Calculating Velocity from Conservation of Momentum: A Science Tutorial

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The discussion focuses on using conservation of momentum to calculate velocity, with a specific example yielding a velocity of 2.86 m/s. The original calculation appears correct, but a participant suggests double-checking the number 400 for potential errors. They recommend using symbolic calculations and maintaining consistent units to avoid mistakes. Acknowledgment of common errors in math is shared, emphasizing that such mistakes can happen to anyone. The conversation highlights the importance of careful calculation and verification in physics problems.
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Homework Statement
My daughter was asked this question:
A person of mass 60 kg running at 8 m/s reaches a wagon of mass 80 kg going at 2 m/s and jumps onto it. Calculate the velocity of the wagon immediately after the person jumps on.
Relevant Equations
total momentum before impact = total momentum after impact
Using conservation of momentum I would say (60 x 8) + (80 x 2) = (60 + 80) v
so v = 400/140 = 2.86 m/s.
However, this is not one of the multiple choice responses. What am I doing wrong?
 
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Setup looks right. But check the number 400.
 
Apart from that, consider using symbolic calculations and only insert your numbers at the end. It often avoids this kind of mistakes. Or at least use units throughout.
 
oh man, I used to be good at maths, honest. Thanks :-)
 
AndrewMorris said:
oh man, I used to be good at maths, honest. Thanks :-)
It was just a careless error. Happens to us all :oldsmile:
 
Thread 'Chain falling out of a horizontal tube onto a table'
My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...

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