A man of 50, kg mass is standing in a gravity | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

Since the man is in gravity free space, $force\,\, on\,\, man + stone\,\, system$  is zero

Vedclass · Accepted Answer

$10.1$

Vedclass · Answer

Since the man is in gravity free space, $force\,\, on\,\, man + stone\,\, system$  is zero.

Therefore center of mass of the system remains at rest. Let tha man goes $x$ m above when the stone reches the floor, then

  ${M_{man}} 	imes x = {M_{stone}} 	imes 10$

$x = \frac{{0.5}}{{50}} 	imes 10$

$x\,\,=\,0.1\,\,m$

Therefore final height of man above floor $\,=\,\,10\,+\,\,x\,\,=\,\,10\,+\,0.1\,\,=\,\,10.1\,\,m$

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 ...... $m$

Similar Questions

Suppose that the angular velocity of rotation of earth is increased. Then, as a consequence.

A spring balance is graduated on sea level. If a body is weighed with this balance at consecutively increasing heights from earth's surface, the weight indicated by the balance

The height at which the weight of a body becomes ${\frac{1}{16}}^{th}$ , its weight on the surface of earth (radius $R$), is

A body weighs $72 N$ on surface of the earth. When it is taken to a height of $h=2 R$, where $R$ is radius of earth, it would weigh ........ $N$

If the change in the value of $‘g’$ at a height $h$ above the surface of the earth is the same as at a depth $x$ below it, then (both $x$ and $h$ being much smaller than the radius of the earth)