- #1

matt-83

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## Homework Statement

I am going through past Olympiad qualifying exams as practice for this year's test. I came upon the following problem in the 2008 F=ma and continue to come up with the wrong answer. I keep coming up with C and that is not the answer provided in the answer key.

22. A bullet of mass m1 strikes a pendulum of mass m2 suspended from a pivot by a string of length

L with a horizontal velocity v0. The collision is perfectly inelastic and the bullet sticks to the

bob. Find the minimum velocity v0 such that the bob (with the bullet inside) completes a circular

vertical loop.

(a) 2√Lg

(b) √5Lg

(c) (m1 + m2)2√Lg/m1

(d) (m1 − m2)√Lg/m2

(e) (m1 + m2)√5Lg/m1

## Homework Equations

m1v1= (m1+m2)v2

1/2 mv ^2 = mgh

## The Attempt at a Solution

m1v1 = (m1+m2)v2

v2 = (m1v1)/(m1+m2)

So the kinetic energy of the block should be:

1/2 (m1+m2)v2^2

Then I set the initial kinetic energy of the block equal to the potential energy it would have at height 2L.

1/2 (m1+m2)v2^2 = 2L(m1+m2)g

Since v2 = (m1v1)/(m1+m2):

1/2(m1+m2)*((m1v1)/(m1+m2))^2 = 2L(m1+m2)g

m1+m2 should cancel:

((m1v1)/(m1+m2))^2 = 4Lg

(m1v1)^2 / (m1+m2) ^2 = 4Lg

m1 ^2 * v1 ^ 2 = 4Lg / ((m1+m2) ^2)

Take the Square Root of the whole thing:

m1 * v1 = 2sqrt(Lg)/(m1+m2)

SO my answer is (c) (m1 + m2)2√Lg/m1, which is apparently incorrect.

If someone could help me correct this that would be great.

Thanks.