Question

A girl 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 her velocity as the cart starts moving? Assume that there is no external unbalanced force working in the horizontal direction . ​

Answer 1

Question

A girl of mass 40 kg jumps with a horizontal velocity of 5ms^-1 onto a stationary cart with frictionless wheels . the mass of the cart is 3 kg what is her velocity of the cart starts moving assume that there is no external unbalanced force working in the horizontal direction .

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 .

Answer 2

no internal force=momentum is constant

momentum of girl=40×5=200kg-ms/Sec

momentum of girl and cart= total momentum of girl momentum of cart.

200+0=200kg m/sec.

Speed =200/(40+3)

=4.65m/Sec.

I hope it's right

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