Physics_114_Practice_Final_Exam_Solutions

Physics 114, Final Exam 3 2. [6 pts] A 500-N person steps on a scale in an elevator. The elevator goes down. During the elevator ride, the person notices the following readings on the scale at different times: 347 N, 500 N, and 653 N. For each of the time intervals described below, identify which of the three scale readings (347 N, 500 N, or 653 N) the person saw. • the time interval during which the elevator speeds up: 347 N • the time interval during which the elevator moved at a constant speed between floors: 500 N • the time interval during which the elevator slowed down: 653 N Explain the reasoning you used to arrive at your answers. There are always two forces acting on the person: the upward normal force from the scale and the downward weight force. According to Newton’s 2 nd (and 1 st law), these two forces must have the same magnitude when the acceleration is 0 m/s 2 , so the scale will read 500 N when it moves at a constant speed between floors. The elevator is moving down, so when it speeds up, the acceleration and the net force must point down. The weight force’s magnitude does not change (since it is always the mass of the person multiplied by the acceleration due to gravity), so the only way for the net force to point down is for the normal force to become smaller in magnitude. Since the scale measures the normal force, the scale will read 347 N while the elevator speeds up. When the elevator is moving down and slowing down, the net force will point up, which means the normal force – and the scale reading – will be bigger than the person’s weight.

Physics 114, Final Exam 6 Wallabies are marsupial mammals native to Australia that are famous for their ability to hop. In the next 4 questions, you will examine various aspects of physics that allow a 10.5-kg red- necked wallaby ( Macropus rufogriseus ) to efficiently hop. Note: Questions 5, 6, 7, and 8 are independent of each other, so you may do them in any order. 6. [4 pts] The gastrocnemius is an important muscle in the wallaby’s leg. For the 10.5-kg red-necked wallaby, scientists measured the force experienced by this muscle as it stretched and relaxed during a hop. Below is a graph of this force versus the change in the length of the muscle. The dashed red lines outline a triangle whose area we will use to approximate the area under the loading curve. The solid blue lines outline a triangle whose area we will use to approximate the area under the unloading curve. Determine the resilience of the wallaby’s gastrocnemius muscle. Show your work. The resilience is equal to the area under the unloading curve (i.e., the work done by the muscle as it relaxes) divided by the area under the loading curve (the work done on the muscle as it is stretched): 𝑟𝑟 = 0.5(21 mm)(380 N) 0.5(19 mm)(480 N) = 0.88 A red-necked wallaby 0 10 20 length change (mm) 100 400 300 200 force (N) 0

Physics 114, Final Exam 9 10. [8 pts] A student was asked to draw a PV diagram and the corresponding energy bar chart for an isothermal process that takes an ideal gas from point B to point A. Point A is at a higher pressure but smaller volume than point B. The number of particles in the gas did not change during this process. The student made four errors in their PV diagram and energy bar chart. Identify all four errors and explain why they are wrong. • Error 1: The path between B and A cannot be a straight line for an isothermal process. Since the gas is ideal, PV = Nk B T. Since the number of particles N and T do not change and since k B is a constant, PV is also constant. Graphing P as a function of V gives a curve corresponding to the relationship P = constant/V, which is not the equation for a straight line. • Error 2: For an isothermal process, the final thermal energy must be the same as the initial thermal energy. Thermal energy depends on the temperature according to E th = 1.5 k B NT, so if the temperature is constant, the thermal energy must also be constant. • Error 3: The gas is being compressed (i.e., the volume gets smaller as it goes from point B to point A), which means the work done on the gas must be positive, not negative as it is currently shown in the energy bar chart. • Error 4: Q cannot be 0 J for an isothermal process. Since the initial and final thermal energies must be the same, the Q must have the same absolute value, but the opposite sign, of W on . volume pressure B A energy E th,i W on Q E th,f Q = 0 J

