Ballistic pendulum and initial speed

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In a ballistic pendulum scenario, an object of mass m is fired at a pendulum bob of mass M, resulting in an inelastic collision where momentum is conserved. After the collision, the combined mass swings to a maximum angular displacement theta, requiring the use of energy conservation principles to find the initial speed v_0. The initial kinetic energy and potential energy at the highest point must be equated, leading to the expression v_0 = ((m+M)/m)(sqrt(2g(L - Lcos(theta)))). The discussion highlights the need to express the height h in terms of theta using trigonometric relationships. The final understanding confirms that h can be represented as L - Lcos(theta).
jaded18
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In a ballistic pendulum an object of mass m is fired with an initial speed v_0 at a pendulum bob. The bob has a mass M, which is suspended by a rod of length L and negligible mass. After the collision, the pendulum and object stick together and swing to a maximum angular displacement theta as shown http://session.masteringphysics.com/problemAsset/1010989/26/1010989A.jpg

Find an expression for v_0, the initial speed of the fired object.
Express your answer in terms of some or all of the variables m, M, L, and theta and the acceleration due to gravity, g.
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i know that there are two events. the first one is the inelastic collision, where momentum is conserved and mv = (m+M)(V) ... and V = mv/(m+M) ... and when the block swings up mechanical energy is conserved so that (K initial + U initial) = (K final + U final) and (0.5(m+M)V^2) = (m+M)g(h) ... and the bullet speed is ((m+M)/m)(sqrt 2gh) ... how do i incorporate this theta variable that the problem is asking for?

i am so confused!
 
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maybe they want me to change h into terms using theta since the problem is not accepting the variable h ... but how do i do this?
 
jaded18 said:
maybe they want me to change h into terms using theta since the problem is not accepting the variable h ... but how do i do this?

use trig. when the pendulum as swung, how far below the pivot is the bob located? you have a right triangle with hypoteneuse L and angle theta.
 
i'm sorry but i don't see it ... i can say cos theta = adj/L and but then what's the 'adj'??
 
oh! wait .. so L-Lcos theta = h?
 
Last edited:
jaded18 said:
oh! wait .. so L-Lcos theta = h?

yes exactly.
 
thanks, it seems like you're always there for me when I'm stuck .. thanks again~
 
jaded18 said:
thanks, it seems like you're always there for me when I'm stuck .. thanks again~

:smile: no prob!
 

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