# Conservation of Energy and Linear Motion

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## Essay title: Conservation of Energy and Linear Motion

Abstract:

I am to determine the speed of the ball as it leaves the ballistic pendulum using conservation of linear momentum and conservation of energy considerations. Students fired a ball out of a spring canon towards a ballistic pendulum and this was knocked upward which left a mark on the protractor at the angle that the pendulum reached. The student used this to calculate how high the pendulum went using basic trigonometry rules. This height was used in the conservation of energy formulas to calculate the velocity of the ball and the total mechanical energy of the system. For the third detent the ball was calculated to have a velocity of 5.09 m/s which is a .7% percent difference from the known value for this canon. The calculated initial and final mechanical energy of the system for the third detent had a .0139% difference and for the first detent a .0000067% difference in the values. This small difference could be from friction of the pivot point of the pendulum or due to air resistance. This difference could also be from uncertainties in the measurements of the mass and length of the pendulum.

Results:

|Detent |ME before |ME after |% Diff |

|3rd |.1800 (.0080 J |.1801 ( .0080 J |.0139 % |

|1st |.03727 ( .0080 J |.03726 ( .0080 J |.0067 % |

|Detent |Vball |

|3rd |5.09 ( .967 m/s |

|1st |2.31 ( 1.82 m/s |

Sample Calculations:

1. (h=Lpa-Lpacos(

.303-.303cos(36)=.058 m

2. ½(mb+MPA)V2PAB =(mb+MPA)g(h

VPAB=((2g(h)

((2(9.8)(.058))=1.066 m/s

3. mbvb = (mb+Mpa)Vpab

vb=(( mb+Mpa)( Vpab))/(mb)

((.0664+.2505)(1.066))/(.0664)=5.09 m/s

4. MEi = KEi

MEi = ½(mb+MPA)V2PAB

½(.0664+.2505)(1.066)2= .1800 J

5. MEf = PEf

MEf = (mb+MPA)g(h

(.0664+.2505)(9.8)(.058)= .1801 J

6. % Diff= ((x1-x2()/(2(x1+x2))(100)

((.1800-.1801()/(2(.1800+.1801))(100)= .00014 %

Discussion:

This lab was a success, all of the objectives were achieved and our results were expected. First the students calculated

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