Question

A boy of mass 50 kg is running with a velocity of 2m/s. He jumps over a stationary cart of 2 kg while running . Find the velocity of cart after jumping of the body.

Answer 1

Answer:

1.92m/s

Explanation:

Given :

• Boy mass ,$\rm{m_1}$  => 50kg
• Initial velocity of the boy ,$\rm{u_1}$ => 2m/s
• Mass of the cart ,$\rm{m_2}$ => 2kg
• Initial velocity of the cart ,$\rm{u_2}$ => 0

Since,boy jumped over cart, thus the Final velocity ,$\rm{v_1}$ of the boy will be equal to that of the cart.

$\rm{v_1=v_2}$

To Find :

• Final velocity ,$\rm{v_2}$ = ?

Solution :

By conversation of momentum :-

$\rm{m_1u_1+m_2u_2=m_1v_1+m_2v_2}$

$\rightarrow\rm{ 50kg \times 2ms -1+ 2kg \times 0 = 50kg\times v_1+2kg \times v_2}$

$\rightarrow\rm{100kg = 50kg\times v_1+2kg \times v_2}$

$\because \rm{v_1=v_2}$

$\rightarrow\rm{100kg = v_2(50kg+2kg)}$

$\rightarrow \rm{ 100 \; kg = v_2 ( 52 \; kg ) }$

$\rightarrow \rm{ \dfrac{ 100 \; kg \; m/s }{ 52 \; kg }} = v_2$

$\because \rm{v_2=1.92m/s }$

Therefore,velocity of cart after jumping of boy over it is equal to 1.92m/s.

[Since it has positive sign,it means cart will go in the same direction of boy.]