# Playing catch in space?

What would happen if two people played catch in outer space?

Lets say person A and B are in space at rest, A throws a mass m at B, what would happen?

Would A move backwards at the same speed? My guess was yes, because of Newton's third law.

Now, what would happen when B catches the ball? My guess was that they would move at the same speed but it doesn't make sense because the mass has increased to the mass of the object (m) plus the mass of B. Can someone help me understand this, thank you.

• Would A move backwards at the same speed? The same speed as what? B? – HDE 226868 Dec 11 '14 at 2:45
• Nope, A and B are at rest initially, A throws the mass m at a speed of v. So, my guess is A would move at -v, or backwards. But what happens when B catches the mass moving at v? – MarsIsPrettyClose Dec 11 '14 at 2:47
• A would move slowed than $v$ because A (presumably) is more massive than $v$. B would move slower than $v$ because the momentum of the ball, now the momentum of the ball-B system, is applied to a greater mass. – HDE 226868 Dec 11 '14 at 2:49
• Won't A move at -v since the initial momentum was 0? – MarsIsPrettyClose Dec 11 '14 at 2:57
• Not unless the mass of A = the mass of the ball. – HDE 226868 Dec 11 '14 at 2:58

Conservation of momentum works here like everywhere else. When A (with mass $m_A$) throws a ball with mass $m_b$ with velocity $v$, then $v_A=-v_b\frac{m_b}{m_A}$ so that after the ball is thrown, the net momentum is zero; note that the ball will not be moving towards $B$ at velocity $v$ but instead at $v-v_A$ since $A$ started moving backwards... When B catches the ball, he/she will start moving backwards - with a velocity that's again given by conservation of momentum.

And so it keeps going. Every time they throw the ball at each other they move away from each other a little faster. But the net momentum of the entire system will continue to be zero at all times, so

$$\Sigma m_i\cdot v_i = 0$$

with the sum taken over A, B and the ball. If both of them can throw the ball with (relative) velocity $v$, then ultimately that will be the speed at which they are moving away from each other (since as they reach that relative velocity, the ball will impart almost no momentum to the person catching; and when their relative velocity becomes $v$, the ball will no longer be caught...). Since this state will take a finite number of throws to reach, it is possible that their final separation velocity will be just greater than $v$ (since the last person throwing the ball may end up going just a little bit faster, even though the ball will never be caught).

• So, when B catches the ball the mass of B + ball would be moving a bit slower than than A, since the net momentum is 0. Is that correct? What would happen when B throws the ball back (at the same velocity)? My guess is, since net momentum is always 0 in this case, if I wanted to find the velocity of B, the momentum the b all has plus the momentum A has would equal to the momentum of B? mB * vB = mb * vb + mA * vA That you for your answer, that's exactly what I was thinking would happen and trying to solve. To see if there was a point where they could never catch the ball. Thanks. :) – MarsIsPrettyClose Dec 11 '14 at 3:31

Would A move backwards at the same speed? no.. Momentum will be conserved,not speed and since A has greater mass than the ball so his speed will be less.. if A throws the ball with speed 'u1',then his speed(in the backward direction) will be m1u1/M1 where m1 is the mass of the ball and M1 is the mass of the person A Now, what would happen when B catches the ball? that will be inelastic collision since the person catches the ball..

if two bodies stick together after the collision and move as a single body with a common velocity then the collision is said to be perfectly inelastic collision.. and here also linear momentum is conserved. So m1u1 + M2 x 0 =(m1 + M2)v

v=m1u1/m1+M2

v is the velocity of (ball + the person B)

M2 is the mass of person B

• What would happen when B throws the ball back? – MarsIsPrettyClose Dec 11 '14 at 3:22
• Same calculations.. If the person B throws the ball with speed u1 then B will move backward with speed m1u1/M2 – Paul Dec 11 '14 at 3:33
• So if I try to find the net momentum to prove that it's zero after B throws the ball would I have to add the momentum of A into the equation? Or is it not part of the system anymore? Thank you for your answers. :) – MarsIsPrettyClose Dec 11 '14 at 3:38
• hyperphysics.phy-astr.gsu.edu/hbase/conser.html – Paul Dec 11 '14 at 4:20