A girl of mass 60 kg jumps with the horizontal velocity of 10ms-1 onto a stationary cart with frictionless wheels. The mass of the cart is 5kg. What is her velocity as the cart starts moving? Assume that there is no external unbalanced force working in the horizontal direction.
Answers
Answer:
9.23m/s
EXPLANATION
let
v be the velocity of girl on cart as starts moving
the total momenta of the girl and cart before interaction
=60kg×10m/s+5kg×0m/s
=600kgm/s
total momenta after interaction
=(60+5)kg×vm/s
=65 v kgm/s
according to the law of conservation of momentum to the total momentum is conserved during the interaction that is,
65v=600
v=600/65
v=9.23m/s
Given : A girl of mass 60 kg jumps with the horizontal velocity of 10m/s onto a stationary cart with frictionless wheels.
The mass of the cart is 5kg.
Assumptions : that there is no external unbalanced force working in the horizontal direction.
To Find : her velocity as the cart starts moving?
Solution :
m₁ = 60 kg v₁ =10 m/s
m₂ = 5 kg , v₂ = 0 m/s
Momentum before jump = m₁v₁ + m₂v₂
= 60(10) + 5(0)
= 600 + 0
= 600 kgm/s
mass after jump = m₁ + m₂ = 60 + 5 =65 kg
Velocity after jump = v m/s
Momentum after jump = 65v kgm/s
Equate momentum :
65v = 600
=> v = 600/65
=> v = 120/13
=> v = 9.231 ms/
Hence her velocity is 9.231 m/s.
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