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Physics

Essay by   •  September 13, 2010  •  Essay  •  432 Words (2 Pages)  •  1,693 Views

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Conservation of Momentum

1.

Trial 1 T1 (s) T2 (s) Vi (m/s) V2 (m/s)

0 0.071 0.351

1 0.111 0.225

2 0.118 0.215

Trial 2

0 0.061 .409

1 0.092 0.272

2 0.101 0.248

Trial 3

0 0.057 0.440

1 0.083 0.300

2 0.088 0.283

Mass of car 1 = 993.0 g

Mass of car 2 = 496.7 g

2. trial 1

Car 1 momentum before collision

P=mv P=(.993kg)(.351m/s) P= .349 kgm/s

Car 2 momentum before collision

P=mv P=(.4967kg)(0m/s) P = 0 kgm/s

Object's (or both cars together) momentum after collision

P=mv P=(1.4897kg)(.225) P = .335 kgm/s

Trial 2

Car 1 momentum before collision P=mv

P= (.993kg)(.409m/s) P= .406 kgm/s

Car 2 momentum before collision P=mv

P= (.4967kg)(0) P= 0 kgm/s

Objects momentum after collision P=mv

P = (1.4897kg)(.272m/s) P= .405 kgm/s

Trial 3

Car 1 momentum before collision

P= mv P = (.993kg)(.440m/s) = .437 kgm/s

Car 2 momentum before collision

P=mv P= (.4967kg)(om/s) = 0 kgm/s

Objects momentum after collision

P=mv P= (1.4897kg)(.300m/s) = .447 kgm/s

3.

Total momentum of the system before the collision (the first car's momentum)

Trial 1= .349 kgm/s

Trial 2= .406 kgm/s

Trial 3= .436 kgm/s

Total momentum of the system after the collision (objects momentum after the collision)

Trial 1= .335 kgm/s

Trial 2= .405 kgm/s

Trial 3= .447 kgm/s

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