Question

A girl of mass 40 kg jumps with a horizontal velocity of 5m/s onto a stationery cart with frictionless wheels. The mass of the cart is 3kg. What is her velocity as the cart starts moving? Assume that there is no external unbalanced force working in the horizontal direction.

Answer 1

Answer

Given :   Mgirl​=40 kg           ugirl​=5 m/s        Mcart​=3 kg

Initial velocity of cart  ucart​=0

Let the final velocity of cart be  V.

Applying conservation of linear momentum :    Pi​=Pf​

Or    Mgirl​ugirl​+Mcart​ucart​=(Mgirl​+Mcart​)V

Or    40×5+3×0=(40+3)V

⟹ V=43200​ m

Answer 2

Answer:

V=43200 m/s

Explanation:

Given:   Mgirl=40 kg           ugirl=5 m/s        Mcart=3 kg

Initial velocity of cart, ucart=0

Let the final velocity of cart be V.

On applying Law of conservation of momentum,

m1u1 = m2mu2, we get,

Mgirl ugirl + Mcart ucart=(Mgirl + Mcart) V

40×5+3×0=(40+3)V

V=43200 m/s

Hope it helps...