1. Unit 5: Conservation of Momentum
Force and Momentum and Conservation of Momentum (9-1,9-2, 9-3)
Collisions (9-4, 9-5, 9-6)
Collisions, Center of Mass (9-7, 9-8)
Center of Mass, Rocket Propulsion, and Quiz Three Review (9-9, 9-10)
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2. Center of Mass and Translational Motion
We can prove that the translational motion of a system can be described by considering the effect of the net force on the center of mass.
In that sense the system is no different than a single force incident on a single particle.
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3. Since all the internal forces cancel by the 3rd Law, the summation involves only the external forces:
This is Newton’s 2nd Law applied to a system: the sum of all the forces acting on a system is equal to the total mass of the system times the acceleration of its CM.
In other words, the CM behaves as if all the mass of a system or object resides at the CM and as if all the external forces act directly upon it!
This makes treatment of the translation motion of a system quite straightforward.
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5. Example: A Two-Stage Rocket
Problem
A rocket shot into space reaches it’s highest point and a horizontal distance d from it’s origin separates into two equal mass segments.
Imagine the first part just stopped and dropped immediately to the earth. Where does part II land?
Answer
Because the explosion is “internal” the CM will continue on it’s path.
The explosion happened at the top of the trajectory and if the rocket had maintained it’s trajectory the CM would have landed 2d from the origin.
Thus the average position of the two equal mass fragments at impact must be 2d.
This can only occur if the second stage travels a total horizontal distance of 3d
Notice how we didn’t need to resort to any fancy equations of motion and just applied the 2nd Law for systems.
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7. Systems of Variable Mass
Up until now we’ve limited ourselves to systems of constant mass such that
But what if the mass of the system varies? Then derivative will have two terms: one proportional to dv/dt and the other to dm/dt.
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8. Two Examples where dm/dt is Nonzero
Rockets where dm/dt < 0.
Conveyer belts where dm/dt>0.
Let’s set up a formalism for such cases.
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9. Time t
Time t+dt
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12. The left side of the equation is just the usual mass times acceleration term.
The summation
refers to all external forces such as gravity, friction, etc…
it does not include the force on M due to dM.
The term which depends on the change of mass with respect to time corresponds to the force exerted on M due to the addition or ejection of mass. It is also the change of momentum due to the addition of mass and often called thrust.
This equation can be used to describe a system subject to changes of mass.
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13. Example 1: A Conveyer Belt
A hopper drops gravel at a rate of 75.0kg/s onto a conveyer belt moving at a constant speed of v=2.20m/s.
Determine
The force needed to keep the belt moving.
The power output of the motor drive.
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15. Example 2: A Rocket!
A vertically fired rocket of total mass 21,000kg carries 15,000kg fuel. The fuel is burned at a rate of 190kg/s and leaves the engine at a rate of 2800m/s relative to the rocket.
Calculate
The rocket thrust
The net force on the rocket at blastoff and at burnout
The velocity as a function of time
Final velocity at burnout.
Assume no friction and g. +X
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19. Same format as first two quizzes. Total of 50 points
Quiz Number Three
Same format as first two quizzes.
Total of 50 points
4 multiple 5 pts each
3 5, 10, 15 points.
Bring a calculator
Bring a 3x5” note card crib sheet.
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20. Definition of Work F is force d is displacement
q is the angle between the force and displacement
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21. Definition of Kinetic Energy and the Work-Energy Theorem
Consider an object of mass m moving initially with speed v1.
Now accelerate it uniformly to speed v2 by applying a constant force F over a distance d. V1 V2 F d
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22. Gravitational Potential Energy
At the surface of the earth
U=mgh
At any radius
U= GMEm/r
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23. Potential Energy of a Spring
U = (1/2)kx2
k = spring constant
x= displacement
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24. Conservation of Mechanical Energy
Let’s just consider a conservative system (earth and object, spring and mass…)
One in which the work does not depend on the path taken
Or equivalently one for which the work around a closed path is zero
One in which energy can be transformed from kinetic energy to potential and back again.
By the work-energy principle we know that
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25. The Classic Example: The Falling Rock
Anywhere along the path the total mechanical energy is E= (1/2)mv2+mgh.
When dropped, the rock has no velocity so K =(1/2)mv2=0 but it has potential energy U=mgh
As it falls, K increases and the potential energy decreases so that the total energy is constant.
At the surface, K is at a maximum and the potential at a minimum, U=mgh=0.
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26. What’s it good for?
If h=3.0m, calculate the speed when the rock is 1.0 m above the ground.
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27. The Law of Conservation of Momentum:
The total momentum of an isolated system of bodies remains constant.
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28. Example: The Kick of a Rifle
What is the recoil velocity of a 5.0-kg rifle that shoots at kg bullet at a muzzle velocity of 120m/s?
What would be the initial velocity of a 75-kg rifleperson?
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30. Center of Mass
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