Bullet & Pendulum Homework: Determine Horiz. Displacement

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The discussion focuses on solving a physics problem involving a bullet embedded in a pendulum, requiring the determination of the pendulum's horizontal displacement. The key concepts include the use of conservation of momentum during the bullet-pendulum collision and conservation of energy as the pendulum swings upward. Participants emphasize the need to apply different conservation principles for each stage of the interaction, noting that mechanical energy is not conserved during the collision. The equations for kinetic and potential energy are discussed, with clarification that different quantities are conserved in each phase of the motion. Understanding these principles is crucial for accurately calculating the pendulum's displacement.
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Homework Statement


An 16.2 g rifle bullet traveling 240 m/s buries itself in a 3.54 kg pendulum hanging on a 2.90 m long string, which makes the pendulum swing upward in an arc. Determine the horizontal component of the pendulum's displacement.


Homework Equations



I used the Y center of mass equation which is:
Ycm= m1y1+m2y2/m1+m2 but i got 2.9


The Attempt at a Solution


Im stuck and don't know what else to do
 
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use conservation of (total) energy and momentum. Neglect the heat that the bullet will "create" when hitting the pendelum. You must use some trigometry also.
 
Consider the interaction as having two stages:
(1) The collision of bullet and pendulum--what's conserved?
(2) The rise of the pendulum+bullet (post collision)--what's conserved?
 
Doc Al said:
Consider the interaction as having two stages:
(1) The collision of bullet and pendulum--what's conserved?
(2) The rise of the pendulum+bullet (post collision)--what's conserved?

so i find the PE+KE=PE+KE for both parts

or in other words

.5MV2=mgh for both parts?
 
Yes.

First the energy is the kinetic energy of bullet. Then all energy will be potential energy of the pendelum and the bullet. (m1+m2)gh
 
BMWPower06 said:
so i find the PE+KE=PE+KE for both parts

or in other words

.5MV2=mgh for both parts?
Not exactly. Different quantities are conserved in each part of the motion. For example: During the collision, mechanical energy is not conserved. What is?
 
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