# 2 people throwing balls at each other on cart – cons of momentum

**1. The problem statement, all variables and given/known data**

Emil and Kerry are standing on a stationary cart with frictionless wheels. Total mass of cart and riders is 130 kg. At the same instant, Emil throws a 1.0 kg ball at 4.5 m/s to Kerry while Kerry throws a 0.5 kg ball to Emil at 1.0 m/s. Emil’s throw is to the right and Kerry’s to the left. a) While balls are in the air, what are the speed and direction of the cart and its riders? b) After balls are caught, what are speed and direction of the cart and riders?

**2. Relevant equations**

m_{cart}v_{cart initial}+m_{Emil ball(Eb)}v_{Eb initial}-m_{Kerry ball(Kb)}v_{Kb initial} = m_{cart}v_{cart air}+m_{Emil ball(Eb)}v_{Eb air}-m_{Kerry ball(Kb)}v_{Kb air}

Conservation of momentum essentially (mv_{initial}=mv_{final})

**3. The attempt at a solution**

So for (a) the entire left hand side goes to zero since the cart is stationary and the balls haven’t been thrown. This allowed me to solve for v_{cart air}, which meant I ended up with [(0.5*1)-(4.5*1)]/130, leaving me with a -0.031 m/s or essentially 0.031 leftwards.

I’m having a bit of trouble with (b) though, provided that (a) is correct. Would I essentially solve m_{cart}v_{cart air}+m_{Emil ball(Eb)}v_{Eb air}-m_{Kerry ball(Kb)}v_{Kb air} = m_{cart}v_{cart final}+0+0–zeroes since the balls are caught and not moving–for v_{cart final}? I would get the answer of -2.31×10^{-4}. Is that the correct way of thinking because what I was also thinking is that I could just set the initial left side equal to the final rather than use the balls in air equal to the final and that would get me an answer of 0 m/s in no particular direction.

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