course PHY201 ”yõù݉O‡Ó̦\‹–Óûå¨Õc‹àÕ¯¥Šû³Ç}assignment #018
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19:06:52 Query intro problem sets Explain how we determine the horizontal range of a projectile given its initial horizontal and vertical velocities and the vertical displacement.
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RESPONSE --> First determine what the time of the fall would be (ignoring any horizontal movement for the time being). We know a, v0, and `ds so we can find `vf by: vf^2 = v0^2 + 2a`ds Once we know vf we can find `dt via: vf = v0 + a`dt And once we know `dt, we can find horizontal displacement by: `dt * horizontal velocity confidence assessment: 2
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19:07:22 ** We treat the vertical and horizontal quantities independently. We are given vertical displacement and initial velocity and we know that the vertical acceleration is the acceleration of gravity. So we solve the vertical motion first, which will give us a `dt with which to solve the horizontal motion. We first determine the final vertical velocity using the equation vf^2 = v0^2 + 2a'ds, then average the result with the initial vertical velocity. We divide this into the vertical displacement to find the elapsed time. We are given the initial horizontal velocity, and the fact that for an ideal projectile the only force acting on it is vertical tells us that the acceleration in the horizontal direction is zero. Knowing `dt from the analysis of the vertical motion we can now solve the horizontal motion for `ds. This comes down to multiplying the constant horizontal velocity by the time interval `dt. **
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RESPONSE --> self critique assessment: 3
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19:36:48 Query class notes #17 Why do we expect that in a collision of two objects the momentum change of each object must be equal and opposite to that of the other?
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RESPONSE --> Because each object is exerting force directly against the other object (assuming a head-on collision). So to the extent that each object exerts its force on the other, the momentum of the other object will react in an equal and opposite manner. confidence assessment: 1
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19:37:41 **COMMON ERROR AND INSTRUCTION CORRECTION: This is because the KE change is going to be equal to the PE change. Momentum has nothing directly to do with energy. Two colliding object exert equal and opposite forces on one another, resulting in equal and opposite impulses. i.e., F1 `dt = - F2 `dt. The result is that the change in momentum are equal and opposite: `dp1 = -`dp2. So the net momentum change is `dp1 + `dp2 = `dp1 +(-`dp1) = 0. **
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RESPONSE --> self critique assessment: 3
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19:43:47 What are the six quantities in terms of which we analyze the momentum involved in a collision of two objects which, before and after collision, are both moving along a common straight line? How are these quantities related?
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RESPONSE --> F m v0 vf `dt `ds They are all part of the work energy theorem equations. confidence assessment: 2
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20:17:00 ** We analyze the momentum for such a collision in terms of the masses m1 and m2, the before-collision velocities v1 and v2 and the after-collision velocities v1' and v2'. Total momentum before collision is m1 v1 + m2 v2. Total momentum after collision is m1 v1' + m2 v2'. Conservation of momentum, which follows from the impulse-momentum theorem, gives us m1 v1 + m2 v2 = m1 v1' + m2 v2'. **
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RESPONSE --> OK, I completely missed the point here. So basically this is the same as saying p1 + p2 = p1' + p2'. This seems pretty straightforward. self critique assessment: 2
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20:28:43 `1prin phy and gen phy 6.47. RR cars mass 7650 kg at 95 km/hr in opposite directions collide and come to rest. How much thermal energy is produced in the collision?
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RESPONSE --> 95000 / 3600 = 26.38 m/s KE = .5mv^2 KE = .5 * 7650 kg * 26.38 m/s^2 KE = 2,663,628 J * 2 trains = 5,327,256 J self critique assessment: 2
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20:30:21 There is no change in PE. All the initial KE of the cars will be lost to nonconservative forces, with nearly all of this energy converted to thermal energy. The initial speed are 95 km/hr * 1000 m/km * 1 hr / 3600 s = 26.4 m/s, so each car has initial KE of .5 m v^2 = .5 * 7650 kg * (26.4 m/s)^2 = 265,000 Joules, so that their total KE is 2 * 265,000 J = 530,000 J. This KE is practially all converted to thermal energy.
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RESPONSE --> self critique assessment: 3
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15:54:40 `1Query* gen phy roller coaster 1.7 m/s at point 1, ave frict 1/5 wt, reaches poin 28 m below at what vel (`ds = 45 m along the track)?
