Dynamics: Cause of Motion Sections 7.1, 7.2. Updates/Reminders Test #2 has been moved from Tuesday...
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Transcript of Dynamics: Cause of Motion Sections 7.1, 7.2. Updates/Reminders Test #2 has been moved from Tuesday...
![Page 1: Dynamics: Cause of Motion Sections 7.1, 7.2. Updates/Reminders Test #2 has been moved from Tuesday next week to Thursday next week due to a need for more.](https://reader036.fdocuments.in/reader036/viewer/2022082818/56649eef5503460f94bffb34/html5/thumbnails/1.jpg)
Dynamics: Cause of Motion
Sections 7.1 , 7.2
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Updates/Reminders
• Test #2 has been moved from Tuesday next week to Thursday next week due to a need for more time to prepare for content.
• Addresses Chapters 4, 6, and 7.• Next reading quiz due prior to class on Tuesday• LAB A4-PM: Projectile Motion due Friday 4pm• Weekly Reflection #7 sent out on Friday morning• Final Exam Thursday, Dec 11, 10am-12:00pm
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Conservation of Momentum
• Demonstration: the total momentum of an isolated system is conserved. That is, pi = pf if and only if we have an “isolated system” – no unbalanced outside forces.
• Example 1 – the collision of carts• Example 2 – the “explosion” of carts• Example 3 – the addition of mass
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Numerical Problem 1
• A truck with a mass of 3,000kg and a velocity of +20m/s collides head on with a car of 1,500kg mass and a velocity of -20m/s. If the vehicles stick together, what is the resulting motion of the pair?
• Solve using conservation of momentum; that is, pi = pf.
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Numerical Problem 2
• Two dynamics carts are separated by a compressed spring. The mass of cart 1 is twice the mass of cart 2; that is, m1=2m2. The spring is released and the carts fly apart. If cart 1 has a velocity of +3m/s, what is the velocity of cart 2?
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Numerical Problem 3
• A train car (mass 11,200kg) is moving along at a speed of +2m/s. A 5,000kg mass of coal with no horizontal motion is dropped into the hopper of the train car. What is the resulting motion of the train car now loaded with coal?
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The Second Law of Motion
• a.k.a. Newton’s Second Law• The net instantaneous force acting on an
object is precisely the instantaneous change of its momentum per unit time. In symbols, the second law can be written as F = Δp/Δt with Δt very small.
• Note that F and Δp are vectors. • Not very enlightening; time for an experiment.
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Newton’s Second Law
• Experiment:– Acceleration as a function of force (constant mass
system)– Acceleration as a function of mass (constant force
system)• Results:– a is proportional to F– a is inversely proportional to m
• Conclusion: F = kma; k=1 if F defined to be in Newtons, N
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Second Law Reprise
• F = Δp/Δt• F = Δmv/Δt• F = mΔv/Δt• F = ma • Both F and a are vectors; m is a scalar.• The sum of forces acting on a body produce an
acceleration inversely proportional to mass.• ΣF=ma (where F is expressed in Newtons)
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2nd Law Problem – Examples
• What is the weight of a person with a mass of 81kg. Note that g = 9.8m/s2.
• An car with a mass of 1,500kg accelerates at a rate of -2m/s2 under a constant force. What are the magnitude and direction of that force?
• How much force would it take to slow a 75kg person riding in a car going +20m/s to a complete stop if the time was 0.03 seconds – the typical time of an auto collision?
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More Examples of Newton’s 2nd Law
• An object accelerates at a rate of 1.5m/s2 under a force of 53N. What is its mass?
• How much upward force does the ground apply to someone with a mass of 75kg to counter balance the pull of gravity? Note that g = -9.8m/s2 and that Fnet = ma and that a = g.
• A force of 7.0N is applied to a 3.5-kg mass for 2.0 seconds. What is the change of velocity?