Question
a man of mass 80kg standing on a stationary frictionless trolley of mass 60kg has in his hands a parcel of mass 10 kg.He throws the parcel horizontally from the rear of the trolley at 4 m.s-1 in relation to the ground.
2.1 what is the total momentum of the system before the parcel is thrown?
2.2 what is the total momentum of the system after the parcel is thrown?
2.3 calculate the speed with which the man and the trolley move after the parcel has been thrown
2.1 what is the total momentum of the system before the parcel is thrown?
2.2 what is the total momentum of the system after the parcel is thrown?
2.3 calculate the speed with which the man and the trolley move after the parcel has been thrown
Answers
2.1 0
2.2 0
2.3 The man+trolley momentum is equal and opposite to the thrown parcel's momentum.
(80+60)V + 10x4 = 0
Solve for V
A negative answer means it moves in the opposite direction from the package.
2.2 0
2.3 The man+trolley momentum is equal and opposite to the thrown parcel's momentum.
(80+60)V + 10x4 = 0
Solve for V
A negative answer means it moves in the opposite direction from the package.
V=180
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