course Phy 231 6/19 7:45 010. `query 10 Question: `qQuery introductory problem set 3 #'s 7-12
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Given Solution: First way: KE change is equal to the work done by the net force, which is net force * displacement, or Fnet * `ds. Second way: KE change is also equal to Kef - KE0 = .5 m vf^2 - .5 m v0^2. ** On one level, KE = 1/2 m v^2 is simply a formula you have to know. It isn't hard to derive that formula, as you'll see soon, and the 1/2 arises naturally enough. A synopsis of the derivation: If force F_net is applied to mass m through displacement `ds then: a = F_net / m, and vf^2 = v0^2 + 2 a `ds It's not difficult to rearrange the result of these two equations to get F_net * `ds = 1/2 m vf^2 - 1/2 m v0^2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique rating #$&* OK ********************************************* Question: `q (This question applies primarily to General College Physics students and University Physics students, though Principles of Physics students are encouraged, if they wish, to answer the question). In terms of the equations of motion why do we expect that a * `ds is proportional to the change in v^2, and why do we then expect that the change in v^2 is proportional to Fnet `ds? Your solution: When you rearrange the 4th equation of motion, you subtract v0^2 and multiply by 1/2. This gives you 1/2(vf^2-v0^2) = a ds. This makes a ds proportional to the change in v^2 with 1/2 as the constant. You also have Fnet = ma, so Fnet is proportional to a with the mass being the constant multiple. Then you put the ds term onto both sides and you keep the proportionality. Confidence rating #$&*: 2
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Given Solution: In a nutshell: • since vf^2 = v0^2 + 2 a `ds, a `ds = 1/2 (vf^2 - v0^2), so a `ds is proportional to the change in v^2 • since F_net = m a, F_net * `ds = m a * `ds so F_net * `ds is proportional to a * `ds • Thus F_net `ds is proportional to a * `ds, which in turn is proportional to the change in v^2. • Thus F_net `ds is proportional to the change in v^2. More detail: It's very important in physics to be able to think in terms of proportionality. • To say that y is proportional to x is to say that for some k, y = k x. • That is, y is a constant multiple of x. To say that a * `ds is proportional to the change in v^2 is to say that • for some k, a * `ds = k * ( change in v^2)--i.e., that • a * `ds is a constant multiple of the change in v^2. In terms of the equations of motion, we know that vf^2 = v0^2 + 2 a `ds so a `ds = 1/2 (vf^2 - v0^2), which is 1/2 the change in v^2. So a `ds is a constant multiple (1/2) of the change in v^2. Formally we have a `ds = k ( change in v^2) for the specific k value k = 1/2. Now since Fnet = m a, we conclude that Fnet * `ds = m a * `ds and since a `ds = k * ( change in v^2) for the specific k value k = 1/2, we substitute for a * `ds to get Fnet `ds = m * k * (change in v^2), for k = 1/2. Now m and k are constants, so m * k is constant. We can therefore revise our value of k, so that it becomes m * 1/2 or m / 2 With this revised value of k we have Fnet * `ds = k * (change in v^2), where now k has the value m / 2. That is, we don't expect Fnet * `ds to be proportional to the change in velocity v, but to the change in the square v^2 of the velocity. The short answer is that by the fourth equation of uniformly accelerated motion, a `ds = 1/2 (vf^2 - v0^2), which is half the change in v^2, so that a `ds is proportional to the change in v^2. (The proportionality constant between a `ds and change in v^2 is the constant number 1/2). F_net = m a, where m is the mass of the object. So F_net is proportional to a. (The proportionality constant between F_net and a is the constant mass m). Thus F_net `ds is proportional to a `ds, which we have seen is proportional to the change in v^2. The conclusion is the F_net `ds is proportional to the change in v^2. (The proportionality constant between F_net `ds and change in v^2 is 1/2 m.) &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique rating #$&* OK Question: How do our experimental results confirm or cause us to reject this hypothesis? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: When you use the motion equation you see that the acceleration times the change in position will equal the change in position times 1/2. You can see the change in velocity from rest will give you just vf^2. This will give you data in a linear pattern, so the graph is a linear line. Confidence rating #$&*: 1