Physics 114, Final Exam 12 c. [4 pts] What is the highest point above the ground reached by the athlete’s COM? Show your work. This occurs at 9.6 s, when v y = 0 m/s. The area under v y up until that time is 0.5(7.0 m/s)(9.6 s – 8.2 s) = 4.9 m. This is the vertical displacement, so the COM’s height is 1.1 m + 4.9 m = 6.0 m. 13. [4 pts] 0.0040 mol of gas follows the isothermal trajectory shown in the figure at right. What is the final temperature? How does the magnitude of the work done on the gas compare to the magnitude of the heat exchange with the gas? Explain your reasoning. Since the path is isothermal, this means that the initial and final temperature of the gas have the same value. Since the thermal energy of the gas is proportional to the temperature of the gas, I can conclude that the thermal energy of the gas remains constant. Since the gas is expanding, the work done on the gas is defined as the negative of the area under the curve. In order for the thermal energy of the gas to remain constant, heat must be added to the gas and the magnitude of the heat added must to be equal to the magnitude of the work done on the gas.

Physics 114, Final Exam 15 17. The pV diagram at right represents the changes two different ideal gases take to get from the same initial state ( P = 3 Pa, V = 3 m 3 ) to the same final state ( P = 1 Pa, V = 1 m 3 ). Both contain 0.08 mol of particles. a. [5 pts] Determine the initial and final temperatures of the gas represented by the solid blue line and the initial and final temperatures of the gas represented by the dashed red line. Since both gases have the same initial volume, pressure, and number of moles, they must also have the same initial temperature according to the ideal gas law: Likewise, they must both have the same final temperature since they have the same final volume, pressure, and number of moles: b. [10 pts] Complete the first law bar chart for the gas represented by the solid blue line. Make sure the bar chart is quantitatively correct. The thermal energy is directly related to the temperature, so we can calculate the initial and final thermal energies: The work done on the gas is positive (since the gas is getting compressed). This work is equal to the area under the pV curve: Once the change in thermal energy and the work done on the gas are known, we can calculate Q: 0 1 2 3 4 0 1 2 3 4 pressure (Pa) volume (m 3 )

Physics 114, Final Exam 18 19. The famous basketball player Michael Jordan once jumped off the court and spent an amazing 0.92 s in the air before landing on the ground. The graphs below represent the x - and y -components of his velocity during his jump. a. [5 pts] Jordan traveled a horizontal distance of 1.08 m during this jump. Determine his initial x - velocity, initial y -velocity, and the magnitude of his initial velocity. Show your work. The horizontal distance he traveled, 1.08 m, is equal to the area under the v x graph, so (v x )(0.92 s) = 1.08 m, which means v x = (1.08 m)/(0.92 s) = 1.17 m/s. The slope of the v y graph is -9.8 m/s 2 since Jordan was in free fall. We can use this to figure out his initial y-velocity: 𝑎𝑎 𝑦𝑦 = 𝑣𝑣 𝑦𝑦 , 𝑓𝑓 − 𝑣𝑣 𝑦𝑦 , 𝑖𝑖 ∆𝑡𝑡 → 𝑣𝑣 𝑦𝑦 , 𝑓𝑓 − 𝑣𝑣 𝑦𝑦 , 𝑖𝑖 = �𝑎𝑎 𝑦𝑦 � ( ∆𝑡𝑡 ) → 𝑣𝑣 𝑦𝑦 , 𝑖𝑖 = 𝑣𝑣 𝑦𝑦 , 𝑓𝑓 − �𝑎𝑎 𝑦𝑦 � ( ∆𝑡𝑡 ) 𝑣𝑣 𝑦𝑦 , 𝑖𝑖 = (0 m/s) − ( − 9.8 m s 2 ⁄ )(0.46 s) = 4.51 m/s Now use the Pythagorean theorem to determine the magnitude of the initial velocity: 𝑣𝑣 = � (1.17 m/s) 2 + (4.51 m/s) 2 = 4.66 m/s b. [3 pts] What was Jordan’s maximum height above the ground? Show your work. This will be equal to the area under the y-velocity curve until t = 0.46 s: Δ y = 0.5(0.46 s)(4.51 m/s) = 1.03 m. 0.46 0.92 time (s) v y 0.46 0.92 time (s) v x

