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KEY NNHS Introductory Physics: MCAS Review Packet #2 2. Conservation of Energy and Momentum Broad Concept: The laws of conservation of energy and momentum provide alternate approaches to predict and describe the movement of objects. 1.) Which of the following objects has the most 2.) Calculate the kinetic energy of a dog, kinetic energy? mass = 10 kg, running at a speed of 2 m/s. A. Toy car with a mass of 1 kg and a speed of A. 10 J 1 m/s. B. 20 J B. Toy car with a mass of 1 kg and a speed of C. 40 J 5 m/s. D. 80 J C. Real car with a mass of 1000 kg and a speed of 1 m/s. D. Real car with a mass of 1000 kg and a speed of 5 m/s. My Answer and Explanation: My Answer and Explanation: D. KE= 1/2 * mass * speed squared. Bigger mass and bigger speed means bigger KE. B. KE=1/2*mass*speed squared 1/2*10*4=20J Correct Answer and Explanation: Correct Answer and Explanation: KEY NNHS Introductory Physics: MCAS Review Packet #2 Broad Concept: The laws of conservation of energy and momentum provide alternate approaches to predict and describe the movement of objects. 3.) A person is sitting at rest at the top of the 4.) You hold a 0.5 kg mass 1 meter above the biggest hill of a rollercoaster. If the person has ground. Its gravitational potential energy is a weight of 600 N, and the hill is 30 m high, approximately: what is the person’s gravitational potential A. .5 J energy? B. 5 J A. 18000 J C. 10 J B. 20 J D. 50 J C. 180000 J D. 9000 J My Answer and Explanation: My Answer and Explanation: A. GPE=Weight*Height GPE = 600 N * 30 m = 18000 J B. GPE=mass * 10m/s2 * height GPE =.5 kg*10m/s2*1 m GPE = 5 J Correct Answer and Explanation: Correct Answer and Explanation: KEY NNHS Introductory Physics: MCAS Review Packet #2 Broad Concept: The laws of conservation of energy and momentum provide alternate approaches to predict and describe the movement of objects. 6.) Which one of the following objects has 5.) Three different boxes are lifted to mechanical energy (KE +GPE) that remains different heights. • Box X weighs 115 N and is lifted to 15 m. constant? • Box Y weighs 210 N and is lifted to 10 m. A. A crate being lifted vertically upwards at a constant velocity. • Box Z weighs 305 N and is lifted to 5 m. B. An apple in free-fall. Which of the following statements best C. A car accelerating on a level (flat) highway. describes the boxes’ change in D. A sky-diver falling to Earth with his mechanical energy? parachute open. A. Box X had the greatest change in mechanical energy. B. Box Z had the smallest change in mechanical energy. C. Boxes X and Y had the same change in mechanical energy. D. Boxes Y and Z had the same change in mechanical energy. My Answer and Explanation: My Answer and Explanation: B. Box X: Change in GPE = weight * height = 115 N * 15 m = 1725 J Box Y: Change in GPE = 210 N * 10 m = 2100 J Box Z: Change in GPE = 305 N * 5 m = 1525 J B. An apple in free-fall does not have air resistance acting on it. Therefore, as it falls the GPE is converted to KE and the total remains constant. Correct Answer and Explanation: Correct Answer and Explanation: KEY NNHS Introductory Physics: MCAS Review Packet #2 Broad Concept: The laws of conservation of energy and momentum provide alternate approaches to predict and describe the movement of objects. 7.) An astronaut drops a 1.0 kg object and a 5.0 8.) The figure below shows a wagon that kg object on the Moon. Both objects fall a total moves from point X to point Y. distance of 2.0 m vertically. Which of the following best describes the objects after they have fallen a distance of 1.0 m? A. They have each lost kinetic energy. B. They have each gained the same amount of potential energy. C. They have each lost the same amount of Which of the following best describes the potential energy. wagon’s change in energy as it coasts from D. They have each gained one-half of their point X to point Y? maximum kinetic energy. A. The wagon has the same kinetic energy at point Y and at point X. B. The wagon has more kinetic energy at point Y than at point X. C. The wagon has the same gravitational potential energy at point Y and at point X. D. The wagon has more gravitational potential energy