course PHY231 ßËâq…èÂIð¼O€Èñèqá¢pȤ|…iË°ŸåÙassignment #017
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14:08:02 `q001. Note that this assignment contains 5 questions. . A mass of 10 kg moving at 5 meters/second collides with a mass of 2 kg which is initially stationary. The collision lasts .03 seconds, during which time the velocity of the 10 kg object decreases to 3 meters/second. Using the Impulse-Momentum Theorem determine the average force exerted by the second object on the first.
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RESPONSE --> 'dv = -2 m/sec 'dt = .03 sec mass = 10 kg. According to the Impulse-momentum theorem p= mv is same quantity as Fave 'dt; average force = mass ('dv/'dt) Substituting given values, Fave = -667 N confidence assessment: 3
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14:08:20 By the Impulse-Momentum Theorem for a constant mass, Fave * `dt = m `dv so that Fave = m `dv / `dt = 10 kg * (-2 meters/second)/(.03 seconds) = -667 N. Note that this is the force exerted on the 10 kg object, and that the force is negative indicating that it is in the direction opposite that of the (positive) initial velocity of this object. Note also that the only thing exerting a force on this object in the direction of motion is the other object.
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RESPONSE --> self critique assessment: 3
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14:13:49 `q002. For the situation of the preceding problem, determine the average force exerted on the second object by the first and using the Impulse-Momentum Theorem determine the after-collision velocity of the 2 kg mass.
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RESPONSE --> The Impulse-Momentum theorem is rearranged and solved for velocity: v = Fave * 'dt / mass: Substituting the values from the last question gives a final velocity for the 2-kg object as 10 m/sec. confidence assessment: 3
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14:15:02 Since the -667 N force exerted on the first object by the second implies and equal and opposite force of 667 Newtons exerted by the first object on the second. This force will result in a momentum change equal to the impulse F `dt = 667 N * .03 sec = 20 kg m/s delivered to the 2 kg object. A momentum change of 20 kg m/s on a 2 kg object implies a change in velocity of 20 kg m / s / ( 2 kg) = 10 m/s. Since the second object had initial velocity 0, its after-collision velocity must be 10 meters/second.
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RESPONSE --> I should have noted that the equation was solved for the change in velocity. It was only correct in this case because initial velocity was zero. self critique assessment: 2
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14:16:03 `q003. For the situation of the preceding problem, is the total kinetic energy after collision less than or equal to the total kinetic energy before collision?
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RESPONSE --> The total kinetic energy of the system remains constant. confidence assessment: 2
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14:18:46 The kinetic energy of the 10 kg object moving at 5 meters/second is .5 m v^2 = .5 * 10 kg * (5 m/s)^2 = 125 kg m^2 s^2 = 125 Joules. Since the 2 kg object was initially stationary, the total kinetic energy before collision is 125 Joules. The kinetic energy of the 2 kg object after collision is .5 m v^2 = .5 * 2 kg * (10 m/s)^2 = 100 Joules, and the kinetic energy of the second object after collision is .5 m v^2 = .5 * 10 kg * (3 m/s)^2 = 45 Joules. Thus the total kinetic energy after collision is 145 Joules. Note that the total kinetic energy after the collision is greater than the total kinetic energy before the collision, which violates the conservation of energy unless some source of energy other than the kinetic energy (such as a small explosion between the objects, which would convert some chemical potential energy to kinetic, or perhaps a coiled spring that is released upon collision, which would convert elastic PE to KE) is involved.
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RESPONSE --> I should have calculated to be sure, but I thought the energy in a collision remained constant. self critique assessment: 2
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14:22:48 `q004. For the situation of the preceding problem, how does the total momentum after collision compare to the total momentum before collision?
