First slide

Motion of centre of mass

Question

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 of 2 ${ms}^{- 1}$ . When the stone reaches the floor, the distance of the man above the floor will be

Moderate

A

9.9 m
B

10.1 m
C

10 m
D

20 m

Solution

As there is no external force acing on man, initially the center of mass of the system was at rest hence there is no displacement of center of mass.
$[(m_{1}) + {(m)}_{2}] ∆ r_{c o m} = 0$
$m_{_{1}} ∆ r_{1} - m_{2} ∆ r_{2} = 0$
$\Rightarrow ∆ r_{2} = \frac{m_{1} ∆ r_{1}}{m_{2}} = \frac{0.5 \times 10}{50} = 0.1 m$
The distance of the man above the floor (total height) = 10+0.1 = 10.1 m

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