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RESPONSE --> v0 = 1.7 m/s y = 35 m `ds = 45 m a = 9.8 m/s^2 PE (grav) = m * g * y PE = m * 9.8 m/s^2 * 35 m = (343m) J KE = .5mv^2 343(m) = .5mv^2 343 m^2/s^2 = .5v^2 v^2 = 686 m^2/s^2 v = 26.19 m/s The previous is with no friction. With friction of .2m it is hard to calculate this because the m's don't cancel. self critique assessment: 2
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16:03:54 **GOOD STUDENT SOLUTION WITH ERROR IN ONE DETAIL, WITH INSTRUCTOR CORRECTION: Until just now I did not think I could work that one, because I did not know the mass, but I retried it. Conservation of energy tells us that `dKE + `dPE + `dWnoncons = 0. PE is all gravitational so that `dPE = (y2 - y1). The only other force acting in the direction of motion is friction. Thus .5 M vf^2 - .5 M v0^2 + M g (y2 - y1) + f * `ds = 0 and .5 M vf^2 - .5M(1.7m/s)^2 + M(9.8m/s^2)*(-28 m - 0) + .2 M g (45m) It looks like the M's cancel so I don't need to know mass. .5v2^2 - 1.445 m^2/s^2 - 274 m^2/s^2 + 88 m^2/s^2 = 0 so v2 = +- sqrt( 375 m^2/s^2 ) = 19.3 m/s.
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RESPONSE --> Wow... that is a good student solution. I have a lot of respect for that student. I ignored the starting velocity as well as the friction. self critique assessment: 2
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16:04:02 Univ. 7.74 (7.62 in 10th edition). 2 kg pckg, 53.1 deg incline, coeff kin frict .20, 4 m from spring with const 120 N/m. Speed just before hitting spring? How far compressed? How close to init pos on rebound?
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RESPONSE --> confidence assessment: 3
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16:04:09 ** The forces acting on the package while sliding down the incline, include gravitiational force, normal force and friction; and after encountering the spring the tension force of the spring also acts on the package. The normal force is Fnormal = 2 kg * 9.8 m/s^2 * cos(53.1 deg) = 11.7 N, approx.. This force is equal and opposite to the component of the weight which is perpendicular to the incline. The frictional force is f = .2 * normal force = .2 * 11.7 N = 2.3 N, approx.. The component of the gravitational force down the incline is Fparallel = 2 kg * 9.8 m/s^2 * sin(53.1 deg) = 15.7 N, approx.. Friction acts in the direction opposite motion, up the incline in this case. If we choose the downward direction as positive the net force on the package is therefore 15.7 N - 2.3 N = 13.4 N. So in traveling 4 meters down the incline the work done on the system by the net force is 13.4 N * 4 m = 54 Joules approx. Just before hitting the spring we therefore have .5 m v^2 = KE so v = +-sqrt(2 * KE / m) = +-sqrt(2 * 54 J / (2 kg) ) = +- 7.4 m/s. If we ignore the gravitational and frictional forces on the object while the spring is compressed, which we really don't want to do, we would conclude the spring would be compressed until its elastic PE was equal to the 54 J of KE which is lost when the object comes to rest. The result we would get here by setting .5 k x^2 equal to the KE loss is x = sqrt(2 * KE / k) = .9 meters, approx.. However we need to include gravitational and frictional forces. So we let x stand for the distance the spring is compressed. As the object moves the distance x its KE decreases by 54 Joules, its gravitational PE decreases by Fparallel * x, the work done against friction is f * x (where f is the force of friction), and the PE gained by the spring is .5 k x^2. So we have `dKE + `dPE + `dWnoncons = 0 so -54 J - 15.7N * x + .5 * 120 N/m * x^2 + 2.3 N * x = 0 which gives us the quadratic equation 60 N/m * x^2 - 13.4 N * x - 54 N m = 0. (note that if x is in meters every term has units N * m). Suppressing the units and solving for x using the quadratic formula we have x = ( - (-13.4) +- sqrt(13.4^2 - 4 * 60 (-54) ) / ( 2 * 60) = 1.03 or -.8 meaning 1.07 m or -.8 m (see previous note on units). We reject the negative result since the object will continue to move in the direction down the incline, and conclude that the spring would compress over 1 m as opposed to the .9 m obtained if gravitational and frictional forces are not accounted for during the compression. This makes sense because we expect the weight of the object (more precisely the weight component parallel to the incline) to compress the spring more than it would otherwise compress. Another way of seeing this is that the additional gravitational PE loss as well as the KE loss converts into elastic PE. If the object then rebounds the spring PE will be lost to gravitational PE and to work against friction. If the object goes distance xMax back up the incline from the spring's compressed position we will have`dPE = -.5 k x^2 + Fparallel * xMax, `dKE = 0 (velocity is zero at max compression as well as as max displacement up the incline) and `dWnoncons = f * xMax. We obtain `dPE + `dKE + `dWnoncons = 0 so -.5 k x^2 + Fparallel * xMax + 0 + 33 N * xMax = 0 or -.5 * 120 N/m * (1.07 m)^2 + 15.7 N * xMax + 2.3 N * xMax = 0 We obtain 18 N * xMax = 72 N m, approx., so that xMax = 72 N m / (18 N) = 4 meters, approx.. This is only 2.93 meters beyond the position of the object when the spring was compressed. Note that the object started out 4 meters beyond this position. **
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RESPONSE --> self critique assessment: 3
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