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PHY 231
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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: ->->->->->->->->->->->-> :
The total momentum of the system is not the same before as it is after; it is not conserved. Some, is probably lost to heat.
m1v1 + m2v2 = m1v1 + m2v2
( .06 kg * .8 m/s ) + ( .02 kg * 0 m/s ( stationary/at rest ) = ( .06 kg * .5 m/s ) + ( .02 kg * .7 m/s )
.048 kg-m/s = .03 kg-m/s + .014 kg-m/s
.048 kg-m/s does not = .044 kg-m/s
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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: ->->->->->->->->->->->-> :
Assuming the steel ball's velocity and the masses are accurate, the marble ball's momentum would have to be .004 kg-m/s, more than what it is. Thus:
m2v2 = .02 kg * x m/s
.018 kg-m/s = .02 kg * x m/s
x = .9 m/s
The marble ball's momentum would have to be .9 m/s.
that's its velocity, not its momentum
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25 Minutes
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&#Good responses. See my notes and let me know if you have questions. &#