Question

A boy of mass 40 kg moving with a velocity 5m/s jumps into skating board of mass 3 kg at rest.If the boy and sking board move together what is the velocity?​

Answer 1

Explanation:

Given :   Mass of boy (M)= 40 kg     

Velocity of Boy (U1) =5 m/s        

Mass of cart (m) = 3 kg

Initial velocity of cart  (U2)=0

Let the final velocity of cart be  V

Applying conservation of linear momentum :  

  Pi=Pf

M U1 + m U2 = (M+m) V

40×5+3×0=(40+3)V

200 = 43 V

V = 200/43

V = 4.65 m/s

Answer 2

Solution :

We have ,

• Mass of the boy , M= 40 kg
• Velocity of boy , u = 5 m/s
• Mass of skating board , m = 3 kg
• Velocity of skating board , v = 0 [ at rest ]

Let the Velocity of cart [ boy and sakting together ] be V .

As no external force acts on the system

So According to the law of conservation of Momentum .

Intital momentum = Final momentum

$\sf\implies\:Mu+mv=(m+M)V$

$\sf\implies\:40\times5+3\times0=(40+3)V$

$\sf\implies\:43V=200$

$\sf\implies\:V=\dfrac{200}{43}$

$\rm\implies\:V=4.65ms^{-1}$

______________

More About the topic :

• Conservation of Momentum :

If no external force acts on the system of constant mass the total momentum of the system remains constant .