Question

A boy of mass 40 kg jumps with a horizontal velocity of 5 ms-1 onto a stationary cart with frictionless wheels. The mass of the cart is 3 kg. What is his velocity as the cart starts moving? Assume that there is no external unbalanced force working in horizontal direction

Don't copy plzzzz.......... no need of copied answers​

Answer 1

Answer:

Solution

Given,

let velocity "v" of the girl on the cat as the cart start moving .

the total momentum of the girl and cart before the interaction

= 40kg × 5 ms^-1 +3kg × 0ms^-1

= 200kgms^-1

Now,

total Momentum after the interaction -

= (40+3)kg × v m s^-1

= 43 v ms^-1

Then,

according to the law of conservation of momentum the total Momentum is conserved during the interaction that is ;

43v = 200

=> v =200/43

=> v = 4.65 ms^-1

the girl on cart move with a velocity 4.65ms^-1 and in the direction in which the girl jumped .