Physics 114, Final Exam 21 22. The graph at right shows the stress- strain curves for the Achilles tendons of rats. Data from two groups of rats are shown: the diabetic group (DG; dashed line) and the control group (CG; solid line). Note that the authors used a comma where we would normally use a decimal point. Also, strains are listed as percentages, so 60,000% is actually 60.000% = 0.60). a. [3 pts] At a stress of 5500 N/m 2 (5,500 MPa on the graph), is the Young’s modulus for the CG’s Achilles greater than, less than, or equal to that of the DG? Explain your reasoning. At this stress, the slope of the DG’s curve is steeper than the slope of the CG’s curve, so the DG must have the larger value of Young’s modulus. b. [4 pts] The unstretched length of a rat’s Achilles tendon 5.5 mm. If an Achilles tendon from a diabetic rat and an Achilles tendon from a non-diabetic rat are both stretched by a stress of 4.000 MPa (4,000 MPa on the graph), which rat (diabetic or non-diabetic) will end up with a longer Achilles tendon and by how much? Show your work. At this stress, the CG’s strain is 0.60 while the DG’s strain is 0.98. For the CG’s, the Achilles tendon is stretched out by (0.60)(5.5 mm) = 3.3 mm, while for the DG the Achilles is stretched out by (0.98)(5.5 mm) = 5.4 mm. The diabetic rats will have Achilles tendons that are longer than their non-diabetic peers by 5.4 mm – 3.3 mm = 2.1 mm.

Physics 114, Final Exam 22 23. A child, starting from rest, sleds down a snow-covered hill. The hill makes an angle of 25 o with respect to the horizontal and the top is 8.0 m above the bottom. The coefficient of kinetic friction between the sled and the snowy hill is 0.05 and the total mass of the child and sled is 35 kg. a. [5 pts] Treat the child and sled as a single object and draw their free-body diagram. Label the types of forces (e.g., normal, friction, weight, etc.) and use appropriate subscripts to indicate the object exerting the force and the object feeling the force. The lengths of your arrows should be qualitatively correct (e.g., if two forces have the same magnitude, then their arrows should be the same length). No explanation is necessary to receive full credit. b. [7 pts] What is the magnitude of the acceleration experienced by the sled and child as they travel down the hill? Show your work. We must apply Newton’s 2 nd law to both the x- and y-components of the forces. Using a rotated coordinate system, this gives me the following relationship for the y-components: F net,y = ma y N HS – W ES cos(25 o ) = 0 N HS = mg cos(25 o ) For the x-components, I get F net,x = ma x W ES sin(25 o ) – f HS = ma W ES sin(25 o ) – μ k N HS = ma mg sin(25 o ) – μ k mg cos(25 o ) = ma a = g sin(25 o ) – μ k g cos(25 o ) = (9.8 m/s 2 )sin(25 o ) – (0.05)(9.8 m/s 2 )cos(25 o ) = 3.7 m/s 2 25 o 8.0 m W ES f HS N HS

Physics 114, Final Exam 23 c. [7 pts] Use work-energy principles to calculate the speed of the child and sled at the bottom of the hill. You may use any system you wish. Show your work. I will use a system that includes the child, sled, and Earth, so friction will be the only external force that does work on the system. The work-energy theorem becomes W net,ext = Δ E W f = Δ K + Δ U g W f = K f – K i + U g,f – U g,i -(f HS )(8.0 m)/sin(25 o ) = K f – U g,i K f = U g,i – (f HS )(8.0 m)/sin(25 o ) 1 2 𝑚𝑚𝑣𝑣 𝑓𝑓 2 = 𝑚𝑚𝑔𝑔𝑔𝑔 𝑖𝑖 − 𝜇𝜇 𝑘𝑘 𝑚𝑚𝑔𝑔 cos(25 𝑜𝑜 )(8.0 m) sin(25 𝑜𝑜 ) 𝑣𝑣 𝑓𝑓 = � 2 𝑔𝑔𝑔𝑔 𝑖𝑖 − 2 𝜇𝜇 𝑘𝑘 𝑔𝑔 cos(25 𝑜𝑜 )(8.0 m) sin(25 𝑜𝑜 ) 𝑣𝑣 𝑓𝑓 = � 2(9.8 m/s 2 )(8.0 m) − 2(0.05)(9.8 m/s 2 )cos(25 𝑜𝑜 )(8.0 m) sin(25 𝑜𝑜 ) = 11.8 m/s

