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10 Pages

### Long-Answer HW 10 - Reference Frames

Course: PHYS 2054, Spring 2008
School: Arkansas
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Word Count: 2706

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for Solution Long-answer Homework 10 Reference Frames problems Solution to Long-answer Homework Problem 10.1(Indy-Comprehensive Inelastic Collision) Problem: Indiana Jones is in a 1000kg train car coasting at 15 m . The bad guys chasing him are in a 1500kg s caboose coasting at 25 m so it rear-ends Indy's train car. The caboose becomes attached to the train car and the s two roll down the track together. (a)What...

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for Solution Long-answer Homework 10 Reference Frames problems Solution to Long-answer Homework Problem 10.1(Indy-Comprehensive Inelastic Collision) Problem: Indiana Jones is in a 1000kg train car coasting at 15 m . The bad guys chasing him are in a 1500kg s caboose coasting at 25 m so it rear-ends Indy's train car. The caboose becomes attached to the train car and the s two roll down the track together. (a)What is the final velocity of the system? (b)What is the minimum amount of kinetic energy required to conserve the momentum of the system? (c)What is the velocity of the zero momentum frame, vcm , for this system? 2 (d)Using Kcm = 1 Mtot vcm , find the center of mass kinetic energy of the system. 2 (e)How much kinetic energy is lost to other forms of energy in this collision? (f)Where does this lost kinetic energy go? (g)Find the kinetic energy for the system in the zero momentum frame before the collision. (h)What are the upper and lower limits on how much kinetic energy can be converted to other forms for this system for inelastic collisions? (i)Please make sure you included your name and lab section.... Solution to Part (a) Conserve momentum, where the mass of Indiana and the train car is mindy and the mass of the bad guys in the caboose is mcab : mindy vindy,i + mcab vcab,i = mindy vindy,f + mcab vcab,f Since they stick together, vindy,f = vcab,f = vindy,f = mindy vindy,i + mcab vcab,i mindy + mcab (1000kg) 15 m + (1500kg) 25 m m s s = 21 (1000kg) + (1500kg) s Grading Key: Part (a) 7 Points 2 point(s) : conserve momentum 1 point(s) : all same sign 1 point(s) : vf same for both 2 point(s) : correct substitutions for each of the initial momenta. 1 point(s) : correct answer with units Solution to Part (b) This would be the least energy required for a totally inelastic collision, which would be K= K= 1 2 Mtot vf 2 2 m 1 (2500kg) 21 2 s 1 = 5.51 105 J Grading Key: Part (b) 5 Points 3 point(s) : correctly identifies as energy associated with center of mass, or totally inelastic collision, or that part of the K equation. 1 point(s) : v from (a) if do it based on reasoning about com motion 1 point(s) : correct answer with units (consistent with (a)). Solution to Part (c) As always with an inelastic collision, the velocity of the center of mass is the velocity of the combination after the collision, so the zero momentum frame has that velocity, m vcm = 21 s Grading Key: Part (c) 2 Points 2 point(s) : same as (a). They could calculate it as well. If the calculate it and get it wrong, full credit if they say they know they made a mistake, since it should have been the same as (a). Solution to Part (d) Since the zero momentum frame has the same velocity as the Indy/Bad Guy combination after the collision (vcm = vf ), the center of mass kinetic energy is the same as in part (b), K = 5.51 105 J Grading Key: Part (d) 4 Points 1 point(s) : it is the same as (b) or they say it should have been the same as (b). 