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Given Solution: The explanation for this result: On a ramp with fixed slope the acceleration is constant so • a `ds is simply proportional to `ds • specifically a `ds = k * `ds for k = a. In the preceding question we saw why • a * `ds = k * (change in v^2), with k = 1/2. In our experiment the object always accelerated from rest. So the change in v^2 for each trial would be from 0 to vf^2. the change would therefore be just • change in v^2 = vf^2 - 0^2 = vf^2. Thus if a `ds is proportional to the change in vf^2, our graph of vf^2 vs. a `ds should be linear. • The slope of this graph would just be our value of k in the proportionality a * `ds = k * (change in v^2), where as we have seen k = 1/2 We wouldn't even need to determine the actual value of the acceleration a. To confirm the hypothesis all we need is a linear graph of vf^2 vs. `ds. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique rating #$&* OK Question: `qGeneral College Physics and Principles of Physics: convert 35 mi/hr to km/hr, m/s and ft/s. Your solution: 35mi/h (1609.344m/1mi)(1km/1000m) = 56.3km/h 35mi/h (1609.344m/1mi)(1hr/3600s) = 15.65m/s 35mi/h (5280ft/1mi)(1hr/3600s) = 51.3ft/s Confidence rating #$&*: 3
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Given Solution: `aWe need a conversions between miles and meters, km and ft, and we also need conversions between hours and seconds. We know that 1 mile is 5280 ft, and 1 hour is 3600 seconds. We also know that 1 inch is 2.54 cm, and of course 1 foot is 12 inches. 1 mile is therefore 1 mile * 5280 ft / mile = 5280 ft, 5280 ft = 5280 ft * 12 in/ft * 2.54 cm / in = 160934 cm, which is the same as 160934 cm * 1 m / (100 cm) = 1609.34 m, which in turn is the same as 1609.34 m * 1 km / (1000 m) = 1.60934 km. Thus 35 mi / hr = 35 mi / hr * (1.60934 km / 1 mi) = 56 (mi * km / (mi * hr) ) = 56 (mi / mi) * (km / hr) = 56 km / hr. We can in turn convert this result to m / s: 56 km/hr * (1000 m / km) * (1 hr / 3600 sec) = 15.6 (km * m * hr) / (hr * km * sec) = 15.6 (km / km) * (hr / hr) * (m / s) = 15.6 m/s. The original 35 mi/hr can be converted directly to ft / sec: 35 mi/hr * ( 5280 ft / mi) * ( 1 hr / 3600 sec) = 53.33 ft/sec. Self-critique (if necessary): OK Self-critique rating #$&* OK ********************************************* Question: `qGen phy and prin phy prob 2.16: sports car rest to 95 km/h in 6.2 s; find acceleration Your solution: Dv = 95km/h = 95000m/h 95000m/3600s = 26.4m/s dt = 6.2s A = 26.4m/s / 6.2s = 4.26m/s^2 Confidence rating #$&*: 3
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Given Solution: `a** 95 km/hr = 95,000 m / (3600 sec) = 26.3 m/s. So change in velocity is `dv = 26.3 m/s = 0 m/s = 26.3 m/s. Average acceleration is aAve = `dv / `dt = 26.3 m/s / (6.2 s) = 4.2 m/s^2. Extension: One 'g' is the acceleration of gravity, 9.8 m/s^2. So the given acceleration is -4.2m/s^2 / [ (9.8 m/s^2) / 'g' ] = -.43 'g'. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique rating #$&* OK ********************************************* Question: univ phy 2.66 train 25m/s 200 m behind 15 m/s train, accel at -.1 m/s^2. Will the trains collide and if so where? Describe your graph. Your solution: The train needs to get down to 15m/s so that it does not wreck Vf= 15m/s V0 = 25m/s A = -.1m/s^2 15m/s = 25m/s + (-.1m/s^2)(dt) -10m/s = -.1m/s^2 dt Dt = 100 seconds to get to 15m/s Ds = 25m/s+15m/s / 2 *100s = 2000 meters to slow down Position of train in front in 100 seconds: Ds = 15m/s * 100s = 1500m Yes, there will be a collision (oops) Time of the collision: Ds = 25m/s(dt)+.5(-.1m/s^2)(dt^2) and ds = 200m+15dt 25dt-.05(dt^2) = 200+15dt .05(dt^2)-10dt +200 = 0 quadratic equation Dt = 22.5s and 178s Time of collision is 22.5 seconds after the initial time. The graph of the distance between the trains versus time is decreasing at a decreasing rate, so it is concave up. At time 0, the distance is 200m. The x intercept for the graph is x=22.5s because that is where the distance equals 0m. Confidence rating #$&*: 3
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Given Solution: If we assume the passenger train is at position x = 0 at clock time t = 0 we conclude that the position function is x(t) = x0 + v0 t + .5 a t^2; in this case a = -.1 m/s&2 and x0 was chosen to be 0 so we have x(t) = 25 m/s * t + .5 * (-.1m/s^2) * t^2 = 25 m/s * t - .05 m/s^2 * t^2. To distinguish the two trains we'll rename this function x1(t) so that x1(t) = 25 m/s * t - .05 m/s^2 * t^2. At t = 0 the freight train, which does not change speed so has acceleration 0 and constant velocity 15 m/s, is 200 m ahead of the passenger train, so the position function for the freight train is x2(t) = 200 m + 15 m/s * t . The positions will be equal if x1 = x2, which will occur at any clock time t which solves the equation 25 t - .05 t^2 = 200 + 15 t(units are suppressed here but we see from the units of the original functions that solutions t will be in seconds). Rearranging the equation we have -.05 t^2 + 10 t - 200 = 0. The quadratic formula tells us that solutions are t = [ - 10 +- sqrt( 10^2 - 4 * (-.05) * (-200) ) ] / ( 2 * .05 ) Simplifying we get solutions t = 22.54 and t = 177.46. At t = 22.54 seconds the trains will collide. Had the trains been traveling on parallel tracks this would be the instant at which the first train overtakes the second. t = 177.46 sec would be the instant at which the second train again pulled ahead of the slowing first train. However since the trains are on the same track, the accelerations of both trains will presumably change at the instant of collision and the t = 177.46 sec solution will not apply. Self-critique (if necessary): OK Self-critique rating #$&* OK "