Physics 114, Final Exam 24 24. The plot at right shows the cumulative number of COVID-19 cases in Canada from February 27 to April 2, 2020. a. [2 pts] Determine how many COVID-19 cases there were in Canada on February 27 (labeled as day 0). Show your work. According to the graph, at day 0 ln(N COVID ) = 2.3. Therefore N COVID = e 2.3 = 10. b. [2 pts] Determine how many COVID-19 cases there were in Canada just 35 days later, on April 2. Show your work. According to the graph, at day 35 ln(N COVID ) = 9.2. Therefore N COVID = e 9.2 = 9900. c. [4 pts] Find the mathematical relationship between the cumulative number of COVID-19 cases N COVID and t , the time in days. There should be no logarithms in your final result. Show your work. The y-intercept of the graph appears to be at 2.3. The slope is (9.2 – 2.3)/35 = 0.20. Therefore the dashed trendline shown on the graph is given by the equation ln(N COVID ) = 0.20t + 2.3 If we make both sides the power to which e is raised, we then get N COVID = e 2.3 e 0.2t N COVID = 10e 0.2t d. [2 pts] Based on the relationship between N COVID and t from part c, how many days would it take, if no restrictive measures had been taken, to reach 330 million cases (which happens to be equal to the population of the United States)? Show your work. 330 x 10 6 = 10e 0.2t (330 x 10 6 )/(10) = e 0.2t ln(330 x 10 5 ) = 0.2t t = ln(330 x 10 5 )/0.2 = 87 days

Physics 114, Final Exam 25 25. A person is using a rope to pull a 60-kg crate up an incline. The incline makes an angle of 35 o with respect to the horizontal. The coefficient of kinetic friction between the crate and the incline is 0.25. The crate accelerates up the incline at 0.5 m/s 2 . a. [4 pts] Draw the crate’s FBD. Make sure to label the forces appropriately using the proper subscripts and that the lengths of your arrows are qualitatively correct (e.g., if two forces have the same magnitude, then their arrows should have the same length). No explanation is necessary to receive full credit. b. [6 pts] What is the magnitude of the tension force that the rope exerts on the crate? Show your work. We can apply Newton’s 2 nd law. Begin by applying Newton’s 2 nd law to the y-components: N IC – W EC cos(35 o ) = 0 N IC = mg cos(35 o ) Now apply Newton’s 2 nd law to the x-components: T RC – W EC sin(35 o ) – f IC = ma T RC – W EC sin(35 o ) – µ k N IC = ma T RC = mg sin(35 o ) + µ k mg cos(35 o ) + ma T RC = (60 kg)(9.8 m/s 2 ) sin(35 o ) + (0.25)(60 kg)(9.8 m/s 2 ) cos(35 o ) + (60 kg)(0.5 m/s 2 ) T RC = 488 N W EC 35 o N IC T RC f IC 35 o

