course Phy 231 7/2 5:30 018. `query 18*********************************************
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Given Solution: `a** 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. ** Self-critique (if necessary): OK Self-critique rating #$&* OK ********************************************* Question: `qQuery class notes #17 Why do we expect that in an isolated collision of two objects the momentum change of each object must be equal and opposite to that of the other? Your solution: You expect the change in the momentum to be equal and opposite because the forces working on the objects are also equal and opposite. The impulse that one object exerts on the other is the same for the other object and since impulses are equal to the change in momentum (impulse-momentum theorem) the changes in momentum will be equal and opposite. Confidence rating #$&*: 2
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Given Solution: `a Briefly, the force exerted on each object on the other is equal and opposite to the force exerted on it by the other, by Newton's Third Law. By assumption the collision is isolated (i.e., this is a closed system); the two objects interact only with one another. So the net force on each object is the force exerted on it by the other. So the impulse F_net `dt on one object is equal and opposite the impulse experienced by the other. By the impulse-momentum theorem, F_net `dt = `d ( m v). The impulse on each object is equal to its change in momentum. Since the impulses are equal and opposite, the momentum changes are equal and opposite. impulse is F * `dt momentum is m v, and as long as mass is constant momentum change will be m `dv by the impulse-momentum theorem impulse is equal to change in momentum (subject, of course, to the conditions of the theorem) &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique rating #$&* OK ********************************************* Question: `qWhat are the six quantities in terms of which we analyze the momentum involved in a collision of two objects which, before and after collision, are both moving along a common straight line? How are these quantities related? Your solution: Mass1, mass2, velocity_before1, velocity_before2, velocity_after1, velocity_after2 M1(v1)+m2(v2) = total momentum before the collision M1(v_a1)+m2(v_a2) = total momentum after the collision Conservation of momentum: M1(v1)+m2(v2) = M1(v_a1)+m2(v_a2) Confidence rating #$&*: 2
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Given Solution: `a** 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'. ** Self-critique (if necessary): OK Self-critique rating #$&* OK Question: `1prin phy and gen phy 6.47. RR cars mass 7650 kg at 95 km/hr in opposite directions collide and come to rest. How much thermal energy is produced in the collision? Your solution: Since the cars come to rest all of the potential energy is turned into thermal energy 95km/h(1000m/1km)(1hr/3600s) = 26.39m/s KE = (1/2)(7650kg)(26.39m/s)^2 KE = 2663853 J (This is for 1 car) Total thermal energy = 2(KE) = 5327706 J confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThere 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 = 2,650,000 Joules, so that their total KE is 2 * 2,650,000 J = 5,300,000 J. This KE is practically all converted to thermal energy. `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)? Your solution: KE+PE+Noncons = 0 (1/2)(m)(vf^2)-(1/2)(m)(v0^2) + m(a)(ds) + f m a (ds) = 0 .5mvf^2 - .5m(1.7m/s)^2 + m(9.8m/s^2)(-28m-0) +.2 m (9.8m/s^2)(45m) = 0 (m’s cancel) .5vf^2 – 1.445m^2/s^2 -274.4m^2/s^2 + 88.2m^2/s^2 = 0 .5 vf^2 = 187.645m^2/s^2 Vf^2 = 375.29m^2/s^2 Vf = 19.4m/s confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Conservation of energy tells us that `dKE + `dPE + `dWnoncons = 0. PE is all gravitational so that `dPE = (y2 - y1). The only other force acting in the direction of motion is friction. Thus .5 M vf^2 - .5 M v0^2 + M g (y2 - y1) + f * `ds = 0 and .5 M vf^2 - .5M(1.7m/s)^2 + M(9.8m/s^2)*(-28 m - 0) + .2 M g (45m) M's cancel so I don't need to know mass. .5v2^2 - 1.445 m^2/s^2 - 274 m^2/s^2 + 88 m^2/s^2 = 0 so v2 = +- sqrt( 375 m^2/s^2 ) = 19.3 m/s. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique rating #$&* OK ********************************************* Question: `q 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? Your solution: Normal Force = 2kg(9.8m/s^2)(cos53.1) = 11.77N Friction = .2(11.77N) = 2.35N Fparallel = 2kg(9.8m/s^2)(sin53.1) = 15.67N Fnet = 13.32N-2.35N = 13.32N Work = 13.32N * 4m = 53.28J 53.28J = (1/2)(2kg)(v^2) V^2 = 53.28m^2/s^2 V = +-7.3m/s (speed right before hitting the spring) dKe+dPE+dnoncons = 0 -53.28J – 15.67N(x)+(1/2)(120N/m)(x^2) + 2.35(x) = 0 60N/m (x^2) – 13.32N(x) – 53.28Nm = 0 quadratic formula-take the positive answer X = 1.06 meters (distance the spring compresses PEgrav+Fparallel+dKE+noncons = 0 (-1/2)(120N/m)(1.06m)^2 + 15.67N(x)+2.35N(x) +0 = 0 -67.416Nm + 15.67N(x) + 2.35N(x) = 0 -67.416Nm+18.02N(x) = 0 X = 3.74m (where the block stops—it is very close to the initial position of the object) confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** 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. 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). 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 + 2.3 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. ** Self-critique (if necessary): OK Self-critique rating #$&* OK "