#$&* course phy 201 Aug 1 4:15pm If your solution to stated problem does not match the given solution, you should self-critique per instructions at
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Given Solution: a** We treat the vertical and horizontal quantities independently. We are given vertical displacement and initial velocity and we know that the vertical acceleration is the acceleration of gravity. So we solve the vertical motion first, which will give us a dt with which to solve the horizontal motion. We first determine the final vertical velocity using the equation vf^2 = v0^2 + 2a'ds, then average the result with the initial vertical velocity. We divide this into the vertical displacement to find the elapsed time. We are given the initial horizontal velocity, and the fact that for an ideal projectile the only force acting on it is vertical tells us that the acceleration in the horizontal direction is zero. Knowing dt from the analysis of the vertical motion we can now solve the horizontal motion for ds. This comes down to multiplying the constant horizontal velocity by the time interval dt. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: qQuery class notes #17 Why do we expect that in an isolated collision of two objects the momentum change of each object must be equal and opposite to that of the other? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Because when 2 objects collide they exert equal and opposite forces upon each other which in turn causes equal and opposite impulse on each of the objects. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. **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 objects 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. ** STUDENT QUESTION Are impulses the same as momentum changes? INSTRUCTOR RESPONSE impulse is F * dtmomentum 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): ------------------------------------------------ Self-critique rating: ********************************************* 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: mass 1 and mass 2, before-collision velocity 1 and velocity 2, and after-collision velocity 1 and velocity 2. Relationships are shown by: Total momentum before collision = m1 v1 + m2 v2 Total momentum after collision = m1 v1' + m2 v2'. Impulse-momentum theorem tells us: m1 v1 + m2 v2 = m1 v1' + m2 v2' confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: a** We analyze the momentum for such a collision in terms of the masses m1 and m2, the before-collision velocities v1 and v2 and the after-collision velocities v1' and v2'. Total momentum before collision is m1 v1 + m2 v2. Total momentum after collision is m1 v1' + m2 v2'. Conservation of momentum, which follows from the impulse-momentum theorem, gives us m1 v1 + m2 v2 = m1 v1' + m2 v2'. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: 1prin phy and gen phy 6.47. RR cars mass 7650 kg at 95 km/hr in opposite directions collide and come to rest. How much thermal energy is produced in the collision? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The initial KE will be lost due to no conservative forces this causing no change in PE. Most of KE will then become thermal energy. Initial speed: 95 km/hr * 1000 m/km * 1 hr / 3600 s = 26.4 m/s Initial KE of each car = .5 m v^2 = .5 * 7650 kg * (26.4 m/s)^2 = 265,000 Joules Total KE = 2 * 265,000 J = 530,000 J (this will nearly all be converted to thermal energy.) confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. STUDENT QUESTIONS Why is the kinetic energy multiplied by two? And why is all of the kinetic energy practically converted to thermal energy? Is thermal energy simply two times the kinetic energy? Is this what happens to all kinetic energy in real life? INSTRUCTOR RESPONSE You've calculated the KE of one of the cars. There are two cars, which is why we multiply that result by 2. Some of the KE does go into producing sound, but loud as the crash might be only a small fraction of the energy goes into the sound. Practically all the rest goes into thermal energy. A lot of the metal in the cars is going to twist, buckle and otherwise deform, and warm up some in the process. They probably won't become hot to the touch, but it takes a lot more thermal energy that that involved in this collision to achieve an overall temperature change we would be likely to notice. If two cars of unequal mass and equal speeds collide they don't come to rest, so they have some KE after the collision. It the cars were perfectly elastic they would rebound with their original relative speed. A perfectly elastic collision is one in which kinetic energy is conserved. No energy would go into thermal energy and there would be no sound. This is an idean and cannot actually be achieved with railroad cars (nor with steel balls, or marbles, or pool balls, etc.). However the collisions of molecules in a gas are perfectly elastic, and analyzing the statistics of those collisions allows us to explain a lot of what we observe about gases. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: Query* 7.53. 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)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: dKE + dPE + dWnoncons = 0 Gravitational PE gives us: dPE = (y2 - y1) Friction will also be a factor in the direction of motion. So we have: .5 M vf^2 - .5 M v0^2 + M g (y2 - y1) + f * ds = 0 .5 M vf^2 - .5M(1.7m/s)^2 + M(9.8m/s^2)*(-28 m - 0) + .2 M g (45m) Masses negate one another and are no longer necessary so we now have: .5v2^2 - 1.445 m^2/s^2 - 274 m^2/s^2 + 88 m^2/s^2 = 0 Solve for v2: v2 = +- sqrt( 375 m^2/s^2 ) = 19.3 m/s confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: a**GOOD STUDENT SOLUTION WITH ERROR IN ONE DETAIL, WITH INSTRUCTOR CORRECTION: Until just now I did not think I could work that one, because I did not know the mass, but I retried it. Conservation of energy tells us that dKE + dPE + dWnoncons = 0. PE is all gravitational so that dPE = (y2 - y1). The only other force acting in the direction of motion is friction. Thus .5 M vf^2 - .5 M v0^2 + M g (y2 - y1) + f * ds = 0 and .5 M vf^2 - .5M(1.7m/s)^2 + M(9.8m/s^2)*(-28 m - 0) + .2 M g (45m) It looks like the M's cancel so I don't need to know mass. .5v2^2 - 1.445 m^2/s^2 - 274 m^2/s^2 + 88 m^2/s^2 = 0 so v2 = +- sqrt( 375 m^2/s^2 ) = 19.3 m/s. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: Query* 7.53. gen phy roller coaster 1.7 m/s at point 1, ave frict 1/5 wt, reaches poin 28 m below at what vel ( ds = 45 m along the track)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: dKE + dPE + dWnoncons = 0 Gravitational PE gives us: dPE = (y2 - y1) Friction will also be a factor in the direction of motion. So we have: .5 M vf^2 - .5 M v0^2 + M g (y2 - y1) + f * ds = 0 .5 M vf^2 - .5M(1.7m/s)^2 + M(9.8m/s^2)*(-28 m - 0) + .2 M g (45m) Masses negate one another and are no longer necessary so we now have: .5v2^2 - 1.445 m^2/s^2 - 274 m^2/s^2 + 88 m^2/s^2 = 0 Solve for v2: v2 = +- sqrt( 375 m^2/s^2 ) = 19.3 m/s confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: a**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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&! ********************************************* Question: Query* 7.53. 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)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: dKE + dPE + dWnoncons = 0 Gravitational PE gives us: dPE = (y2 - y1) Friction will also be a factor in the direction of motion. So we have: .5 M vf^2 - .5 M v0^2 + M g (y2 - y1) + f * ds = 0 .5 M vf^2 - .5M(1.7m/s)^2 + M(9.8m/s^2)*(-28 m - 0) + .2 M g (45m) Masses negate one another and are no longer necessary so we now have: .5v2^2 - 1.445 m^2/s^2 - 274 m^2/s^2 + 88 m^2/s^2 = 0 Solve for v2: v2 = +- sqrt( 375 m^2/s^2 ) = 19.3 m/s confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: a**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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!#*&!