A 60 kg man stands on a 120 kg cart which moves
horizontally with speed 6 m/s on a smooth floor. The
man jumps horizontally onto another stationary cart of
mass 100 kg. What will be the common velocity of the
man and the second cart if the first cart now moves
with speed 3 m/s in the same direction?
Given : A 60 kg man stands on a 120 kg cart which moves horizontally with speed 6 m/s on a smooth floor. The man jumps horizontally onto another stationary cart of mass 100 kg.
the first cart now moves with speed 3 m/s in the same direction
To Find : What will be the common velocity of the man and the second cart
Step-by-step explanation:
Solution :
Initial Momentum = ( 60 + 120) * 6 = 1080 kgm/s
Final momentum = ( 60 + 100) * v + 120 * 3 kgm/s
= 160v + 360 kgm/s
Equate both
=> 160v + 360 = 1080
=> 160v = 720
=> 16v = 72
=> 2v = 9
=> v = 4.5
common velocity of the man and the second cart = 4.5 m/s
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