1 point(s) : correct mass 1 point(s) : velocity from c. 1 point(s) : correct answer with units Solution to Part (e) E = -KE = - (KEf - KEi ) = KEi - KEf 1 1 1 2 2 2 E = mindy vindy,i + mcab vcab,i - M vf 2 2 2 1 m 2 1 m 2 1 m E = (1000kg) 15 + (1500kg) 25 - (2500kg) 21 2 s 2 s 2 s 4 E = 3.0 10 J Grading Key: Part (e) 7 Points 1 2 2 point(s) : KE = KEf - KEi or KE = 2 vrel,i 2 2 point(s) : correct mass substitutions or calculates 2 point(s) : correct velocity substitutions or calculates e 1 point(s) : correct answer with units, do not worry about sign 2 Solution to Part (f) Some of the lost energy goes into heating, some of it goes into deforming various parts of the train cars, some of it goes into sound, etc. Grading Key: Part (f) 2 Points Solution to Part (g) We could reason that since the zero momentum frame has the same velocity as the final velocity of the vehicles, and in that frame the train cars have zero KE after the collision, the zero-p-frame kinetic energy has to be precisely that energy which is converted to other forms, the E found in part (g). We can prove this by going through the math. The initial velocity of Indy in the zero-p frame is vindy,i = vindy,i - vcm = 15 m m m - 21 = -6 s s s For the caboose, m m m - 21 = 4 s s s 1 1 KE = mindy (vindy,i )2 + mcab (vcab,i )2 2 2 1 m 2 1 m 2 KE = (1000kg) -6 + (1500kg) 4 = 3.0 104 J 2 s 2 s vcab,i = vcab,i - vcm = 25 Grading Key: Part (g) 3 Points 3 point(s) : can actually calculate it, but if so, must recognize should have been same as (e). If they got (e) wrong and get it right here, 2 points out of 3, if no comment that it should be the same. Solution to Part (h) The most kinetic energy that can be converted to other forms in this collision is the amount of kinetic energy in excess of the amount needed to conserve momentum, i.e. the amount measured in the zero momentum frame. The least energy would be something just barely greater than zero, something just barely not elastic. Grading Key: Part (h) 2 Points 1 point(s) : least: zero (close to or approximately zero or zero are all okay answers) 1 point(s) : most: all of the convertible kinetic energy-in words, or can use their own numbers Solution to Part (i) look at top Grading Key: Part (i) 3 Points 1 point(s) : legible name 1 point(s) : lab section 1 point(s) : was in the right box 3 Total Points for Problem: 35 Points Solution to Long-answer Homework Problem 10.2() Problem: An object moving at 10 m with a mass of 5m strikes a stationary object with mass m. s (a)What is the velocity of the zero-momentum reference frame for this system? (b)If the heavier object's final velocity in the lab frame is 2 m , what is the lighter object's final velocity? s (c)Repeat part (b) in the zero-momentum reference frame. (d)Classify the collision as elastic, explosive, inelastic, or totally inelastic, and verify that the coefficient of restitution is the same in both reference frames. Solution to Part (a) The zero-momentum frame is found by psys M (5m)(10 m ) + (5m)(0 m ) s s = 5m + m m = 8.33 s V = Grading Key: Part (a) 2 Points Solution to Part (b) Use conservation of momentum: m1 v1i + m2 v2i = m1 v1f + m2 v2f m m (5m)(10 ) + (m)(0) = (5m)(2 ) + (m)v2f s s m v2f = 40 s Grading Key: Part (b) 2 Points Solution to Part (c) We need to find the velocities in this reference frame. m m m v1i = 10 - 8.33 = 1.67 s s s m m v2i = 0 - 8.33 = -8.33 s s m m m v1f = 2 - 8.33 = -6.33 s s s Now use conservation of momentum to find v2f . (5m)(-8 m m ) + mv2f - m(-8.33 ) = 0 s s m v2f = 31.67 s m m = 8.33 s + 31.67 s = 40 m . s 4 We can check our answer: v2f = V + v2f Grading Key: Part (c) 2 Points Solution to Part (d) The coefficient of restitution is given by e=- vrel,f vrel,f =- vrel,i vrel,i Relative velocity is the same in all reference frames. - 40 m - 2 m 31.67 m + 6.33 m s s s s =- -10 m -8.33 m - 1.67 m s s s 38 38 = >1 10 10 The collision is explosive. Grading Key: Part (d) 2 Points Total Points for Problem: 8 Points Solution to Long-answer Homework Problem 10.3(Adventures in Mechanics) Problem: A rather rotund arctic explorer and his young assistant are scouting out