course Phy 121 fɪڞI}c|CStudent Name:
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18:06:45 `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 --> m1 = 10kg vo = 5m/s vf = 3m/s `dv = -2m/s m2 = 2kg `dt = 0.03s `ds = vAve * `dt `ds = (4m/s)(0.03s) = 0.12m The Impulse-Momentum Theorem tells sus that m * `dv = F * `dt. F1 = (m1*`dv1)/`dt = (10kg * -2m/s)/0.03s = -667 Newtons. This is the average force exerted on the first object by the second object. Since it is negative, we see that the first object probably exerted more force and the second object would have been moved in the opposite direction of motion after the collision.
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18:07:21 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 --> I understand this problem.
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18:09:55 `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 --> m1 = 10kg `dv = -2m/s F = -667N Since the second object exerts -667 Newtons of force on the first object, the first object will exert an equal and opposite amount of force on the second. Therefore, the average force exerted on the second object by the first is 667 Newtons. m2 = 2kg `dt = 0.03s `ds = vAve * `dt `ds = (4m/s)(0.03s) = 0.12m Impulse-Momentum Theorem: F * `dt = m * `dv We know F and `dt, so we can find the impulse and therefore, change in momentum (m*`dv). F * `dt = 667N * 0.03s = 20Ns = 20 kg*m/s Given the momentum change and the mass of the object, we can find the change in velocity. (m2 * `dv)/ m2 = `dv (20 kg*m/s)/2kg = 10 m/s
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18:10:20 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 understand this problem.
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18:12:04 `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 --> Before collision: We only have information for the first object because the initial velocity of the second object is zero. Therefore, the only kinetic energy before the collision is that of the first object. m1 = 10kg v1 = 5m/s KE = 1/2mv^2 = 1/2(10kg)(5m/s)^2 = 125 Joules After collision: m1 = 10kg v1 = 3m/s m2 = 2kg v2 = 10m/s KE1 = 1/2(10kg)(3m/s)^2 = 45 Joules KE2 = 1/2mv^2 = 1/2(2kg)(10m/s)^2 = 100 Joules Total kinetic energy after the collision is 145 Joules, and is obviously larger than the initial kinetic energy.
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18:12:21 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) is involved.
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RESPONSE --> Ok.
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18:13:20 `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 total momentum of an object is equal to the mass times the change in velocity. BEFORE COLLISION: Momentum 1 = m * vo = 10kg * 5m/s = 50 kg*m/s Momentum 2 = m * vo = 2kg * 0 = 0 Total momentum = 50 kg*m/s AFTER COLLISION: Momentum 1 = 10kg * 3m/s = 30 kg*m/s Momentum 2 = 2kg * 10m/s = 20 kg*m/s Total momentum = 50 kg*m/s The total momentums before and after the collision are equal.
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18:13:40 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 --> Ok
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18:14:08 `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 --> The Impulse-Momentum Theorem: F * `dt = m * `dv To see how the theorem ensures this equality, we must examine the variables of the theorem. F: The force exerted by each object is equal and opposite to the other. `dt: The collision occurs over the same time interval for each object. Since the impulse (F * `dt) is equal to the momentum (m *`dv) we can see that the momentum changes are equal and opposite of each other. If we add these changes together we see that there is no change in total momentum.
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18:14:31 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 --> I understand this concept.
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[_k«k|ս`xV assignment #017 017. `query 17 Physics I 06-28-2006
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22:03:24 prin phy and gen phy 6.33: jane at 5.3 m/s; how high can she swing up that vine?
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RESPONSE --> V = 5.3m/s The vine is hanging vertically. We want to know how high she can swing upward. To find this answer we must consider the mass, gravitational acceleration and altitude of swing. These are all variables of gravitational potential energy: PE = m * g * h Since the kinetic energy is equal and opposite to potential energy, we can say KE = 1/2mv^2 = m * g * h 1/2mv^2 = m*g*h We are trying to find the altitude of the swing. h = (m*v^2)/(2m*g) mass will cancel out. h = (5.3m/s)^2/(2m*9.8m/s^2) = (2.87 m)/2 = 1.43 m To answer the second part, no we do not need to know the length of the vine.
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22:04:46 Jane is going to convert her KE to gravitational PE. We assume that nonconservative forces are negligible, so that `dKE + `dPE = 0 and `dPE = -`dKE. Jane's KE is .5 M v^2, where M is her mass. Assuming she swings on the vine until she comes to rest at her maximum height, the change in her KE is therefore `dKE = KEf - KE0 = 0 - .5 M v0^2 = - .5 M v0^2, where v0 is her initial velocity. Her change in gravitational PE is M g `dy, where `dy is the change in her vertical position. So we have `dKE = - `dPE, or -5 M v0^2 = - ( M g `dy), which we solve for `dy (multiply both sides by -1, divide both sides by M g) to obtain `dy = v0^2 / (2 g) = (5.3 m/s)^2 / (2 * 9.8 m/s^2) = 1.43 m.
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RESPONSE --> I understand
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22:07:40 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 --> stiffness constant (k) = 950N/m distance compressed (x)= 0.15m mass = 0.3 kg With the variables we are given, we can find the elastic potential energy and the gravitational potential energy. PE elastic = 1/2*k*x^2 = 1/2(950N/m)(0.15m)^2 = 10.69 Joules. When the object is restored to its original position, the restoring force is negative. So, the elastic potential energy change is -10.69 Joules. PE grav = m * g * h = (0.3kg)(9.8m/s)(0.15m) = 0.441 Joules `dPE = PE elastic + PE grav = -10.69J + 0.441J = -10.25 Joules We know that `dPE is equal and opposite to `dKE. So `dKE = 10.25 Joules. Upward speed can be found now using `dKE = 1/2m`dv^2 vo = 0m/s, so KEo = 0 Joules `dKE = KEf = 1/2mvf^2 10.25J = 1/2(0.3kg)vf^2 vf^2 = 68.33m^2/s^2 vf = 8.3 m/s Maximum distance from the original position PE grav = m * g * h h = PE grav/(m*g) = 10.69J / (0.3kg*9.8m/s^2) = 3.63 m
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22:09:11 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 --> My numbers were different but my technique is the same.
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22:09:37 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 --> I am not required to answer this problem.
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22:10:17 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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22:10:40 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 --> I am not required to solve this problem.
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22:10:44 ** 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 -->
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22:10:59 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 --> I am not required to answer this question.
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22:11:04 ** 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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