anonymous
  • anonymous
Prove that the fractional loss in kinetic energy for the ball-pendulum system is given by equation n=1-m/m+M
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
This doesn't make sense without context.
anonymous
  • anonymous
frictional loss=0.8099 and 0.7963. for silver and brass respectively
anonymous
  • anonymous
Context, meaning you need to explain what the system is.

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anonymous
  • anonymous
how would i explain it. that was the question on my lab book. I apologize for being vague
anonymous
  • anonymous
I would assume the question pertains to a system that was described somewhere...
anonymous
  • anonymous
does it apply to the prior question?
anonymous
  • anonymous
1. A 15 g bullet is fired horizontally into a block of wood with mass of 2.5kg and embedded in the block. Initially the block of wood hangs vertically and the impact causes the block to swing so that its center of mass raises 15 cm. Find velocity of bullet just before impact.
anonymous
  • anonymous
There you go. 1) Use the velocity you found to calculate the kinetic energy of the bullet before it strikes the block 2) Use the conservation of momentum to find the kinetic energy of the block and bullet together right after it hits 3) Compare them
anonymous
  • anonymous
thank you
anonymous
  • anonymous
this is what i got so far Kinetic Energy of ball just before collision is Ki = (1/2)mv^2 Kinetic Energy of Ball+Pendulum just after collision is; Kf = (1/2)(m+M)u^2 Fractional ratio is; Kf/Ki = [(m+M)/m](u^2/v^2) Kf = [(m+M)/m](u^2/v^2)Ki Fractional Loss is; (Ki - Kf)/Ki = 1 - [(m+M)/m](u^2/v^2) Using conservation of momentum just before & just after the collision to find relation between v & u; mv = (m+M)u u = mv/(m+M) u/v = m/(m+M) u^2/v^2 = m^2/(m+M)^2 what should the answer be. help!
anonymous
  • anonymous
The answer is in the question..... And if you plug what you just calculated in, then that would be correct.

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