an icy glacier. The explorer 1 is standing at rest enjoying the view, 1m from the edge of the water. His hapless young assistant, who has 4 of the explorer's inertia, trips and begins sliding hopelessly out of control and collides with the explorer, bouncing back at 1 his original speed. The explorer yelps as he hits the frigid water, 0.6s later. the Consider direction of 8 the young assistant's initial velocity to be positive. (a)How fast was the young assistant going when he hit the explorer? (b)To their eskimo guide, paddling directly toward them in a canoe at 2 m , what is the explorer's s momentum before and after the collision, in terms of mexplorer ? (c)What kind of collision is this? Carefully justify your answer. Solution to Part (a) Use conservation of linear momentum. masst vasst,i + mexpl vexpl,i = masst vasst,f + mexpl vexpl,f We know that vexpl,i = 0, masst = mexpl 4 , and vasst,f = - vasst,i , 8 so mexpl mexpl (vasst,i ) = 4 4 A little algebra yields vasst,i = - vasst,i + mexpl vexpl,f 8 32 vexpl,f 9 The explorer's final velocity can be found easily: x = 1.0m and t = 0.6s, so use the relation vexpl,f = So vasst,i = 5.93 1.0m m x = = 1.67 t 0.6s s m s 5 Grading Key: Part (a) 3 Points Solution to part (b) Since the assistant's initial velocity was taken to be positive, and the eskimo is approaching them, V = -2m/s. To the paddling eskimo, the momentum is given in terms of p = p - mV Since the explorer is originally at rest, p = 0, so p expl = 0 - mexpl (-2m/s) p expl = mexpl (2m/s) His velocity after the collision is 1.67m/s, so p expl = mexpl 1.67m/s - mexpl (-2m/s) p expl = mexpl (3.67m/s) Grading Key: Part (b) 1 Points Solution to part (c) To determine what kind of collision this is, we can caclulate the initial and final relative velocities, and thus the coefficient of restitution: vrel,i = vasst,i - vexpl,i = 5.93 vrel,f = vasst,f - vexpl,f and e=- So this is an inelastic collision. Grading Key: Part (c) 1 Points Total Points for Problem: 5 Points m s -5.93 m m m = - 1.67 = -2.41 8 s s s -3.15 m vrel,f s = 0.41 =- vrel,i 5.93 m s Solution to Long-answer Homework Problem 10.4() Problem: A dump truck (m = 4000kg) collides with an empty, stationary car (m = 1000kg). The dump truck was parked on a hill. The parking brake disengaged and the truck rolled down the hill, then coasted briefly along a flat stretch of road at 36 km/hr before hitting the car. The car sticks to the grill of the truck and the two continue moving forward. (a)If there were NO friction or energy loss of any kind, how far vertically would the truck have to have km travelled to go from rest to 36 hour ? (b)What is the speed of the truck-car immediately after the collision? 6 (c)How fast, and in what direction, would a jogger have to run for the momentum of the system to appear to him to be zero? (d)What is the momentum of the truck-car in the frame of reference of the jogger if he is approaching the collision at 2 m , heading in the same direction as the truck was going when it hit the car? s (e)How much energy is convertible in this collision? How does this compare to the amount of energy converted in this collision? Why? (f)If the truck and car had collided completely elastically, what would have been the final velocities of the two vehicles? Just set up the equations you would need, carefully labelling them, for full credit. Solve them correctly for extra credit. Solution to Part (a) The best and easiest way to do this problem is to use conservation of energy. At the top of the hill, the dump truck is at rest, and it has no Kinetic energy. All its energy is in the form of potential, U = mgh. If we define the bottom of the hill as zero, h is the vertical height of the hill. So, at the bottom of the hill, there is no potential energy; all of the truck's energy is in the form of Kinetic energy, KE = 1 mv 2 . Note that the velocity 2 m is 36 km = 36000 hr = 10 m . Ignoring energy losses, we can relate kinetic and potential energies: hr s U = KE mgh = (4000kg)(9.8 1 mv 2 2 m m )h = (0.5)(4000kg)(10 )2 2 s s h = 5.1m Grading Key: Part (a) 4 Points 2 point(s) : conserve energy 1 point(s) : correct conversion for m/s 1 point(s) : answer with units Solution to Part (b) To find the velocity of the combined truck and car immediately after the collision, use conservation of momentum. mc vc,i + mt vt,i = (mc + mt )vf The initial velocity of the car is zero. After the collision, the car and truck move together at the same velocity. (4000kg)(10 m ) = vf (5000kg) s m s vf = 8 Grading Key: Part (b) 4 Points 2 point(s) : conserve momentum 1 point(s) : totally inelastic; vf the same 1 point(s) : answer with units 7 Solution to Part (c) Use the equation for momentum in a different reference frame: p = p - mV where V is the velocity of the runner. We want to find the velocity of the runner such that p = 0; that is, to him, the velocity of the system appears to be zero. kg m - (5000kg)V s m V =8 s This is in the same direction as the truck is moving. I doubt if the jogger is this fast! 0 = 40000 Grading Key: Part (c) 4 Points 2 point(s) : expression for momentum in different reference frame (or could do individually) 1 point(s) : same momentum as in part (a) 1 point(s) : correct answer w/ units. Full credit for just recognizing it must be the same answer as (b) Solution to Part (d) Using the same equation, with the jogger moving 2 m in the same direction as the truck, s p = p - mV = 40000 m kg m - (5000kg)(2 ) s s kg m = 30, 000 s Grading Key: Part (d) 4 Points 2 point(s) : correct approach 1 point(s) : correct sign 1 point(s) : answer with units Solution to Part (e) The convertible kinetic energy is the amount of kinetic energy that can be converted to another form withouth violating conservation of momentum. In a totally inelastic collision, the maximum energy that can be converted is 1 2 Ksys = - vrel,i 2 1 mc mt (vt,i - vc,i )2 =- 2 mc + m t = -(0.5) m (1000kg)(4000kg) (10 - 0)2 1000kg + 4000kg s 8 = -40, 000J This is also the same as Kf - Ki . (0.5)(5000kg)(8 m m 2 ) - (0.5)(4000kg)(10 )2 = -40, 000J s s Grading Key: Part (e) 7 Points 2 point(s) : correct equation or recognizes E since totally inelastic 1 point(s) : correct substitution 1 point(s) : correct answer with units 1 point(s) : converted and convertible same 2 point(s) : same because totally inelastic Solution to Part (f) In an elastic collision, vrel,i = -vrel,f . Since vrel,i = 10 m , we know that vrel,f = vt,f - vc,f = -10 m . Solving s s for one variable in terms of the other, m vt,f = vc,f - 10 s Now, use conservation of momentum and replace vt,f with this expression. mc vc,i + mt vt,i = mc vc,f + mt vt,f 0 + mt vt,i = mc vc,f + mt (vc,f - 10 m ) s m mt vt,i = vc,f (mc + mt ) + mt (-10 ) s mt vt,i - mt (-10 m ) s vc,f = mc + m t = (4000kg)(10 m ) + (4000kg)(10 m )) s s 1000kg + 4000kg m = 16 s Now we can use this to find the final velocity of the dump truck. vt,f = vc,f - 10 = 16 m s m m - 10 s s m =6 s Another way to find both final velocities is to use both conservation of momentum and conservation of energy. mc vc,i + mt vt,i = mc vc,f + mt vt,f Solve for vc,f . 0 + (4000kg)(10 m ) = vc,f (1000kg) + vf,t (4000kg) s 9 vc,f = (4000kg)(10 m ) - vf,t (4000kg) s 1000kg Now, write the conservation of energy equation. 1 1 1 1 2 2 2 2 mc vc,i + mt vt,i = mc vc,f + mt vt,f 2 2 2 2 m 2 2 2 ) = (1000kg)vc,f + (4000kg)vt,f s From here, you can plug in the expression for vc,f , but you'll get a quadratic that won't be easy to solve. However, you should arrive at the same answers as above. 0 + (4000kg)(10 Grading Key: Part (f) 6 Points 2 point(s) : conserves relative velocity or kinetic energy 2 point(s) : conserves momentum 2 point(s) : correctly carrying out the calculations Total Points for Problem: 29 Points 10
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