Question

A boy 60 kg hopes with the velocity of 5 m/s on a stationary skateboard. The skateboard weighs 4 kg. Find out the velocity when the board starts rolling? The condition given here is the any external unbalanced force is absent.

Answer 1

Let v be the velocity of the girl on the cart as the cart starts moving.

The total momenta of the girl and cart before the interaction = (50 x 5 m s-1) + (4 x 0 m s-1)

                                                                            = 250 kg m s-1

The total momenta after the interaction = (50 + 4) kg x v m s-1

                                                               = 54 v kg m s-1

According to the law of conservation of momentum,

Total momentum before the interaction = Total momentum after the interaction

→ 54 v = 250

→ 

PLZ MARK AS BRAINLIEST

Answer 2

Answer: The ans we will get is 4.68 m/s

Explanation:

Taking the case of the boy

Mass of boy (M1)= 60kg

Initial velocity (u1) = 5m/s

Final velocity (v1)= ?? (To find)

Taking the case of skateboard

Mass of skateboard (m2) =4 kg

Initial velocity (u2) = 0 m/s. (the board is stationary)

Final velocity (v2)= ?? (To find)

Now v1 = v2 ( both are moving in the same direction and the boy is standing on the

skate board so velocity will be same)

Using law of conservation of momentum

(mu)1+(mu)2=(mv)1 + (mv)2

(60*5)+(4*0)= v(M1+m2). V1 = V2 proved above

300+0=v(60+4)

300=64v

300÷64= v

V=4.68 m/s

Thank you

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