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the science of collision
MOMENTUM the science of collision
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= “Keep Goingness” of an object.
p = mv where p = momentum m = mass in kg v = velocity in m/s Momentum
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They can stop dead. i.e. mv = mv So if a fast moving little object collides with a slow moving big object head on
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Example: What is the momentum of a 300 g swallow going 15 m/s?
p = mv = (.300 kg)(15 m/s) p = 4.5 kgm/s Duh! I can tell by the mass.
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What is the momentum of a 0.25 kg swallow going 5.2 m/s ?
1.3 kgm/s p = mv p = (.25 kg )(5.2 m/s) = 1.3 kgm/s W
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What velocity must a 275 gram bullet have for its momentum to be 126
What velocity must a 275 gram bullet have for its momentum to be kg-m/s? 460. m/s p = mv 126.5 kgm/s = (.275kg)v v = (126.5 kgm/s)/(.275kg) = 460 m/s W
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A bowling ball has a momentum of 43
A bowling ball has a momentum of 43.6 kg-m/s when it is moving at 12 m/s. What is its mass? 3.6 kg p = mv 43.6 kgm/s = m(12 m/s) m = (43.6 kgm/s)/ (12 m/s) = 3.6 kg W
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LAW OF CONSERVATION OF MOMENTUM
The sum of the momentum before and after the collision remains the same. m1v1 = m2 v2
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Elastic and inelastic collisions in one dimension
Momentum is conserved in any collision, elastic and inelastic. Mechanical Energy is only conserved in elastic collisions. Perfectly inelastic collision: After colliding, particles stick together. There is a loss of energy (deformation). Elastic collision: Particles bounce off each other without loss of energy. Inelastic collision: Particles collide with some loss of energy, but don’t stick together.
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Perfectly inelastic collision of two particles
(Particles stick together) Notice that p and v are vectors and, thus have a direction (+/-) There is a loss in energy Eloss
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Elastic collision of two particles
(Particles bounce off each other without loss of energy. Momentum is conserved: Energy is conserved:
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The momenta must be equal but opposite:
60 kg Fran is running at 4 m/s when she collides with 80 kg Joe. They hit and stop dead, so how fast was Joe going? 3 m/s The momenta must be equal but opposite: Fran: p=mv=(60 kg)(4 m/s)=240 kgm/s Joe: p=mv, 240 kgm/s = (80 kg)v, v = (240 kgm/s)/ (80 kg) = 3 m/s. W
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Consider the previous problem:
Since momentum is really a vector, the true initial momenta were: Fran: p=mv=(60 kg)(+4 m/s)=+240 kgm/s Joe: p=mv=(80 kg)(-3 m/s)=-240 kgm/s The total momentum was the same before and after the collision. This is true for every collision. This is very useful
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