at point Y than at point X. My Answer and Explanation: My Answer and Explanation: D. As the objects fall, their PE is converted to KE. When they have fallen 1.0m they are halfway to the ground. So half of their GPE has been converted to KE. So they have gained half of their KE. D. At point Y the wagon is higher above the ground. Thus it has more GPE than it did at point X. Correct Answer and Explanation: Correct Answer and Explanation: KEY NNHS Introductory Physics: MCAS Review Packet #2 KEY NNHS Introductory Physics: MCAS Review Packet #2 Broad Concept: The laws of conservation of energy and momentum provide alternate approaches to predict and describe the movement of objects. 9.) At a weightlifting competition, two 10.) An archer pulls back the bowstring to competitors lifted the same weight to the same prepare to shoot an arrow as shown below. height. The second competitor accomplished the lift 2 seconds faster than the first competitor. This demonstrated that the second competitor had more A. energy than the first. B. inertia than the first. C. power than the first. D. work than the first. My Answer and Explanation: She uses an average force of 40 N, moving the bowstring 0.2 m. How much energy is stored in the bow? A. 8 J B. 16 J C. 24 J D. 36 J My Answer and Explanation: C. The second competitor has more power since the same amount of work was done in less time: Power = Work/time. A. The amount of energy stored is based on the work done. W = F*d = 40 N * 0.2 m = 8 J. Correct Answer and Explanation: Correct Answer and Explanation: KEY NNHS Introductory Physics: MCAS Review Packet #2 KEY NNHS Introductory Physics: MCAS Review Packet #2 Broad Concept: The laws of conservation of energy and momentum provide alternate approaches to predict and describe the movement of objects. 11.) The Watt is the unit for which quantity: 12.) One Joule is equal to A. energy A. One Watt B. work B. One kg•m/s C. force C. One Newton-meter D. power D. One Newton My Answer and Explanation: My Answer and Explanation: D. The Watt is the unit for power. C. Work is measured in Newtonmeters. Energy is measured in Joules. Work is a transfer of energy. Thus you know that Joules and Newton-meters are the same. Correct Answer and Explanation: Correct Answer and Explanation: KEY NNHS Introductory Physics: MCAS Review Packet #2 Broad Concept: The laws of conservation of energy and momentum provide alternate approaches to predict and describe the movement of objects. 13.) What is the mass of an asteroid with a 14.) A bowling ball with a mass of 8.0 kg rolls speed of 200 m/s and a momentum of down a bowling lane at 2.0 m/s. What is the 2,000 kg•m/s? momentum of the bowling ball? A. 10 kg A. 4.0 kg • m/s B. 1,800 kg B. 6.0 kg • m/s C. 2,200 kg C. 10.0 kg • m/s D. 400,000 kg D. 16.0 kg • m/s My Answer and Explanation: My Answer and Explanation: A. m = = p = m v, so = p / v (2000 kg m/s) / 200 m/s 10 kg Correct Answer and Explanation: D. p = m v p = 8 kg * 2 m/s p = 16 kg m/s Correct Answer and Explanation: KEY NNHS Introductory Physics: MCAS Review Packet #2 Broad Concept: The laws of conservation of energy and momentum provide alternate approaches to predict and describe the movement of objects. 15.) A student is standing on a skateboard that 16.) You are at an ice skating rink and are is not moving. The total mass of the student gliding towards a friend who is initially at rest. and the skateboard is 50 kilograms. The When you reach your friend, you grab your student throws a ball with a mass of friend around the waist and the two of you 2 kilograms forward at 5 m/s. Assuming the continue gliding forward. Which one of the skateboard wheels are frictionless, how will the following is true: student and the skateboard move? A. Your speed after the collision is greater than A. forward at 0.4 m/s your speed before the collision. B. forward at 5 m/s B. Your speed after the collision is the same as C. backward at 0.2 m/s your speed before the collision. D. backward at 5 m/s C. Your speed after the collision is smaller than your speed before the collision. D. Not enough information has been provided. My Answer and Explanation: My Answer and Explanation: C. Here are two ways to answer: C. When you grab