Question

Astronaut Mr X (mass 120kg) and Mr Y (mass 90kg) float in a negative free space with no relative velocity to one another. Mr Y throws a mass of 5kg toward Mr X moving with a speed of \[2m{{s}^{-1}}\]. If Mr X catches it, the change in velocity of X and Y are:
\[\begin{align}
  & \text{A) 0}\text{.21 and 0}\text{.80}m{{s}^{-1}} \\
 & \text{B) 0}.8\text{0 and }0.21m{{s}^{-1}} \\
 & \text{C) 0}.1\text{2 and }0.08m{{s}^{-1}} \\
 & \text{D) 0}\text{.08 and }0.12m{{s}^{-1}} \\
\end{align}\]

Answer 1

Hint: When the Astronaut Mr y throws a mass of 5kg, there will be a motion and due to the motion there will be momentum. Similarly when Mr X catches the mass of 5 kg then there will be a motion and momentum. We know that momentum is given by the product of mass and velocity and by using law of conservation of momentum we can find the change in velocity of Mr X and Mr Y.

Formula used:
\[p=mv\]

Complete answer:
It is given that two astronauts Mr X and Mr Y float in a negative free space with no relative velocity to one another, that is they don't experience any gravitational pull or any other forces due to each other.
When Mr Y throws a mass of 5kg, let say a ball, toward Mr X then there will be a momentum of the ball and the momentum produced while throwing the ball. According to the law of conservation these momentums should be equal to each other. We have given the mass of ball and its velocity which are \[m=5kg\text{ and }v=2m{{s}^{-1}}\] and the mass of the Mr Y, \[{{M}_{Y}}=90kg\] and we have to find change in velocity of Mr Y due to throwing the ball, let say the changed velocity of Mr Y is \[{{v}_{Y}}\], then according to law of conservation of momentum
\[({{M}_{Y}}-m){{v}_{Y}}=mv\]
As Mr Y is throwing the ball the mass of the ball is subtracted from the mass of Mr Y.
Substituting the given values, we get
\[\begin{align}
  & (90-5){{v}_{Y}}=5\times 2 \\
 & 85{{v}_{Y}}=10 \\
 & {{v}_{Y}}=\dfrac{10}{85} \\
 & {{v}_{Y}}=0.117\simeq 0.12m{{s}^{-1}} \\
\end{align}\]
Similarly, when Mr X catches the ball there will be change in the momentum, the momentum of the ball will be equal to the momentum of Mr X when he catches the ball. Let say the change in velocity of Mr X is \[{{v}_{X}}\]and given mass of Mr X is \[{{M}_{X}}=120kg\], then according to law of conservation of momentum, we have
\[({{M}_{X}}+m){{v}_{X}}=mv\]
As Mr X catches the ball the mass of the ball and Mr X is added.
Substituting value of m, v and \[{{M}_{X}}\]in the above equation, we get
 \[\begin{align}
  & (120+5){{v}_{X}}=5\times 2 \\
 & 125{{v}_{X}}=10 \\
 & {{v}_{X}}=\dfrac{10}{125} \\
 & {{v}_{X}}=0.08m{{s}^{-1}} \\
\end{align}\]
Hence the change velocity of Mr X and Mr Y will be \[\text{ 0}\text{.08}m{{s}^{-1}}\text{ and }0.12m{{s}^{-1}}\]respectively so correct option is D.

Note: Just for convenience and for better understanding I have used “ball” for the body of mass 5 kg. Don’t get confused that the ball was mentioned in the question and it is something different.
In case there was downward gravitational pull experienced by the bodies then there should be an extra term added for it.