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 $2 \,m/s$. When the stone reaches the floor, the distance of the man above the floor will be
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
Updated On: May 25, 2022
- 9.9 m
- 10.1 m
- 10 m
- 20 m
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The Correct Option is B
Solution and Explanation
Since the man is in gravity free space, force on man + stone system is zero.
Therefore centre of mass of the system remains at rest. Let the man goes $x \,m$ above when the stone reaches the floor, then
$ M_{man} \times x = M_ {stone} \times 10 $
$ x= \frac{ 0.5}{50} \times 10 $
$ x= 0.1 m $
Therefore final height of man above floor $= 10 + x$
$= 10 + 0.1 = 10.1 \,m$
Therefore centre of mass of the system remains at rest. Let the man goes $x \,m$ above when the stone reaches the floor, then
$ M_{man} \times x = M_ {stone} \times 10 $
$ x= \frac{ 0.5}{50} \times 10 $
$ x= 0.1 m $
Therefore final height of man above floor $= 10 + x$
$= 10 + 0.1 = 10.1 \,m$
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Concepts Used:
Gravitation
In mechanics, the universal force of attraction acting between all matter is known as Gravity, also called gravitation, . It is the weakest known force in nature.
Newton’s Law of Gravitation
According to Newton’s law of gravitation, “Every particle in the universe attracts every other particle with a force whose magnitude is,
- F ∝ (M1M2) . . . . (1)
- (F ∝ 1/r2) . . . . (2)
On combining equations (1) and (2) we get,
F ∝ M1M2/r2
F = G × [M1M2]/r2 . . . . (7)
Or, f(r) = GM1M2/r2
The dimension formula of G is [M-1L3T-2].