course Phy 231 ”鎂仰ˆa®Ü¶ÛÎছçñì¿í¦‚–assignment #018
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07:55:10 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 --> In order to determine horizontal displacement we must first determine the time elapsed, which is determined by the vertical velocity and displacement; therefore, once we use the equation 'ds = v0 + 1/2 * a * 'dt^2, which is rearranged to 0 = - 'ds + v0 + 1/2a * 'dt^2, we can find the time elapsed. Once we determine the time elapsed we can determine by multiplying the horizontal velocity by the time interval to find the horizontal displacement. confidence assessment: 3
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08:02:35 ** 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 --> Ok self critique assessment: 3
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08:05:29 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 --> We expect the momentum of each object to be equal and opposite of each other due to Newtons third law that for every reaction there is an equal but opposite reaction. Also, the Conservation of Momentum theory which derives from this law states that the total momentum of a system does not change; therefore if m1 * v1 goes up by a positive 'dp then m2 * v2 must go down by negative 'dp equal to the change in the first. confidence assessment: 3
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08:31:36 **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 --> Ok self critique assessment: 3
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08:42:55 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 --> We can analyze displacement, velocity, acceleration, time elapsed, force, and mass. We relate these quantities by several equations that can be derived to find the other. We have F * 'dt = 'dp or change in momentum, we have m * 'dv = 'dp to find the change in momentum of single mass, we have 'dv = a 'dt = (F * 'dt)/ m, and we also have 'ds = (vf^2 - v0^2)/ 2a. These equation show us how to find the change in momentum by the first direct eqation, or by working our way back to the values of force and time or mass and velocity, which can be used to determine the momentum before and after collisions. confidence assessment: 3
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08:44:27 ** 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 see that you were referring to the six quantites of m1, m2, v1, v2, v1', and v2', which are involved in the impulse-momentum theorem. self critique assessment: 3
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08:44:34 `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 --> Ok self critique assessment: 3
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08:44:40 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 --> Ok self critique assessment: 3
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08:44:45 `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 --> Ok self critique assessment: 3
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08:45:36 **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 --> Ok self critique assessment: 3
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21:11:50 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 --> We must find the force acting upon the package as it moves down the slope towards the spring, which is found by the Fnet = Fg - Fs, where we have the force of gravity - force of kinetic friction. This gives us Fnet = [2 kg * 9.8 m/s^2 * sin 53.1] - [ .2 * 2 kg * 9.8 m/s^2 * sin 53.1] = 15.68 N - 3.14 N = 12.54 N acting on the package downwards; therefore, we can find the change in velocity equal to the change in KE since it initially had no KE and the KE found by vf will give us the change. This means that with a KE = 12.54 N * 4 m = 50.16 J, which means that 'sqrt [( 2 * 50.16 J )/2 kg] = 7.08 m/s = vf. If the spring is met with the force of the pacakage of 12.54 N we find the compression distance by 12.54 N/ (120 N/m) = 0.1045 m. The spring will then store all of the KE the package originally had; however, as the package moves back up the slope it will lose KE to gravity and friction with a Fnet = 15.68 N + 3.14 N = 8.82 N against the motion. With the energy provided this will allow the package to move 'ds = 50.16 J/ 18.82 N = 2.67 m up the incline. confidence assessment: 3
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07:36:46 ** 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 --> My determination of forces acting upon the package agrees with yours, but I found the stopping distance of the spring in a different manner. Was my method not correct, because even though it was different our values were not to far apart for the distance traveled back up the incline. self critique assessment: 3
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