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 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 of 2 m/s. When the stone reaches the floor the distance of the man above the floor will be

A) 9.9 m   B) 10.1 m  C) 10 m   D) 20 m

Concepts: Conservation of Linear momentum. Similar to the recoiling of gun. When the man pushes the stone downward the man will get pushed upward. Since there is no external force acting on the system, conservation of linear momentum is valid and center of mass of the whole system remains in the same location.

Method 1: Using Law of Conservation of Momentum :

\[Momentum_{\underline{ }before}=Momentum_{\underline{ }after}\]

\[m_1.v_1+m_2.v_2= m_1.v_1^{'}+m_2.v_2^{'}\]

\[\left(m_{man}+m_{stone}\right)v_{total\hspace{1mm} system}=m_{man}. v_{man} + m_{stone}. v_{stone}\]

\[\left(50 + 0.5\right) 0 = 50 .v_{man}+0.5 . \left(-2\right) \]

\[v_{man}=\frac{1}{50} m/s\]

Time taken for the stone to reach the floor:

\[v=\frac{d}{t}\Rightarrow 2=\frac{10}{t}\Rightarrow t=5\hspace{2mm} sec\]

Distance travelled by the man in the same time:

\[v=\frac{d}{t}\Rightarrow \frac{1}{50}=\frac{d}{5}\Rightarrow d=0.1\hspace{2mm}m\]

So the distance of the man above the floor when the stone reaches the floor will be 10.1 m.

Method 2: Using Center of Mass of a system:

10 m from the floor is taken as Zero reference.

\[M.R_{c.m}=m_1.r_1+m_2.r_2\]

\[\left(m_{man}+m_{stone}\right)r_{cm\underline{  }system}=m_{man}. r_{cm\underline{  }man} + m_{stone}. r_{cm\underline{  }stone}\]

\[\left(50 + 0.5\right) 0 = 50 .r_{cm\underline{  }man}+0.5 . \left(-10\right) \]

\[r_{cm\underline{  }man}=0.1 \hspace{2mm}m\]

So the distance of the man above the floor when the stone reaches the floor will be 10.1 m.

Method 3: Using Center of Mass of a system:

The floor is taken as Zero reference.

\[M.R_{c.m}=m_1.r_1+m_2.r_2\]

\[\left(m_{man}+m_{stone}\right)r_{cm\underline{  }system}=m_{man}. r_{cm\underline{  }man} + m_{stone}. r_{cm\underline{  }stone}\]

\[\left(50 + 0.5\right) 10 = 50 .r_{cm\underline{  }man}+0.5 . \left(0\right) \]

\[r_{cm\underline{  }man}=10.1 \hspace{2mm}m\]

So the distance of the man above the floor when the stone reaches the floor will be 10.1 m.

Concept:

How can we move around in space?

In earth we push the ground backward and the ground pushes us forward.  

When you are sitting in an office chair with wheels, only if you put your feet to the ground and push it backward, we will be able to move forward.

The car tires pushes the ground backward, so that the ground pushes the tires forward. 

We push the water backward, so that the water pushes us forward in swimming or for a boat. 

The air plane sucks the air by the jet engine and pushes it backward, so that the air pushes the plane forward. 

The rocket pushes the exhaust gases from the combustion of hydrogen and oxygen, so that the gases push the rocket forward.

Moving around in gravity free, friction free space could be very challenging. Unless you have a rail to hold on, or some other objects to throw away, we cannot move. Jet packs carried by astronauts are of same principle. You must have used Jet packs in video games.

Law of conservation of momentum is a direct consequence of Newton's Third Law of Motion.

See the interesting videos by astronauts.