Physics 114, Final Exam 26 26. A 2.0-kg mass is attached to a spring of spring constant k = 7.0 N/m. The mass is oscillating back- and-forth on a horizontal plane between the positions x = -15 cm and +15 cm (note that x = 0 corresponds to the position at which the spring is at its equilibrium length). The plane is frictionless, so there is no damping. The mass is at x = -6.0 m at t = 0 s and is headed toward x = +15 cm. a. [2 pts] What is the amplitude of this oscillatory motion? Explain/show your work. Since the mass oscillates between -15 cm and 15 cm, the amplitude must be 15 cm. b. [2 pts] What is the period of this oscillatory motion? Explain/show your work. The period of a mass on a spring is given by 𝑅𝑅 = 2 𝜋𝜋� 𝑚𝑚 𝑘𝑘 = 2 𝜋𝜋 � 2.0 kg 7.0 N/m = 3.4 s c. [3 pts] What is the phase constant (in radians) of this oscillatory motion? Explain/show your work. The equation for the oscillating mass is 𝑥𝑥 ( 𝑡𝑡 ) = (15 cm ) cos � 2 𝜋𝜋 3.4 s 𝑡𝑡 ± 𝜑𝜑� Solve for the phase constant by setting x = -6 cm at t = 0 s: − 6 cm = (15 cm ) cos � 2 𝜋𝜋 3.4 s (0 s ) ± 𝜑𝜑� 𝜑𝜑 = ± cos −1 ( − 6/15) = ±2 rad Since the mass is heading up toward x = +15 cm at t = 0 s, the peak of the oscillation must have occurred after t = 0 s, so we must choose the negative value for the phase constant: -2 rad. d. [3 pts] Write down an equation for the position x of the mass as a function of time. 𝑥𝑥 ( 𝑡𝑡 ) = (15 cm ) cos � 2 𝜋𝜋 3.4 s 𝑡𝑡 − 2 rad �

Physics 114, Final Exam 27 27. A 2500-kg rocket is sent up in the atmosphere carrying an instrument to study the outer Earth’s atmosphere. While the engine is firing, the rocket speeds up with a constant acceleration. After the engine shuts off, the rocket is in free fall. The rocket’s motion is entirely in the vertical direction. Define vectors that point up to be positive and vectors that point down to be negative. a. [4 pts] Draw a qualitatively correct velocity-time diagram for the motion of the rocket, from the instant it is launched until the instant it reaches its maximum height. You do not need to put any numbers on your graph in order to receive full credit. b. [3 pts] Knowing that the rocket reaches a maximum speed of 500 m/s, how far upward did it travel after the engine shut off? Show your work. At its maximum height, the rocket will have slowed down to 0 m/s. To find how far upward the rocket traveled after the engine shut off, we need to find the area under the curve from the peak of the velocity versus time graph until the velocity = 0 m/s. This area is given by 𝐴𝐴 = 1 2 ( ∆𝑡𝑡 )(500 m/s ) We need to figure out how much time Δt passes from the instant when the engines shut off until the rocket’s speed becomes 0 m/s. During this time, the rocket is in free fall, so it’s acceleration must be -9.8 m/s 2 . From the definition of acceleration, we can figure out Δt: ∆𝑡𝑡 = 𝑣𝑣 ⃗ 𝑓𝑓 − 𝑣𝑣 ⃗ 𝑖𝑖 𝑎𝑎 ⃗ = (0 m/s) − (500 m/s) − 9.8 m/s 2 = 51 s Therefore, the distance the rocket travels upward after its engines shut off is 0.5(51 s)(500 m/s) = 13,000 m. This result can also be found from the following kinematics equation: 𝑣𝑣 ⃗ 𝑓𝑓 2 = 𝑣𝑣 ⃗ 𝑖𝑖 2 + 2 𝑎𝑎 ⃗ ( ∆𝑔𝑔 ) ∆𝑔𝑔 = 𝑣𝑣 ⃗ 𝑓𝑓 2 − 𝑣𝑣 ⃗ 𝑖𝑖 2 2 𝑎𝑎 ⃗ = (0 m/s) 2 − (500 m/s) 2 2( − 9.8 m/s 2 ) = 13,000 m c. [3 pts] If it took 600 s to reach the 500 m/s speed, what was the highest altitude reached by the rocket? Show your work. From the time it is launched until the engine’s shut off, the rocket travels a distance of 0.5(600 s)(500 m/s) = 150,000 m. The highest altitude reached is therefore 150,000 m + 13,000 m = 163,000 m. time velocity

