- #1
compwiz3000
- 17
- 0
A bullet of mass [tex]m_1[/tex] strikes a pendulum of mass [tex]m_2[/tex] suspended from a pivot by a string of length [tex]L[/tex] with a horizontal velocity [tex]v_0[/tex]. The collision is perfectly inelastic and the bullet sticks to the bob. Find the minimum velocity [tex]v_0[/tex] such that the bob (with the bullet inside) completes a circular vertical loop.
What did I do wrong:
Using conservation of momentum, we have
[tex]m_1 v_0 = \left(m_1+m_2\right) v_n.[/tex]
Then
[tex]v_n=\frac{m_1 v_0}{m_1+m_2}[/tex]
and
[tex]0=v_n^2+2a dx.[/tex]
Then
[tex]v_n^2=-2 \int_0^\pi -g \sin \theta L d \theta?[/tex]
I think that's where I screwed up?
If I follow my wrong steps, I end up with
[tex]v_0=\left(m_1+m_2\right)2 \sqrt{Lg}/m_1,[/tex]
but that's not the answer.
What did I do wrong:
Using conservation of momentum, we have
[tex]m_1 v_0 = \left(m_1+m_2\right) v_n.[/tex]
Then
[tex]v_n=\frac{m_1 v_0}{m_1+m_2}[/tex]
and
[tex]0=v_n^2+2a dx.[/tex]
Then
[tex]v_n^2=-2 \int_0^\pi -g \sin \theta L d \theta?[/tex]
I think that's where I screwed up?
If I follow my wrong steps, I end up with
[tex]v_0=\left(m_1+m_2\right)2 \sqrt{Lg}/m_1,[/tex]
but that's not the answer.