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course Phy 201
`q001. Give your data from today's lab activities.The farthest point that the steel ball does not go backwards when collided with the marble was at 35cm.
The farthest point that the marble does not go backwards when collided with the steel ball was at 46cm.
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The length of the ramp is also necessary to make sense of these quantities. The positions at which the balls were released would also be very relevant..
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`q002. Sketch an elliptical orbit representing the path of a satellite about a planet, with the orbit 4 times as 'long' as it is 'wide', with the 'long' side in the direction of the x axis. The planet should be much closer to one end of the ellipse than the other.
The perigee of the orbit is the point at which the satellite makes its closest approach, and the apogee is the point at which it is furthest from the planet.
On your sketch mark the point of the orbit that is 1/4 of the way from the perigee to the apogee. At that point sketch a vector which shows the direction in which the satellite would be moving at the instant it reaches the point. What do you estimate to be the angle of that vector relative to the direction of the positive x axis? 70deg
Also sketch a vector that represents the force experienced by the satellite at that point. This force will be in the direction of the center of the planet. What angle does this force make with the positive x axis? 45 deg
Sketch the projection of the force vector on the direction vector. What is the length of this projection vector, as a percent of the length of the force vector?
About 60%
Is the projection vector in the direction of motion or opposite to the direction of motion? Opposite
Is the gravitational force therefore doing positive or negative work on the satellite? neg
Is the gravitational potential energy of the satellite increasing or decreasing? dec
Is the kinetic energy of the satellite increasing or decreasing? inc
Is the satellite speeding up or slowing down? Speeding up
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If gravitational force is doing negative work on the satellite, then since the change in PE is equal and opposite to the work done by the net force, the PE will be increasing. Since the only force present is the gravitational force, it follows that the kinetic energy must be decreasing and the satellite slowing.
If gravity is pulling in the direction opposite motion it will be slowing the satellite.
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`q003. Repeat the exercise of the preceding problem for a satellite in a circular orbit about a planet, with the center of the orbit at the center of the planet.
45 deg, 25deg, 50%, opposite, neg, dec, inc, stay the same
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The velocity will be tangent to the circular path, the force directed toward planet at the center of the circle. So the force and the velocity will be perpendicular, and the force component in the direction of the velocity will be zero.
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`q002. A BB has a mass of about 1 gram. A certain locomotive has a mass of about 10 000 kg. A BB is shot at the locomotive, which is approaching at 30 m/s. The BB strikes the locomotive head-on with a speed of about 100 meters / second and bounces straight back with a velocity of 130 meters / second.
Choose whether you want the direction of the BB's velocity or the direction of the locomotive to be positive or negative.
What are the velocities of the BB before and after the collision? Be sure your velocities have the correct signs.
What is the momentum of the BB before collision, and what is its momentum after collision? Again be sure you have the correct signs.
What therefore is the change in the BB's momentum (remember, to get the change you subtract the initial value from the final value; be careful of signs)?
Approaching pos 100m/s
Bounces off neg 130m/s
What is the initial momentum of the locomotive?
Before M = 0.001kg(100m/s) = 0.1N
After M = 0.001kg(-130m/s) = -0.13N
What must be the change in the momentum of the locomotive?
0.1N- -0.13N = 0.23N
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You have found the momentum of the BB before and after, and the change in its momentum.
You haven't yet found the momentum of the locomotive.
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What therefore is the change in its velocity?
Momentum = mv
10,000kg(30m/s) = 300,000N
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This is the momentum of the locomotive, not its change in velocity.
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The change in the locomotive's momentum will be equal and opposite to that of the BB. What therefore will be the locomotive's new momentum, and by how much will its velocity change?
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Assuming that the locomotive traveled with this velocity for a day, how much distance would it lose compared to the distance it would travel had it maintained its 30 m/s velocity?
-30m/s a days’ worth of travel
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Good start, but you'll need a few revisions.
&#Please see my notes and submit a copy of this document with revisions, comments and/or questions, and mark your insertions with &&&& (please mark each insertion at the beginning and at the end).
Be sure to include the entire document, including my notes. &#
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