FALL 2015 Pre-Test Solution for Exam 3 10/29/15 Time Limit: 75 Minutes

Transcription

1 PHYC FALL 2015 Pre-Test Solution for Exam 3 10/29/15 Time Limit: 75 Minutes Name (Print): This exam contains 10 pages (including this cover page) and 8 problems. Check to see if any pages are missing. Enter all requested information on the top of this page, and put your initials on the top of every page, in case the pages become separated. You may use your cheat sheet of equations (which should be one side of an paper and turned in at the end of the exam), and a calculator on this exam. You are required to show your work (write down the equations you use, etc.) on each problem on this exam. You might find the following useful: Useful Equations U g = mgh K = 1 2 mv2 W = F d cos θ ΣF = ma U s = 1 2 kx2 F sp = kx J = p = F ave t p = mv f mv i τ = rf sin φ what you get e = what you put in Unit Conversion & Constants 1 hour = 3600 seconds g e = 9.8 m/s 2 Mysterious or unsupported answers will not receive full credit. A correct answer, unsupported by calculations, explanation, or algebraic work will receive no credit; an incorrect answer supported by substantially correct calculations and explanations might still receive partial credit. Problem Points Score Total: 100 Do not write in the table to the right.

2 PHYC Pre-Test Solution for Exam 3 - Page 2 of 10 10/29/15 1. (10 points) While walking through the county fair with his family, John notices a sign hanging from the wall for the petting zoo (see figure). He is told by the owner of the petting zoo that if he guesses the weight of the sign being only given that the tension in the cable is 30 lbs and that the angle between the cable and the pole that it is attached to is 30 o, then he can win a stuffed animal for his kids. The pole that the sign is hanging on is free to pivot about the end where it touches the wall. John decides to take on the challenge. What is the weight of the sign? (Hint: This is a static equilibrium problem.) Consider the right end of the rod from which the sign hangs and assume the length of the rod is l. Simply apply the fact that the system is in static equilibrium, meaning not only that ΣF = 0, but also that Στ = 0. The two forces are the downward weight force, w sign, which is at an angle of 90 to the rod, and the tension in the cable, T, at an angle of 30 to the rod. The vertical component of the tension in the cable that counteracts the weight is T y = T sin 30. In order to find the tension, T, in the cable, John computes the sum of the torques around the left end of the rod as follows Στ y = T y l w sign l = 0 w sign l = T l sin 30 w sign l = T l sin 30 w sign = T sin 30 ( ) 1 = (30 lb) 2 = 15 lb The limiting cases that exist assure the correctness of the answer; if the angle were smaller the tension and the weight of the sign would be greater and vice-versa.

3 PHYC Pre-Test Solution for Exam 3 - Page 3 of 10 10/29/15 2. (10 points) To win a prize at the carnival, John the Hippy has to try to knock down a heavy bowling pin by throwing an object and hitting it. He has three options: a rubber ball, a hacky sack, and a ball of clay, all of equal size and weight. Which should he choose? Explain. He should throw the rubber ball instead of the hacky sack or ball of clay, to increase his chances of knocking down the bowling pin. This is because the rubber ball will exert more impulse on the pin. The impulse on the bowling pin equals the negative of the impulse on the projectile, which in turn equals the negative of the change in momentum of the projectile: J on pin = p projectile. The rubber ball, hacky sack, and clay ball would start with the same momentum, but the rubber ball would have a greater change in momentum because it would bounce off and so its direction would change. The hacky sack and clay ball would continue to move in the same direction. Thus the rubber ball would exert a greater impulse on the bowling pin. This result makes sense because we see that a collision with the rubber ball is more violent than a collision with the hacky sack or clay ball in that the collision turns the projectile around.

4 PHYC Pre-Test Solution for Exam 3 - Page 4 of 10 10/29/15 3. (10 points) The force versus time graph for the rubber ball after it is thrown is shown below. What is the total impulse that it experiences from the force described by the graph? Using the knowledge that impulse, J, equals the area under a force versus time graph we have that J x = area under the force curve = F Ave t = F x,max ( 1 2 s ) = (2 N) ( ) 1 2 s = 1 kg m/s The force is to the right and from this we see that the change in momentum of the rubber ball is also positive.

5 PHYC Pre-Test Solution for Exam 3 - Page 5 of 10 10/29/15 4. (10 points) Mary stands at the edge of a small merry go round in the park that is slowly rotating. Assume that it is rotating on a frictionless axle so that the angular momentum, L, is conserved and the merry go round can be treated as an isolated system. As Mary walks toward the center, the moment of inertia, I, of the merry go round decreases which means that the angular velocity, ω, of the system increases, because the angular momentum, L = Iω, stays constant. Once she passes the center and gets closer to the opposite edge of the merry go round as it spins, the moment of inertia, I, increases, until it is the same as initially. Hence the angular velocity, ω, decreases until it is the same as initially. increases; decreases; stays constant; the same as initially NOTE: This is because the merry go round s moment of inertia is dependent on how the mass is distributed. Just like the ice skater whose spin speed decreases with his arms out wide and increases when he brings his arms in due to the change in moment of inertia (see example 9.11 from lecture), as Mary walks towards the center, more mass is brought to the center, increasing the spin speed of the merry go round and decreasing its moment of inertia. As she walk out, the moment of inertia is increases as the mass is redistributed and the spin speed decreases.

