a man of mass 50 kg is standing in a gravity free space at a height of 10 m above the floor. he throws a stone of mass 0.5 kg downwards with a speed 2 m/s. when the stone reaches the floor, the distance of the man above the floor will be?
Asked by | 2nd Mar, 2012, 10:11: PM
Expert Answer:
From the law of conservation of momentum and we obtain the recoil speed v of the man from the equation, 50×v = 0.5×2, from which v = 0.02 ms1.
The time taken by the stone to reach the floor is 10/2 = 5 s. During this time the man will move up through a distance 0.02×5 = 0.1 m so that when the stone reaches the floor, the man will be at a distance 10 + 0.1 = 10.1 m above the floor
From the law of conservation of momentum and we obtain the recoil speed v of the man from the equation, 50×v = 0.5×2, from which v = 0.02 ms1.
The time taken by the stone to reach the floor is 10/2 = 5 s. During this time the man will move up through a distance 0.02×5 = 0.1 m so that when the stone reaches the floor, the man will be at a distance 10 + 0.1 = 10.1 m above the floor
Answered by | 4th Mar, 2012, 11:59: AM
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