:of\:the\:trolly\:(M)=200kg$

• $\sf\:speed\:of\:the\:trolly\:(v)=10m/s$

• $\sf\:Mass\:of\:the\:boy\:(m)=20kg$

$\sf\star\bold\purple{\underline{{SOLUTION:-}}}$

Initial momentum of the system of the boy and the trolly

$\sf\:=(M+m)v$

$\sf\:=(200+20)\times\:10$

$\sf\:=2200kg\:m/s$

let v' be the final velocity of the trolley with respect to the ground.

final velocity of the boy with respect to the ground$\sf\: =v'-4$

$\sf\:Final\: momentum=Mv'+m(v'-4)$

$\sf\:=200v'+20v'-80$

$\sf\:=220v'-80$

we know that law of conservation of momentum:-

$\sf\boxed\star\pink{\underline{\underline{{Initial\: momentum=final\: momentum}}}}$

$\sf\implies\:2200=220v'-80$

$\sf\therefore\:v'=\dfrac{2280}{220}$

$\sf\green{{\implies\:10.36m/s}}$

• $\sf\:length\:of\:the\:trolley\:(l)=10m$

• $\sf\:speed\:of\:the\:boy\:(v'')=4m/s$

$\sf\:Time\:taken\:by\:the\:boy\:to\:run$

$\sf\implies\:t=\dfrac{10}{4}$

$\sf\red{{\implies\:2.5s}}$

$\sf\therefore\: Distance\:moved\:by\:the\: trolley=$

$\sf\:v''\times\:t=10.36\:\times\:2.5$

$\sf\purple{{\implies\:25.9m}}$