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A man of 50 kg mass is standing in a gravity free space at a height of 10 m above the floor. He throws a stone of 0.5 kg mass downwards with a speed 2 m/s. When the stone reaches the floor, the distance of the man above the floor will be
A
20 m
B
9.9 m
C
10.1 m
D
10 m
Detailed Solution
There is no gravity and no external force, so the total momentum of man + stone stays zero.
When the stone is thrown down at 2 m/s, the man recoils upward: $m_1v_1 = m_2v_2$, so $v_2 = \frac{0.5\times2}{50} = 0.02$ m/s
Time taken by the stone to reach the floor (it moves with constant speed): $t = \frac{10}{2} = 5$ s
Distance moved up by the man in this time = $0.02\times5 = 0.1$ m
(Equivalently, $m_1x_1 = m_2x_2$ gives $x_2 = \frac{0.5\times10}{50} = 0.1$ m.)
Height of the man above the floor = 10 + 0.1 = 10.1 m
When the stone is thrown down at 2 m/s, the man recoils upward: $m_1v_1 = m_2v_2$, so $v_2 = \frac{0.5\times2}{50} = 0.02$ m/s
Time taken by the stone to reach the floor (it moves with constant speed): $t = \frac{10}{2} = 5$ s
Distance moved up by the man in this time = $0.02\times5 = 0.1$ m
(Equivalently, $m_1x_1 = m_2x_2$ gives $x_2 = \frac{0.5\times10}{50} = 0.1$ m.)
Height of the man above the floor = 10 + 0.1 = 10.1 m