Physics 114, Final Exam 28 28. The PV diagram at right shows two different processes (Process A in blue and Process B in red) that a gas can take. Note that both processes have the same initial point and the same final point. Process A is an isothermal process. a. [4 pts] Sketch a first-law bar chart for Process A. You are not required to calculate any numbers, but your chart should be qualitatively correct. The dark black line represents 0; above this line are positive values of energy and below the line are negative values of energy. b. [5 pts] Is the heat added to the gas in Process B greater than, less than, or equal to the heat added to the gas in Process A? Explain your reasoning. Since both processes start at the same point and end at the same point, they both correspond to the same change in thermal energy (which is 0, since Process A is an isothermal process). Negative work is done on the gas in both processes since both correspond to an expanding gas. However, the work done in Process B is a larger negative number than the work done in Process A since Process B has a larger area under the curve. That means that the heat added in Process B must be larger than the heat added in Process A since, for both processes, W + Q = 0 since there is no change in thermal energy. Volume Pressure Process B E th,i W on Q E th,f Energy

Physics 114, Final Exam 29 29. A 4.5-kg cat is on a horizontal, uniform beam that is 0.75 m above a roof. The cable that helps attach the beam to a building makes an angle of 25 o with respect to the beam. The cat is walking with a constant speed of 1.0 m/s away from the building. The beam has a mass of 18 kg and a length of 3.0 m. a. [10 pts] The cable will break when the tension force exceeds 300 N. How far from the building can the cat walk before the cable breaks? Show your work. Place the pivot point at the location where the beam touches the building, since we know neither the direction nor the magnitude of the force that the building exerts on the beam. The beam is in static equilibrium, so we can set the net torque = 0 N-m. If x represents the maximum distance from the building that the cat can walk, then the condition of net torque = 0 N-m becomes (T CB )(3.0 m)sin(25 o ) – (W EB )(1.5 m) – (N CB )(x) = 0 𝑥𝑥 = ( 𝑅𝑅 𝐶𝐶𝐵𝐵 )(3.0 m ) sin(25 𝑜𝑜 ) − ( 𝑊𝑊 𝐸𝐸𝐵𝐵 )(1.5 m ) 𝑁𝑁 𝐶𝐶𝐵𝐵 𝑥𝑥 = (300 N )(3.0 m ) sin(25 𝑜𝑜 ) − (18 kg ) � 9.8 m/s 2 � (1.5 m ) (4.5 kg )(9.8 m/s 2 ) = 2.6 m b. [4 pts] The cable breaks as the cat is walking, causing the cat to fall. Use work-energy principles to determine the speed of the cat the instant before it reaches the roof. Ignore air resistance. Show your work. Below are two different, but equally correct, ways to solve the problem, based on how the system is defined (specifically, whether or not Earth is included in the system). 25 o building beam System: cat and Earth W net,ext = Δ E 0 = Δ K + Δ U grav 0 = K final – K initial + U grav,final – U grav,initial 0 = K final – K initial + 0 – U grav,initial K final = K initial + U grav,initial 1 2 𝑚𝑚𝑣𝑣 𝑓𝑓 2 = 1 2 𝑚𝑚𝑣𝑣 𝑖𝑖 2 + 𝑚𝑚𝑔𝑔ℎ 𝑣𝑣 𝑓𝑓 = �𝑣𝑣 𝑖𝑖 2 + 2 𝑔𝑔ℎ 𝑣𝑣 𝑓𝑓 = � (1.0 m/s) 2 + 2(9.8 m/s 2 )(0.75 m) 𝑣𝑣 𝑓𝑓 = 4.0 m/s System: cat only W net,ext = Δ E W grav = Δ K W grav = K final – K initial W grav = K final – K initial K final = K initial + W grav 1 2 𝑚𝑚𝑣𝑣 𝑓𝑓 2 = 1 2 𝑚𝑚𝑣𝑣 𝑖𝑖 2 + 𝑚𝑚𝑔𝑔ℎ 𝑣𝑣 𝑓𝑓 = �𝑣𝑣 𝑖𝑖 2 + 2 𝑔𝑔ℎ 𝑣𝑣 𝑓𝑓 = � (1.0 m/s) 2 + 2(9.8 m/s 2 )(0.75 m) 𝑣𝑣 𝑓𝑓 = 4.0 m/s

