Query Assignment 10

Your work on this assignment is good.

Let me know if anything is unclear, and include specifics about what you do and do not understand.

You are incidentally correct about the University Physics problem. You don't need to do it, and that is a bit of a scary problem.

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19:44:45 Query introductory problem set 3 #'s 7-12 Describe two ways to find the KE gain of an object of known mass under the influence of a known force acting for a given time, one way based on finding the distance the object moves and the other on the change in the velocity of the object, and explain why both approaches reach the same conclusion.

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RESPONSE --> KE = .5mv^2 is used for the change in the velocity of the object. By finding the distance the object moves, use, 'ds = vAve'dt = .5 m (F^2/m) * dt^2 Both equations reach the same conclusion, they only use different variables that are given. One uses the change in the velocity and the other uses the distance that the object moves.

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19:46:52 ** 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. **

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RESPONSE --> I found these equations in the notes but they're for 'dW, which I'm now guessing is the same thing as KE. I hadn't realized that previously.

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19:47:59 Query what is our experimental evidence that the acceleration of a constant mass is proportional to the net force exerted on the mass?

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RESPONSE --> By looking at the graph constructed from the data, since nearly all the points lie on the line or very close to it (depending on experimental error) and the line passes through the origin, this shows that the acceleration of a constant mass is directly effected by the net force that is exerted on the mass.

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19:48:24 ** STUDENT RESPONSE: acceleration is proportional because if given the mass and force you can find the acceleration and also if given the acceleration and mass you can get the force INSTRUCTOR COMMENT In exp 12 we see that net force is proportional to effective slope. In a previous experiment we found that acceleration is proportional to effective slope. Since net force and acceleration are both proportional to slope, we conclude that at least in the case of a cart rolling down a ramp the net force is proportional to acceleration. **

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RESPONSE --> ok.

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20:01:19 Query gen phy prob 2.16 sports car 90 km/h stops in 50 m; g force.

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RESPONSE --> vf = 95km/h v0 = 0 'dt = 6.2s 95km/h (1000m / 1km) ( 1hr / 3600s) = 95000/3600 vf = 26.389m/s vf = v0 + a 'dt a = vf / 'dt = 26.389m/s / 6.2s = 4.256m/s^2

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20:06:21 ** 90 km/hr = 90,000 m / (3600 sec) = 25 m/s. So change in velocity is `dv = 0 m/s - 25 m/s = -25 m/s. Average velocity is vAve = (vf + v0) / 2 = (25 m/s + 0 m/s) / 2 = 12.5 m/s. So the time to come to a stop is `dt = `ds / vAve = 50 m / (12.5 m/s) = 4 s. Acceleration is rate of velocity change = change in velocity / change in clock time = -25 m/s / (4 s) = -6.25 m/s^2. One 'g' is the acceleration of gravity, 9.8 m/s^2. So the given acceleration is -6.25 m/s^2 / [ (9.8 m/s^2) / 'g' ] = -.61 'g'. The sign of the g-force is usually omitted so we may say that the acceleration is .61 'g'. **

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RESPONSE --> My book gave a different problem for number 2.16. It said a car accelerates from rest to 95km/h in 6.2s. What is its average acceleration in m/s^2... which is what I solved... I know I converted the km/hr to m/s correctly. I'm not sure how the change in velocity is -25m/s when the question said that the car slowed down in 50m. I'm not sure I could have figured out the 'g force regardless though. I'm actually wondering if I had to do this problem at all...

I didn't notice the change in the problem with the new edition, which has the car accelerating from rest rather than coming to a stop, and uses 95 m/s rather than 90 m/s.

However the two problems are pretty much equivalent, and your solution was good.

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20:07:04 univ phy 2.54 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.

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RESPONSE --> Haha I'm pretty sure this problem doesn't apply to me. Thank God for that. Scary.

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20:07:32 ** 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. GOOD STUDENT SOLUTION: for the two trains to colide, the 25 m/s train must have a greater velocity than the 15 m/s train. So I can use Vf = V0 + a('dt). 15 = 25 + (-.1)('dt) -10 = -.('dt) 'dt = 100 so unless the displacement of the 25 m/s train is greater than the 15 m/s train in 100 s, their will be no colision. 'ds = 15 m/s(100) + 200 m 'ds = 1700 m 'ds = 25 m/s(100) + .5(-.1)(100^2) = 2000 m. The trains collide. **

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RESPONSE --> ok

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