6 PHYC Pre-Test Solution for Exam 3 - Page 6 of 10 10/29/15 5. Remember the roller coaster ride that John and Mary (the Hippies) decided to take at the amusement park, called the Roller Coaster of Doom. This roller coaster was given its name due to its 600 foot ( 183 meter) drop. In order to get to the top of this drop, the cart that they ride (which has a maximum weight limit of 750 lbs 340 kg) must first go up an incline of 70o at a constant speed of 4.5 mph ( 2.0 m/s). The cart is pulled at this constant speed by an electric winch which winds carts to the top of the ramp. Once the cart reaches the top, the electric winch releases it, and the cart coasts to a stop (v = 0 m/s) at the edge of the hill because of the friction on the cart s wheels due to the track, before it makes its way down the other side of the hill. (a) (5 points) What is the energy transformation that takes place while the cart goes from the bottom of the hill to the top? (HINT: The winch is doing to get the cart up to a height of 183 meters at a constant speed.) A. B. C. D. E. F. W Ug W K W Eth All of the above Both A and B Both A and C (b) (5 points) What is the energy transformation that takes place while at the top, where the cart slows down at a constant acceleration due to friction? (HINT: What energy is a result of friction?) A. K Ug B. Ug K C. K Eth D. Ug Eth

7 PHYC Pre-Test Solution for Exam 3 - Page 7 of 10 10/29/15 (c) (10 points) Ignoring friction, what is the amount of work that the winch has to do to displace the cart to the top of the hill? (HINT: What energy transformation is taking place?) Since we have an energy transfer from the work done by the winch, W, to gravitational potential energy, U g, and thermal energy, E th, we can use the work-energy theorem. Now, the problem says to ignore friction, so we can just assume that all the work, W, that the winch does on the cart is transformed into gravitational potential energy, U g. Using the work-energy theorem we have that W = U g = U g,f U g,i = m cart gh f m cart gh i = m cart g (h f h i ) = (340 kg) (9.8 m/s) (183 m) J (d) (10 points) Ignoring friction, what is the cart s final speed at the bottom? (HINT: What energy transformation is taking place?) Since we have an energy transfer from the gravitational potential energy, U g, to kinetic energy, K, we can use the work-energy theorem where we know that the initial velocity is 0 m/s and the final height is 0 m. Now, the problem says to ignore friction, so we can just assume that all the gravitational potential energy, U g, of the cart is transformed into kinetic energy, K. Using the work-energy theorem we have that U g = K mgh = 1 2 mv2 v = 2gh = 2 (9.8 m/s) (183 m) 60 m/s

8 PHYC Pre-Test Solution for Exam 3 - Page 8 of 10 10/29/15 6. (10 points) Mary and John decide to go and play with bumper cars. In order to get the bumper cars going they have to put quarters into the cars themselves to get them going. The maximum speed of each bumper car is 5 m/s. They are all identical cars, just different colors, each having the same mass of 120 kg. John has a problem with his car and it will not start (v = 0 m/s). Mary, reaching deep inside and finding her inner, competitive child decides to floor it straight towards John. They collide in such a way that their collision can be treated as perfectly elastic, and Mary flies backward at a speed of 2.5 m/s. What is John s speed after the collision and in what direction? Treat this as a one-dimensional collision. (HINT: What is conserved in a perfectly elastic collision?) ANSWER to HINT: Mechanical energy and momentum are conserved. Here we can use either of the conservation laws to solve the problem. If consider the system as the two bumper cars with John and Mary, we can treat it as an isolated system. We can take the masses of the two cars with John and Mary in them as m J = 120 kg and m M = 120 kg, respectively. Using conservation of momentum, we have p total = 0 p final = p initial m J (v J ) f + m M (v M ) f = m J (v J ) i + m M (v M ) i m J (v J ) f = m J (v J ) i + m M (v M ) i m M (v M ) f (v J ) f = (v J ) i + m M m J ((v M ) i (v M ) f ) = (0 m/s) + = (7.5 m/s) = 7.5 m/s 120 kg (5 m/s ( 2.5 m/s)) 120 kg

9 PHYC Pre-Test Solution for Exam 3 - Page 9 of 10 10/29/15 7. (10 points) While walking through the carnival, John decides to test his strength on the Strongman Game. This is a game, where a mallet is used to propel a puck vertically up a long tower by hitting the heavy base with the mallet. At the top of the tower a bell is hung. The base is composed of the rubber-padded, circular plate that is attached to a spring of spring constant k = 5000 N/m. In order to get the puck to hit the bell, the spring has to be displaced from equilibrium by 10 cm. How much force must John apply to hit the bell and how much work does he do? (HINT: You can either use conservation of energy of energy or Hooke s law.) In order to compress the spring by 10 cm downward (i.e., in the negative direction), the force that John applies is equal and opposite in direction to that calculated using Hooke s law, which is the spring s restoring force. This leads to F = kx = (5000 N/m) ( 0.10 m) = 500 N We know that the work, W, that John puts into compressing the spring transforms into the elastic potential energy of the spring, U s, so we can use the work-energy theorem to write: W = U s This is the work that John puts in. = 1 2 kx2 = 1 (5000 N/m) ( 0.10 m)2 2 = 25 J

10 PHYC Pre-Test Solution for Exam 3 - Page 10 of 10 10/29/15 8. (10 points) After all the work that John the Hippy did displacing the spring to hit the bell on the Strongman Game, he gets hungry and decides to order some food from one of the vendors. Assuming that his efficiency for converting the energy in the food to workable energy is 25%, e =.25, which food (or combinations of food) would be more practical to eat?(hint: The food is what you put in!) We see from the last problem that the amount of work needed to compress the spring is 25 Joules. Using the efficiency equation of e = what you get what you put in, where we have that what he gets is the 25 Joules, and that the efficiency is 0.25, we find that the amount that he has to eat is what you get e = what you put in what you get what you put in = e = 25 J 0.25 = 100 J Seeing that the food listed above is in kilo-joules (kj) = 10 3 Joules (J), we see that for 100 J of energy, John would only have to eat 1/1250 of a large carrot or of a fried egg. Not a whole lot of food.