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Phy 241
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.
Is the total momentum of the system after collision the same as the total momentum before?
answer/question/discussion: ->->->->->->->->->->->-> :
m_b1 = 60 g
v_b1 = 80 cm/s
v_m1 = 0 m/s
m_m2 = 20 g
v_b2 = 50 cm/s
v_m2 = 70 cm/s
m1v1 + m2v2 = m1v1 +m2v2
60g * 80 cm/s + 20 g * 0 m/s = 60 g * 50 cm/s + 20 g * 70 m/s
4800 g-cm/s = 4400 g-cm/s
No the momentum is not the same
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You could use v1' and v2' for the after-collision velocities, or if you prefer u1 and u2. Best not to use the same symbol on both sides. However your meaning is clear.
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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: ->->->->->->->->->->->-> :
60g * 80 cm/s + 20 g * 0 m/s = 60 g * 50 cm/s + 20 g * v
v = (4800 g-cm/s - 3000 g-cm/s) / 20 g
= 90 cm/s
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&#This looks good. See my notes. Let me know if you have any questions. &#