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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.