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RESPONSE --> The momentum before the collision is due to the 10-kg object only and is equal to 10 kg (5 m/sec) = 50 kg * m/sec. The momentum after the collision is the sum of the momenta of both objects: 10 kg * 3 m/sec = 30 kg*m/sec 2 kg * 10 m/sec = 20 kg * m/sec Sum to 50 kg * m/sec. It's the momentum that stays constant in a collision. confidence assessment: 3
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14:22:56 The momentum of the 10 kg object before collision is 10 kg * 5 meters/second = 50 kg meters/second. This is the total momentum before collision. The momentum of the first object after collision is 10 kg * 3 meters/second = 30 kg meters/second, and the momentum of the second object after collision is 2 kg * 10 meters/second = 20 kg meters/second. The total momentum after collision is therefore 30 kg meters/second + 20 kg meters/second = 50 kg meters/second. The total momentum after collision is therefore equal to the total momentum before collision.
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RESPONSE --> self critique assessment: 3
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14:26:43 `q005. How does the Impulse-Momentum Theorem ensure that the total momentum after collision must be equal to the total momentum before collision?
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RESPONSE --> Momentum of an object is equal to the impulse of the force acting on the object. Momentum can therefore be changed by changing either of the factors that Impulse depends on. These factors are average force and the time during which the force acts. Since a collision involves an equal and opposite force acting on each object for a single time interval, the change in momentum for each object must be equal and opposite. confidence assessment: 3
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14:26:56 Since the force is exerted by the 2 objects on one another are equal and opposite, and since they act simultaneously, we have equal and opposite forces acting for equal time intervals. These forces therefore exert equal and opposite impulses on the two objects, resulting in equal and opposite changes in momentum. Since the changes in momentum are equal and opposite, total momentum change is zero. So the momentum after collision is equal to the momentum before collision.
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RESPONSE --> self critique assessment: 3
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½¢½´ÏíâÓ€™ýöÍù«Õc‘q{£¿ assignment #017 017. `query 17 Physics I 06-27-2007
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20:40:59 prin phy and gen phy 6.39: 950 N/m spring compressed .150 m, released with .30 kg ball. Upward speed, max altitude of ball
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RESPONSE --> confidence assessment:
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20:41:03 We will assume here that the gravitational PE of the system is zero at the point where the spring is compressed. In this situation we must consider changes in both elastic and gravitational PE, and in KE. The PE stored in the spring will be .5 k x^2 = .5 ( 950 N/m ) ( .150 m)^2 = 107 J. When released, conservation of energy (with only elastic and gravitational forces acting there are no nonconservative forces at work here) we have `dPE + `dKE = 0, so that `dKE = -`dPE. Since the ball is moving in the vertical direction, between the release of the spring and the return of the spring to its equilibrium position, the ball t has a change in gravitational PE as well as elastic PE. The change in elastic PE is -107 J, and the change in gravitational PE is m g `dy = .30 kg * 9.8 m/s^2 * .150 m = +4.4 J. The net change in PE is therefore -107 J + 4.4 J = -103 J. Thus between release and the equilibrium position of the spring, `dPE = -103 J The KE change of the ball must therefore be `dKE = - `dPE = - (-103 J) = +103 J. The ball gains in the form of KE the 103 J of PE lost by the system. The initial KE of the ball is 0, so its final KE is 103 J. We therefore have .5 m vv^2 = KEf so that vf=sqrt(2 KEf / m) = sqrt(2 * 103 J / .30 kg) = 26 m/s. To find the max altitude to which the ball rises, we return to the state of the compressed spring, with its 107 J of elastic PE. Between release from rest and max altitude, which also occurs when the ball is at rest, there is no change in velocity and so no change in KE. No nonconservative forces act, so we have `dPE + `dKE = 0, with `dKE = 0. This means that `dPE = 0. There is no change in PE. The initial PE is 107 J and the final PE must also therefore be 107 J. There is, however, a change in the form of the PE. It converts from elastic PE to gravitational PE. Therefore at maximum altitude the gravitational PE must be 107 J. Since PEgrav = m g y, and since the compressed position of the spring was taken to be the 0 point of gravitational PE, we therefore have y = PEgrav / (m g) = 107 J / (.30 kg * 9.8 m/s^2) = 36.3 meters. The ball will rise to an altitude of 36.3 meters above the compressed position of the spring.
