course Phy 231 σaՀz`assignment #018
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20:04:40 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 --> With the initial vertical velocity, the vertical displacement, and the assumption of acceleation being near the surface of the earch at 9.8m/s, the time that the object takes to fall can be found with `ds = v0`dt + .5a`dt. This equation can be solved for `dt by using the quadratic formula. `dt can then be used to find the horizontal range by multiplying `dt by the constant horizontal velocity. confidence assessment: 3
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20:07:32 ** 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 --> This is an alternate method for finding the time. Using vf^2 = v0^2 + 2 a `ds and finding vf, then using the average velocity to find the time over the displacement would also work. self critique assessment: 2
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20:11:27 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 --> This is because of Newton's third law which tells us that the forces acting on two objects are equal and opposite, therefore the Impulse (F* 'dt), and thus momentum due to the Impulse-Momentum theorem will be equal and opposite. confidence assessment: 3
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20:11:50 **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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20:15:34 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 --> m1, m2, v1, v2, `dt, F. These quantities are related through the Impulse-Momentum Theorem: Fnet * `dt = m2 * v2 - m1 * v1 confidence assessment: 3
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20:15:54 ** 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 --> self critique assessment: 3
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20:16:03 `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 --> n/a self critique assessment: 3
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20:16:08 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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20:16:17 `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 --> n/a self critique assessment: 3
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20:16:26 **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 --> self critique assessment: 3
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20:41:51 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 --> The normal force is 9.8m/s^2 * 2.00kg cos 53.1 degrees = 11.8N. The kinetic frictional force will therefore be 11.8N * .20 = -2.36N. As the package moves through the 4.00m, the work done against friction is 2.36N * 4.00m = 9.44J The PE of the package is 4.00m * 2.00kg * 9.8m/s^2 sin 53.1 degrees = 62.7J. When the packgage reaches the point just before hitting the spring, all of the PE of the 4.00m will have turned into KE or dissipated against the friction. `dKE + `dPE + `dW = 0 `dKE = -`dPE + -`dW `dKE = 62.7J - 9.44J = 53.26J The speed just before the package hits the spring will therefore be v = sqrt(2*KE/m) = sqrt(2*53.26J/2.00kg) = 2.27m/s The 53.26J of energy that the package has will be directed toward the spring, but also frictional forces will still be acting on the package. The work done on the spring will be .5*120N/m*x^2 and the energy put into the spring will be 53.26J - 2.36Nx since we do not know how much work will be done against friction because we do not know how far the spring will compress. The quadratic formula should help find x though because 53.26J - 2.36Nx = .5 * 120N/m * x^2 can be arranged to be: 60N/mx^2 + -2.36Nx - 53.26 Using the quadratic formula: x = (-(2.36N) + sqrt((2.36N)^2 - 4*(60N/m)*(-53.26J)))/(2*60N) x = .92m or x = (-(2.36N) - sqrt((2.36N)^2 - 4*(60N/m)*(-53.26J)))/(2*60N) x = -.96m We are moving in the direction of motion so we chose .92m as the maximum compression of the spring. The work done against friction is -2.36J so the total PE stored in the spring is 53.26-2.36J = 50.9J After release, friction still acts on the block so: `dKE + `dPE + `dW = 0 -50.9J + `dPE + `dW = 0 `dW = -2.36Nx and `dPE = -9.8m/s^2 * 2.00kg * x * sin 53.1 degrees, so: -50.9J = `dPE + `dW 50.9J =x * ( -9.8m/s^2 * 2kg * sin 53.1 deg - 2.36N) x = 50.9J/-18.0N = 2.83m So the package will go back up the incline 2.83m and will therefore be 1.17m short of where it started. confidence assessment: 3
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20:54:07 ** 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 --> I must have punched something into the calculator wrong to have gotten 2.27m/s from sqrt(2 * 53.26J / 2kg) = 7.30m/s. I missed the PE decrease of the package during compression which reduced the compression distance I had. Had I done this, I belive my calculations would have been correct. self critique assessment: 2
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