Phy 231
Your 'cq_1_24.2' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
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A steel ball of mass 60 grams, moving at 80 cm / sec, collides with a stationary marble of mass 20 grams. As a result of the collision the steel ball slows to 50 cm / sec and the marble speeds up to 70 cm / sec.
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Is the total momentum of the system after collision the same as the total momentum before?
answer/question/discussion: ->->->->->->->->->->->-> scussion:
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Yup, it has to be based on the conservation of momentum.
However, we can go ahead and check it.
(.06kg)(.8m/s) + (.02kg)(0m/s) = (.5m/s)(.06kg) + (.7m/s)(.02kg)
.048kg*m/s = .044kg*m/s
Well, actually, it's not exactly the same, but I think it's supposed to be.
If the system is isolated, with no forces except the contact forces, they should be exactly the same. If these results come from experimental data, we would not expect exact agreement.
What would the marble velocity have to be in order to exactly conserve momentum, assuming the steel ball's velocities to be accurate?
answer/question/discussion: ->->->->->->->->->->->-> scussion:
Guess it's not supposed to be, then.
So, .048kg*m/s = .06kg*(.5m/s) + .02kg(v)
and v = .9m/s
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15 min
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&#Good responses. See my notes and let me know if you have questions. &#