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