A girl of mass 60kg jumps with a horizontal velocity of 6m/s onto a stationary cart with frictionless wheels. The mass of the cart is 6kg. What is her velocity as the cart starts moving? Assume that no external force is acting in horizontal direction.
Answers
the total momenta of the girl and cart before the interaction
=60kg×6 m/s +6kg×0m/s
=360kg m/s
total momenta after interaction
=(60+6)kg × v m/s
=66 v kg m/s
According to the law of conservation of momentum,the total momentum is conserved during the interaction
that is,
66v=360
v=360/66
v=5.45m/s
the girl on cart would move with a velocity of 5.45m/s
hope it will help u!!!
Given : A girl of mass 60 kg jumps with the horizontal velocity of 6m/s onto a stationary cart with frictionless wheels.
The mass of the cart is 6kg.
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 =6 m/s
m₂ = 6 kg , v₂ = 0 m/s
Momentum before jump = m₁v₁ + m₂v₂
= 60(6) + 6(0)
= 360 + 0
= 360 kgm/s
mass after jump = m₁ + m₂ = 60 + 6 = 66 kg
Velocity after jump = v m/s
Momentum after jump = 66v kgm/s
Equate momentum :
66v = 360
=> v = 360/66
=> v = 60/11
=> v = 5.45 ms/
Hence her velocity is 5.45 m/s.
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