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RESPONSE --> self critique assessment:
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20:41:07 gen phy problem A high jumper needs to be moving fast enough at the jump to lift her center of mass 2.1 m and cross the bar at a speed of .7 m/s. What minimum velocity does she require?
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RESPONSE --> confidence assessment:
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20:41:10 FORMAL SOLUTION: Formally we have `dPE + `dKE = 0. `dPE = M * g * `dy = M * 20.6 m^2 / sec^2, where M is the mass of the jumper and `dy is the 2.1 m change in altitude. `dKE = .5 M vf^2 - .5 M v0^2, where vf is the .7 m/s final velocity and v0 is the unknown initial velocity. So we have M g `dy + .5 M vf^2 - .5 M v0^2 = 0. Dividing through by M we have g `dy + .5 vf^2 - .5 v0^2 = 0. Solving for v0 we obtain v0 = sqrt( 2 g `dy + vf^2) = sqrt( 2* 9.8 m/s^2 * 2.1 m + (.7 m/s)^2 ) = sqrt( 41.2 m^2/s^2 + .49 m^2 / s^2) = sqrt( 41.7 m^2 / s^2) = 6.5 m/s, approx.. LESS FORMAL, MORE INTUITIVE, EQUIVALENT SOLUTION: The high jumper must have enough KE at the beginning to increase his PE through the 2.1 m height and to still have the KE of his .7 m/s speed. The PE change is M * g * 2.1 m = M * 20.6 m^2 / sec^2, where M is the mass of the jumper The KE at the top is .5 M v^2 = .5 M (.7 m/s)^2 = M * .245 m^2 / s^2, where M is the mass of the jumper. Since the 20.6 M m^2 / s^2 increase in PE must come at the expense of the initial KE, and since after the PE increase there is still M * .245 m^2 / s^2 in KE, the initial KE must have been 20.6 M m^2 / s^2 + .245 M m^s / s^2 =20.8 M m^s / s^2, approx. If initial KE is 20.8 M m^s / s^2, then .5 M v0^2 = 20.8 M m^s / s^2. We divide both sices of this equation by the jumper's mass M to get .5 v0^2 = 20.8 m^2 / s^2, so that v0^2 = 41.6 m^2 / s^2 and v0 = `sqrt(41.6 m^2 / s^2) = 6.5 m/s, appprox.
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RESPONSE --> self critique assessment:
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20:59:29 query Univ. 7.42 (7.38 in 10th edition). 2 kg block, 400 N/m spring, .220 m compression. Along surface then up 37 deg incline all frictionless. How fast on level, how far up incline?
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RESPONSE --> The block slides along a frictionless surface at a constant speed that is the result of the work done on it by the spring. The work done on it by the spring is: v = sqrt(k/m * x^2) = x* sqrt(k/m) Substituting k = 400N/m m = 2.0kg and x = .220 gives v = 3.11 m/sec The block starts traveling up the frictionless incline with an intial kinetic energy of m/2 * v^2 = 9.68J. It will travel up the incline until all of its kinetic energy has been converted to gravitational potential energy. At that point, its PE = mg 'dy 'dy = 9.68J / mg = .4938 We are given the degrees of inclination, and can use this to solve for the 'ds along the incline: .4938 / sin (37degrees) = .821 KE0 confidence assessment: 3
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21:17:10 ** The spring exerts a force of 400 N / m * .220 m = 84 N at the .220 m compression. The average force exerted by the spring between equilibrium and this point is therefore (0 N + 84 N) / 2 = 42 N, so the work done in the compression is `dW = Fave * `ds = 42 N * .220 m = 5.0 Joules, approx. If all this energy is transferred to the block, starting from rest, the block's KE will therefore be 5.0 Joules. Solving KE = .5 m v^2 for v we obtain v = sqrt(2 KE / m) = sqrt(2 * 5.0 Joules / (2 kg) ) = 2.2 m/s, approx.. No energy is lost to friction so the block will maintain this speed along the level surface. As it begins to climb the incline it will gain gravitational PE at the expense of KE until the PE is 5.0 J and the KE is zero, at which point it will begin to slide back down the incline. After traveling through displacement `ds along the incline the height of the mass will be `ds sin(37 deg) = .6 `ds, approx., and its gravitational PE will be PE = m g h = m g * .6 `ds = .6 m g `ds. Setting this expression equal to KE we obtain the equation .6 m g `ds = KE, which we solve for `ds to obtain `ds = KE / (.6 m g) = 5.0 Joules / (.6 * 2 kg * 9.8 m/s^2) = .43 meters, approx. **
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RESPONSE --> Using the average force and distance still gives a quantity of work similar to the one I calculated: 42 N * .220 m = 9.24 J Is my calculation incorrect, or are these results supposed to have small discrepancies?
