Bullet & Pendulum Homework: Determine Horiz. Displacement

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Homework Help Overview

The problem involves a bullet embedding itself in a pendulum, leading to a discussion on determining the horizontal displacement of the pendulum after the collision. The subject area includes concepts from mechanics, specifically conservation of momentum and energy.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss using conservation of energy and momentum, with some suggesting the need for trigonometry. There are considerations of the collision dynamics and the subsequent motion of the pendulum.

Discussion Status

Some participants have provided guidance on the stages of the interaction, noting the conservation laws applicable during the collision and the subsequent rise of the pendulum. There appears to be an exploration of different conservation principles, but no consensus has been reached on the specific approach to take.

Contextual Notes

Participants are navigating the complexities of energy conservation in different phases of the motion, questioning the assumptions about energy conservation during the collision.

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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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