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 ~ms^ -1 . When the stone reaches the floor, the distance of the

Correct answer: B. 10.1 ~m m r= constant aligned m_1 r_1 & =m_2 r_2 r_2 & =(m_1 r_1)/(m_2) & =(0.5 × 10)/(50)=0.1 aligned The distance of the man above the floor (total height)=10+0.1=10.1 ~m .

NEET 2010

Exam: NEET
Year: 2010
Subject: Physics
Chapter: Center of Mass, Momentum and Collision

NEET 2010 · Physics · Center of Mass, Momentum and Collision

A man of 50 ~kg mass is standing in a gravity free space at a...

A man of

50 \mathrm{~kg}

mass is standing in a gravity free space at a height of

10 \mathrm{~m}

above the floor. He throws a stone of

0.5 \mathrm{~kg}

mass downwards with a speed

2 \mathrm{~ms}^{-1}

. When the stone reaches the floor, the distance of the man above the floor will be

Options

A $9.9 \mathrm{~m}$
B $10.1 \mathrm{~m}$
C $10 \mathrm{~m}$
D $20 \mathrm{~m}$

Correct answer

B. $10.1 \mathrm{~m}$

Step-by-step solution

m r=

constant

\begin{aligned}m_1 r_1 & =m_2 r_2 \\ r_2 & =\frac{m_1 r_1}{m_2}\\ & =\frac{0.5 \times 10}{50}=0.1 \end{aligned}

The distance of the man above the floor (total height)=10+0.1=10.1 \mathrm{~m}

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