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21:49:19 query univ phy 7.50 62 kg skier, from rest, 65 m high. Frict does -10.5 kJ. What is the skier's speed at the bottom of the slope? After moving horizontally over 82 m patch, air res 160 N, coeff frict .2, how fast is she going? Penetrating 2.5 m into the snowdrift, to a stop, what is the ave force exerted on her by the snowdrift?
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RESPONSE --> Using the 7.50 problem from text: A 2.8-kg block slides over the smooth, icy hill shown. The ope of the hill is horizontal and 70 m higher than the base. What minimum speed must the block have at the base of the hill so that it will not fall into the pit on the far side of the hill? The pit at the far side of the hill is 40 m wide and its far side is 20 m lower than the top of the hill. The block must leave the top of the hill with a horizontal velocity of 20 m/sec: The time of 20 m fall with initial vertical velocity of zero: -20 = 1/2 * 9.8 m/sec^2 * 'dt^2 'dt = positive sqrt(-40 m/ 9.8 m/sec^2) = 2.02 s The block must travel the 40 meters between hilltop and far side of pit in 2.02 s: 40 m/2.02 s = 19.79 s, approximately 20 m/sec The block needs to have 560 J of kinetic energy when it leaves the top of the hill, from substituting known values into the m/2 * v^2 = KE equation. Going up the hill, the block will lose KE in the same amount of PE it gains, which is m * g * 'dy = 1921 J So the initial KE of the block must be 1921 J+ 560 J = 2481 J For a 2.8-kg block, this correlates to 42 m/sec: v = sqrt ( 2 * KE / m ). Substituting known values gives 42.1 m/sec confidence assessment: 3
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21:49:29 ** The gravitational PE of the skier decreases by 60 kg * 9.8 m/s^2 * 65 m = 38 kJ, approx. (this means 38 kiloJoules, or 38,000 Joules). The PE loss partially dissipated against friction, with the rest converted to KE, resulting in KE = 38 kJ / 10.5 kJ = 27.5 kJ. Formally we have `dKE + `dPE + `dWnoncons = 0, where `dWnoncons is the work done by the skier against friction. Since friction does -10.5 kJ of work on the skier, the skier does 10.5 kJ of work against friction and we have `dKE = -`dPE - `dWnoncons = - (-38 kJ) - 10.5 kJ = 27.5 kJ. The speed of the skier at this point will be v = sqrt( 2 KE / m) = sqrt( 2 * 27,500 J / (65 kg) ) = 30 m/s, approx. Over the 82 m patch the force exerted against friction will be .2 * 60 kg * 9.8 m/s^2 = 118 N, approx., so the force exerted against nonconservative forces will be 118 N + 160 N = 280 N approx.. The work done will therefore be `dWnoncons = 280 N * 82 m = 23 kJ, approx., and the skier's KE will be KE = 27.5 kJ - 23 kJ = 4.5 kJ, approx. This implies a speed of v = sqrt( 2 KE / m) = 12 m/s, approx. To stop from this speed in 2.5 m requires that the remaining 4.5 kJ of KE be dissipated in the 2.5 m distance. Thus we have `dW = Fave * `ds, so that Fave = `dW / `ds = 4500 J / (2.5 m) = 1800 N. This is a significant force, about 3 times the weight of the skier, but distributed over a large area of her body will cause a good jolt, but will not be likely to cause injury.**
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RESPONSE --> self critique assessment:
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