your friend, Conceptual Reasoning: Due to your total momentum remains conservation of momentum, if constant but the amount of mass the ball is thrown forward, the that moves is increasing. Thus student will go backward. Since your speed must be decreasing. the boy has much more mass he will have much less speed in order to have the same momentum. Mathematical Reasoning: The total momentum in the system must remain to be zero so the momentum of the ball forward must be equal and opposite to the momentum of the boy: pball = 2 kg (5 m/s) = 10 kg m/s therefore pboy = -10 kg m/s = (50 kg) v therefore v = (-10 kg m/s)/(50 kg) = -0.2m/s, or 0.2m/s backward. Correct Answer and Explanation: Correct Answer and Explanation: KEY NNHS Introductory Physics: MCAS Review Packet #2 KEY NNHS Introductory Physics: MCAS Review Packet #2 Broad Concept: The laws of conservation of energy and momentum provide alternate approaches to predict and describe the movement of objects. 17.) An engineering student is gathering 18.) Mike, who has a mass of 75 kg, is data on the motion of a model car traveling running north at 2.6 m/s. He accidentally down a ramp. If energy is conserved, the collides with Tom, who has a mass of 125 potential energy of the car at the top of the kg and is not moving. ramp should equal the kinetic energy of Which of the following statements the car at the bottom of the ramp. After the describes how much momentum each first trial, the student calculates that the person has before the collision? kinetic energy at the bottom of the ramp is A. Mike has a momentum of 130 kg • m/s less than the potential energy at the top of north, and Tom has no momentum. the ramp. B. Mike has a momentum of 195 kg • m/s Which of the following can best explain north, and Tom has no momentum. this difference? C. Both Mike and Tom have a momentum A. The car gained a small amount of mass of 130 kg • m/s north. as it moved down the ramp. D. Both Mike and Tom have a momentum B. The student accidentally accelerated of 195 kg • m/s north. the car at the top of the ramp. C. The measured height of the ramp was less than the actual height. D. The student did not include the effect of frictional force in the calculation. My Answer and Explanation: My Answer and Explanation: D. Two ways to answer: Rule out wrong choices: Choices A, B, and C would all result in more KE of the car at the bottom of the ramp. KE depends directly on mass and velocity, and “A” and “B” would increase those quantities. Choice “C” would mean the car lost more GPE, therefore having more KE at the bottom. Why D is right: Friction would do negative work, which would reduce the kinetic energy of the car at the bottom. Correct Answer and Explanation: B. We can calculate the momentum each Person has before the collision: Mike: p = m v = 75 kg * 2.6 m/s = 195 kg m/s Tom: p = m v = 125 kg * 0 m/s = zero kg m/s Correct Answer and Explanation: KEY NNHS Introductory Physics: MCAS Review Packet #2 KEY NNHS Introductory Physics: MCAS Review Packet #2 Practice: Open-response Question #1 BE SURE TO ANSWER AND LABEL ALL PARTS OF THE QUESTION. Show all your work (diagrams, tables, or computations) If you do the work in your head, explain in writing how you did the work. Start End The above diagram shows a simple roller coaster track and one roller coaster car. The car, when full of passengers, has a mass of 500kg. A.) If the first hill has a height of 20 meters, calculate the amount of work that must be done to get the full car to the top of the first hill. B.) If there were no friction, where on this track would the car be going the fastest? Place an x on the spot on the diagram. Explain. C.) In reality, there is friction on a roller coaster track. Explain why this means that the first hill must be the highest hill on a roller coaster track. A.) The work done to get the car to the top of the first hill is equal to the amount of energy that the car has at the top of the hill. The amount of energy at the top of the first hill is GPE=mgh=500*10*20=100,000Joules B.) If there were no friction, then the total amount of energy will remain constant. The car will be going the fastest when it has the most KE. It has the most KE when it has the