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course Phy 242
March 1 around 11am. I had a few questions on some near the end, but I commented on what I didn't understand and such.
110216`q001. A BB of mass .12 g is shot at 80 m/s into the space between two tiles. The tiles are separated by 15 cm.
If the BB continues bouncing back and forth between the tiles for 2 seconds, without losing any of its speed, what average force does it exert on each tile?
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30cm = one roundtrip
Fave*dt = m dv
Fave = (m dv)/(dt)
(m dv) = (0.12 g)*(160m/s) = 0.02 kg
(dt) = (ds/v) = (0.3 m/80 m/s) = 0.004 seconds
Fave = (0.02 kg) / (0.004 seconds) = 5 N
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How would the result change if the tile separation was reduced to 5 cm?
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The roundtrip would be different. It would be 10cm instead of 30cm, which would change the change in time calculation.
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@& You can be more specific. Time interval would become 1/3 as great, tripling the force to 15 N.*@
How would the result change if the BB was fired into a slightly irregular tile-lined 'box' approximately 5 cm on a side?
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The BB would be going in all sorts of directions, hitting each tile.
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@& The directions would be randomly distributed in 3-dimensional space, splitting the KE (originally all directed along a single axis) into 3 equal parts, one in the original direction and the other two in two remaining perpendicular directions. This would result in 1/3 the average force on each tile.*@
What does this question have to do with the kinetic theory of gases?
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Molecules are in constant random motion, frequent collisions between the molecules and sides of container.
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`q002. A wave is observed to travel down a rubber band chain of length 2.5 meters, making seven round trips in 10 seconds. What is the propagation velocity of that wave?
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7 * [(2*2.5m)/(10 sec)]
=3.5 m/s
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By sending alternate pulses down the chain on alternate sides, a standing wave is created with a single antinode, located halfway down the chain. How much time must elapse between the
ulses?
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Wave travels 2.5 meters, makes 7 roundtrips in 10 seconds.
5 seconds?!
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@& In 5 seconds a pulse would complete about 7 round trips. You need to reinforce each reflected pulse, so you're going to need to do that a number of times in 5 seconds.
A pulse takes only about .7 seconds to travel the length of the chain, at which time a pulse must be sent down the other side to reinforce the reflected pulse.*@
How many pulses are sent by the time the first disturbance reaches the end of the chain? What is the shape of the chain at that instant?
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Is the next 4 questions where I use the “N A N A N” and lambda methods?!
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The pulse rate is increased until the standing wave contains two antinodes, with a node in the middle. How much time must elapse between the pulses?
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@& To get two antinodes into the chain at the same time, you need to send two opposite pulses down the chain by the time the initial disturbance reaches the other end. You'll need two pulses, one to the right and one to the left, in about .7 seconds, so it will be about .35 seconds between a right and a left pulse.*@
How many pulses are sent by the time the first disturbance reaches the end of the chain? What is the shape of the chain at that instant?
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@& There will be a 'peak' and a 'valley' in the chain, separated by half the length of the chain. *@
If a complete cycle consists of two alternate pulses, then in each case, how many complete cycles of the wave correspond to its length?
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@& In the preceding the length of the chain corresponds to a complete cycle.*@
`q003. Suppose the tension in a rubber band chain is 1 Newton for every 10% change in its length. If our chain has length 2.5 meters when under a tension of 1 Newton, then what will be the tension when its length is 2.7 meters, what will be the tension at length 3 meters, and at what length will the tension be 5 Newtons?
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2.7 meters: 3 Newtons
3 meters: 6 Newtons
Length at tension 5 Newtons: 2.9 meters
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@& At 2.7 meters, the length has increased by .2 meters over its 2.5 meter length, which is .2 / 2.5 = .08 or 8% of its length. If a 10% increase adds a Newton, an 8% increase adds .8 Newton so the tension will be 1.8 N.
At 3.0 m the length is 20% greater than the original 2.5 m length so the tension is 2 N greater, or 3 N.
To get 5 N the length must be 50% greater, or 2.5 m + 50%(2.5 m) = 3.75 meters.*@
If the speed of the wave is c = sqrt( T / (m/L) ), then what is (m / L) for our chain (use also the information you obtained in the preceding question)?
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m/L = T/c^2
Wouldn’t you need “c” or the propagation speed in order to solve these problems? Or would I use the previous “c” which was 3.5m/s?!
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Using the 3.5 m/s speed from the previous question, which corresponds to length 2.5 m and therefore to the 1 N tension given here, we would rearrange
c = sqrt( T / (m / L)
to get
m / L = T / c^2 = 1 N / (3.5 m/s)^2 = .1 kg / m,
very approximately.*@
What will be the wave speed at length 3 meters?
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c = sqrt (T/(m/L))
c = sqrt (6 N/(m/L))
c =
See above question…
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@& RIght equation.
The tension at the 3 meter length would be 3 N.
If m / L = .1 kg / m when the chain is 2.5 m long, then since the same mass is now distributed over 3 meters of length we have
m / L = .01 kg / m * (2.5 m / 3 m) = .08 kg / m.
Thus
c = sqrt( T / (m / L) ) = sqrt( 3 N / (.08 kg / m) ) = 6 m/s, approx..*@
At length 3 meters, how frequently should pulses be sent in order to create the fundamental mode of vibration (the one with a single antinode)?
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L= (lambda/2)
3 meters = (lambda/2)
Every 6 meters?!
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@& For the fundamental mode of vibration, one wavelength is double the length of the string, or 6 meters.
At length 3 meters the wave speed is roughly 6 m / s.
Two pulses, one to the right and one to the left, are required to create a full cycle, so two pulses would be sent out per second, and the result would be one complete cycle per second.*@
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@& Check my notes and be sure you understand. Ask if you dont.*@