A bullet of mass m1 is fired with a speed v into the bob of a ballistic pendulum
ID: 1695901 • Letter: A
Question
A bullet of mass m1 is fired with a speed v into the bob of a ballistic pendulum of mass m2. The bob is attached to a very light rod of length L that is pivoted at the other end. The bullet is stopped in the bob. Find the minimum v such that the bob will swing through a complete circle. (Use any variable or symbol stated above along with the following as necessary: g.)v0 = ?
I thought the answer was ((M1+M2)/M1)SQRT2gL but it says that it is incorrect, so I guess I am stumped on what equation they are looking for.
Explanation / Answer
The mass of the bullet is m1
the mass of the block is m2
let the initial speed of the bullet is vi = v
To complete the ciclre the combine block should contain speed at the bottom is 5gr
so from law of conservaiton of momentum
m1v1i = (m1+m2)vf
m1v = (m+m2) 5gL
since 5gl is the minimum speed at the bottom to the complete the circle.
therefore the minimum speed of the bullet
v = (m1 + m2)m1 (5gL)
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