Question

A girl of mass 40 kg jumps
with a horizontal velocity of 5 m sonto
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

Step-by-step explanation:

we need to apply momentum conservation to get the final speed of girl who jumped on the cart

Intial momentum = mass of girl × intial horizontal speed = 40×5 = 200 kg m/ s

momentum of girl and cart after jumping (40+3) v kg m/ s

by conservation of momentum , ( 40+3) × v = 200 or v = (200/43)m/s =4.65 m/s

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Answer 2

Answer:

$v = \frac{200}{23} m | s$

Step-by-step explanation:

$\huge\fbox\red{Given}$

Mgirl = 40 kg. Ugirl = 5m/s

Mcart = 3kg

initial velocity of cart Ucart = 0

final velocity of cart be V

Applying conservation of linear momentum : P1 = Pf

Or Mgirl Ugirl + Mcart Ucart = (Mgirl+ Mcart) V

Or 40×5+3×0 = (40+3) V

Hence,The $\mathfrak{Answer:-}$

$v = \frac{200}{23} = \frac{m}{s}$

$\huge\fbox\red{Thank you}$