least GPE. It has the least GPE at the lowest height. It has the lowest height at the end. Thus it is going the fastest at the end. C.) Since there is friction, energy will continually be lost to heat and sound. Therefore the total amount of energy that the car has will continually decrease. In order to make it over a hill, a car needs both GPE and KE. If the total is KEY NNHS Introductory Physics: MCAS Review Packet #2 less, the height of a hill will need to be less that the first hill in order to have any KE. KEY NNHS Introductory Physics: MCAS Review Packet #2 Practice: Open-response Question #2 BE SURE TO ANSWER AND LABEL ALL PARTS OF THE QUESTION. Show all your work (diagrams, tables, or computations) If you do the work in your head, explain in writing how you did the work. The illustrations below show an air track with two carts before and after a collision. The mass and the initial velocity of each cart are shown below. Before Collision: First Cart: mass = 0.20 kg and velocity = 0.10m/s Second Cart: mass = 0.30 kg and velocity = 0.050m/s The first cart slides on the air track and collides with the second cart. The two carts stick together upon impact and move together along the track, as shown below. a. What is the momentum of the first cart before it collides with the second cart? Show your calculations and include units in your answer. b. What is the momentum of the second cart before the collision? Show your calculations and include units in your answer. c. Describe two changes that could be made initially to either one or both carts that would result in an increase in the momentum of the combined carts after the collision. p mv A.) p 0.2 kg * 0.1m/s p 0.02 kg m/s p mv B.) p 0.3 kg * 0.050 m/s p 0.015 kg m/s C.) You could increase the mass of either one or both of the carts. You could increase the initial velocity of either one or both of the carts. Any one of these four changes would increase KEY NNHS Introductory Physics: MCAS Review Packet #2 the total amount of momentum before the collision and thus the total amount of momentum after the collision. KEY NNHS Introductory Physics: MCAS Review Packet #2 Practice: Open-response Question #3 BE SURE TO ANSWER AND LABEL ALL PARTS OF THE QUESTION. Show all your work (diagrams, tables, or computations) If you do the work in your head, explain in writing how you did the work. In the diagram below, the falling water turns the waterwheel. The turning waterwheel generates electricity. The water moves slowly at point A and then falls rapidly past point B. a. Describe the changes in kinetic and gravitational potential energy of the water as it travels from point A to point B. b. Explain why not all of the energy of the moving water available at point A is captured by the waterwheel to generate electricity. c. Describe two ways the system can be changed so that more energy from the falling water is converted into electrical energy. A.) As the water travels from point A to point B, its gravitational potential energy decreases. While falling from A to B, its kinetic energy increases. B.) The waterwheel does not capture all of the energy of the moving water at point A because some of the energy is used to turn the wheel itself, and that energy is not available to generate electricity. Also, it appears that some of the water is splashing and spilling over the water wheel, so that water would not transfer its energy to the wheel. Also, some of the energy goes into turbulence and heating of the water. C.) To get more energy from the falling water converted into electrical energy, you could build a dam above the wheel, raising the water level above point A, or move the wheel lower, so the water would fall from a greater height. You could build a sluice or channel that would direct all KEY NNHS Introductory Physics: MCAS Review Packet #2 the water smoothly onto the wheel, reducing losses due to spilling and turbulence. If the wheel is made less massive, with less friction, then less of the water’s energy would be used to make the wheel turn. Another possibility is to make the paddles of the wheel more of a “U-shape” rather than flat, which lengthens the time the water is in contact, transferring more momentum to the wheel.