roastedchickenwings 2 years ago Prove that the fractional loss in kinetic energy for the ball-pendulum system is given by equation n=1-m/m+M

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1. Jemurray3

This doesn't make sense without context.

2. roastedchickenwings

frictional loss=0.8099 and 0.7963. for silver and brass respectively

3. Jemurray3

Context, meaning you need to explain what the system is.

4. roastedchickenwings

how would i explain it. that was the question on my lab book. I apologize for being vague

5. Jemurray3

I would assume the question pertains to a system that was described somewhere...

6. roastedchickenwings

does it apply to the prior question?

7. roastedchickenwings

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.

8. Jemurray3

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

9. roastedchickenwings

thank you

10. roastedchickenwings

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!

11. Jemurray3

The answer is in the question..... And if you plug what you just calculated in, then that would be correct.