course 201
11/5 9:30 pm
Class 091102Things you need to know for the test:
Here is an overview of things you really need to understand for the upcoming test:
• Definitions of rate, velocity, acceleration.
• Equations of uniformly accelerated motion.
• Newton's three laws.
• Definitions of work/energy, KE, impulse, momentum.
• Work-KE theorem, definition of PE, work-energy theorem
• Impulse-momentum theorem, conservation of momentum.
• Vector components from magnitude and angle; magnitude and angle from vector components.
and, of course
• Ability to apply the above to various situations.
Using your notes from class, information given in class notes, information given in the text, tt is suggested that you carefully write out what you know, and are supposed to know, about each of the first seven topics above.
Short-answer questions
`q001. What are the magnitude and the standard angle of a vector whose components are -7 m/s and +5 m/s?
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A^2+b^2=c^2
-7m/s^2+5m/s^2=c^2
C=8.6 m/s
Theta=arctan(y/x)=arctan(5/-7)= -35.5, but since this vector is in the first quadrant, we add 180 degrees to compensate for arctan’s capabilities to calculate this value. -35.5+180=144.5 degrees
Right, but note that -35.5 deg is in the fourth quadrant, and the reason you add 180 deg is that the x component is negative.
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`q002. If you add a vector with magnitude 3 N and angle 70 degrees to a vector with magnitude 5 N and angle 210 deg, what are the components of the resultant?
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VectorAxcomp=3N*cos(70)=1 N
VectorAycomp=3N*sin(70)=2.8 N
VectorBxcomp=5N*cos(210)=-4.3 N
VectorBycomp=5N*sin(210)= -2.5 N
Total x component= 1 N+-4.3 N= -3.3 N
Total y component=2.8 N+-2.5 N=.3 N
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What are the magnitude and angle of the resultant? (note: the resultant of a set of vectors is their sum, i.e., what you get when you add them).
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To find the magnitude of the vector, the Pythagorean theorem can be used with each of its components as the sides of the triangle and the magnitude as the hypotenuse.
A^2+b^2=c^2
-3.3N^2+.3N^2=c^2
C=3.3 N
Theta=arctan(.3/-3.3)=-5.2 degrees, again this angle is in the wrong quadrant so we add 180 degrees, -5.2+180=174.8 degrees
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`q003. What are the x and y components of the initial projectile velocity of a pendulum mass if you let go of the string when the mass is traveling at 80 cm/s at an angle of 15 degrees above horizontal (i.e., what are the x and y components of a velocity vector with magnitude 80 cm/s and angle 15 degrees as measured from a horizontal x axis)?
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X component=80 cm/s*cos(15)=77.3 cm/s
Y component= 80 cm/s*sin(15)=20.7 cm/s
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Additional problems relevant to preparation for the test:
`q004. An ideal projectile is initially at a known distance `ds_y above the floor, with vertical velocity v0_y and horizontal velocity v0_x. What quantities are known for its vertical motion?
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`ds of the vertical motion is known, it is equal to `ds_y
We can also find the magnitude and angle of the resultant force using vectors. Magnitude can be found using the Pythagorean theorem, and the angle theta can be found using the arctangent function.
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Can these quantities be used to figure out the time interval `dt between the initial instant and contact with the floor? If so, how?
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Yes. We can find average velocity of the horizontal motion. Since we know initial velocity, and we know that final velocity is zero, we then know that the average of the two will simply be half the initial velocity. Since Vave=.5V0 and Vave=`ds/`dt, we can set the two equations equal to each other. .5V0=`ds/`dt and we can then plug in our givens and solve for `dt.
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Having found `dt, is it possible to find the distance the projectile moves in the horizontal direction during this interval? If so, how?
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Yes. Since the `dt of the vertical motion ends in Vf=0, we know that the horizontal vf must also be zero at this point. The `dt can then be used to calculate distance traveled in the horizontal direction by simply multiplying `dt by the initial velocity in the horizontal direction.
There is no uniform-acceleration interval on which the final velocity is zero.
&#See the appended document for details and/or discussion of the solution.
&#
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`q005. A ball of mass .4 kg is tossed vertically upward at 12 m/s and a point which is 3 meters above the ground. It rises a certain distance then falls back down, where it is caught at a point 1 meter above the ground.
What is its KE at the initial instant?
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KE=1/2*m*v^2=.5(.4kg)(12m/s)^2=28.8 kgm^2/s^2=28.8 J
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What is `dPE between the initial and final instant?
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`dPE is equal and opposite to the work done by the conservative forces (in this case gravity) on the system.
Work=Fnet`ds=(9.8 m/s^2)(.4kg)(3m-1m)=7.8 J
`dPE is therefore -7.8 J
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What therefore is the KE at the final instant, and how fast is it therefore traveling? Assume that air resistance is negligible. Solve this problem using energy considerations only, without analyzing the ball's uniformly accelerated motion.
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`dKE=-`dPE+Fnonconservative*`ds
`dKE=7.8 J+0=7.8 J
`dKE=KEf- KE0
28.8 J=KEf-7.8 J
KEf=36.6 J
To find how fast it was traveling:
KE=1/2*m*v^2
36.6 J=.5(.4 kg)(v^2)
V^2=183 m^2/s^2
V=13.5 m/s
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`q006. At the instant it stops at the top of its path, by how much has the gravitational PE of the ball in the preceding exercise changed?
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I am unsure how to solve this problem.
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What therefore is its height above the ground at that instant? Solve this problem using energy considerations only, without analyzing the ball's uniformly accelerated motion.
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If `dW_nc_on = 0, then `dPE + `dKE = 0.
KE_f = 0 so `dKE = -28.8 J.
&#See the appended document for details and/or discussion of the solution.
&#
`q007. Sketch the forces acting on a steel ball as it rolls down a steel ramp which is slope in such a way that its velocity remains constant. What is the net force on the ball as it rolls down this ramp, and how does your sketch depict this fact?
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There are no net forces acting on the ball. Fnormal is equal and opposite to the y component of the gravitational force, so the net force in the perpendicular direction is zero. Ffriction is equal and opposite to the x component of gravity so there is then no net force in the parallel direction either.
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Sketch the forces acting on the same ball as it rolls up this ramp. What is the net force on the ball as it rolls up the ramp, and how does your sketch depict this fact?
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The net force in this situation is parallel to the ramp and in the negative direction. The difference in this case is that since the ball is rolling up the ramp, the x component of gravity and friction are working in the same direction. They are no longer equal and opposite, so they have a net force in the negative direction, slowing the ball down as it rolls up the ramp.
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`q008: Argue for the truth or falsity of the following statement: 'When rolling down a constant-velocity ramp, the rolling friction and the 'parallel component' of the gravitational force are equal and opposite'
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This statement is true. It can be proven because there is no change in velocity as the ball rolls on the constant velocity ramp. Then acceleration is equal to zero because a=`dv/`dt and the numerator is zero. When acceleration is zero, net force is equal to zero as well because of Newton’s second law: fnet=mass*acceleration. When analyzing the system, we know that the forces, if any, must act in the parallel direction because the horizontal forces are already equal and opposite, netting zero. Since we found the net force overall to be zero, it can be proven that the parallel forces (friction and the x component of gravity) are equal and opposite, netting zero as well.
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Good work. See my notes, as well as the appended document.