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course PHY 232
11:09 am 3/15
110216 Physics 232`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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Imp/Mom
F_ave* `d t = m `d v
m= 0.00012 kg
`d v = 160 m/s
m `d v = 0.0192 kg m/s
`d t = 0.3/80 = 0.00375 s
F_ave = 0.0192/0.00375 = 5.12 N
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How would the result change if the tile separation was reduced to 5 cm?
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Greater Pressure because of greater frequency of collisions,which is due to the fact that `d t gets much smaller.
`d t= 0.05/80 = 0.000625
F_ave = 0.0192/0.000625 = 30.72 N
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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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Now the distance through which the BB moves varies so the frequency of collisions varies depending on which part is struck. It depends on how much of the area is irregular to determine the exact amount of P change there is and it seems you could work out a ratio based on that coverage.
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What does this question have to do with the kinetic theory of gases?
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The BB could be replaced by the word particle. This process continues on to arrive at the translational KE expression used to determine internal energy.
@& In an irregular box the bb will quickly find its velocity randomly distributed among the 3 directions. *@
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Optional problem: If the BB loses 10% of its momentum with each collision, what average force does it exert over the first 10 round trips? General College Physics students can use estimates, as can University Physics students. However University Physics students should consider applying calculus and/or differential equations.
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v_n= v_0- 0.1*n*v_0
Thus the momentum change is found my multiplying this expression by mass:
p_n= m(v_0-0.1nv_0) = p_0 - 0.1*n*p_0
The total momentum can be found by integrating this eq over the interval 0-10, 10 since that is how many strikes before the BB has lost all v and thus p.
P= p_0*n-0.05n^2*p_0 = 5*p_0
@& Very good.*@
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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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total distance travelled= 14 trips * 2.5 m/trip = 35 m
c= 35 m / 10 s = 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 pulses?
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The second pulse should occur when the first pulse is just arriving at the far node so that the two will interfere constructively.
t_delay = x/c = 2.5 m/ 3.5 m/s = 0.7 s
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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?
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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At the instant the first pulse is at the far node 2 pulses will be sent and the chain is entirely in the dir of first pulse.
For 2As the 2nd pulse should be sent when the 1st pulse is halfway down the chain so that the 2 pulses meet at the 3/4 mark and interfere constructively.
1.25/ 3.5 = 0.4 s
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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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The shape is such that the 1st 1/2 of the chain is in dir of 2nd pulse, the 2nd 1/2 is in dir of 1st pulse.
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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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For the 1A the wavelength is double the length of the chain and for the 2A the wavelength is the length of the chain.
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`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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The init length is 2.27m and so 2.7 is 19% of that and so the tension is 1.9 N
2.5/1.1 = 2.27 --> 2.7/2.27 = 1.19 = +19% --> 19/10 = 1.9--> 1.9 * 1 = 1.9 N
3/2.27 = 1.32 --> +32% --> 32/ 10= 3.2 --> T= 3.2 N
5* 10 = 50 ---> 50% --> L = 2.27*0.5 + 2.27 = 3.405
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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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c= 3.5 m/s T= 1 N
m/L = 0.0816 kg/m
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What will be the wave speed at length 3 meters?
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c= sqrt (3.2/0.0816) = 6.26 m/s
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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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`d t = x/c = 3/ 6.26 = 0.48 s
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`q004. BB data: Shot at the same trajectory the BB follows a path similar to that of a good sand wedge shot in golf, but I believe it starts out with greater velocity and loses speed more quickly. In the sunlight against a blue sky you can track the thing for just about its entire flight. Pretty neat. Shot at the Moon, without correcting for gravity, and didn't really miss it by all that much. Don't want to steal NASA's thunder to I'll leave it at that. Shot into water at a low enough trajectory it will skip without losing too much speed (note: you generally don't want to shoot a gun at water because the bullet could 'skip' with dangerous consequences; I have a high bank on the opposite side). Shot at a nearly vertical trajectory the water stops it within a few centimeters. All this appears to be consistent with the gun's rating of 0.3 Joules and the BB's .12 gram mass (which is in turn consistent with the 250 ft/sec muzzle velocity). At that speed the BB should climb for about 1000 feet, if shot in a vacuum, which is what leads me to believe it's losing velocity.
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I don't see a question here.
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@& Very good. I guess that last question was so obvious I didn't bother to ask it. My fault.*@