Physics 114, Final Exam 30 30. Two objects of different mass are dropped, object A (blue circle on the left) is dropped straight down (with zero initial speed), and object B (red circle on the right) is shot off to the right with a constant horizontal speed. Define velocity components that point up or to the right to be positive and components that point down or to the left to be negative. A velocity is 0 m/s when it crosses the time axis. a. [4 pts] Sketch qualitatively correct graphs for the x - and y -components of object A’s velocity as a function of time. You do not need to put any numbers on your graph nor do you need to provide any explanation in order to receive full credit. Explanation (not required): Object A starts from rest. It is in free fall, so the only acceleration it experiences is the acceleration due to gravity, which points straight down. Since there is no acceleration in the x-direction, its x-velocity will not change. b. [4 pts] Sketch qualitatively correct graphs for the x - and y -components of object B’s velocity as a function of time. You do not need to put any numbers on your graph nor do you need to provide any explanation in order to receive full credit. Explanation (not required): Object B does not start from rest; it initially has a velocity that only points to the right. It is in free fall, so the only acceleration it experiences is the acceleration due to gravity, which points straight down. Since there is no acceleration in the x-direction, its x-velocity will not change. c. [3 pts] Based on your graphs, which object will reach the ground first (or do they reach the ground at the same time)? Explain your reasoning using the velocity vs. time diagrams you just drew. The distance each object will travel to the ground is the same for both and is equal to the areas under their y-velocity curves. Since these areas are the same, as are the initial y-velocities (0 m/s) and the slopes on the y-velocity graphs (-9.8 m/s 2 ), the amount of time that passes must also be the same for both objects, so they hit the ground at the same time. time x -velocity time y -velocity time x -velocity time y -velocity A B

Physics 114, Final Exam 31 d. [4 pts] Which object will hit the ground with greater speed (or will they have the same speed)? Explain. Object B will hit with the greater speed. This is because the decrease in gravitational potential energy is the same for both situations, since both objects start the same height above the ground. This potential energy is all converted into kinetic energy, so both objects experience the same increase in kinetic energy. Since Object B has some initial kinetic energy, while Object A has none, Object B will end up with more kinetic energy when it hits the ground. Alternatively, one can reason that the y-component of the velocity with which they will hit the ground is the same, but the x-component of the velocity is zero for A and non-zero for B. Therefore, the magnitude of the velocity with which they will hit the ground is going to be greater for B. 31. The figure at right is the position versus time graph of a 150 g baseball right before and right after being hit by a baseball bat. a. [5 pts] Calculate the change in momentum of the ball. Show your work. To get the change in momentum of the ball, we must determine its velocity both before and after the collision by looking at the slopes on the position versus time graphs at right: 𝑣𝑣 ⃗ 𝑖𝑖 = 0 m − 0.15 m 0.002 s = − 75 m/s 𝑣𝑣 ⃗ 𝑓𝑓 = 0.1 m − 0 m 0.005 s − 0.003 s = 50 m/s The change in momentum is therefore ∆𝑝𝑝⃗ = 𝑚𝑚�𝑣𝑣 ⃗ 𝑓𝑓 − 𝑣𝑣 ⃗ 𝑖𝑖 � = (0.150 kg)(50 m/s + 75 m/s) = 19 N-s b. [3 pts] Knowing that the contact time between the baseball and the bat is 1 ms, calculate the magnitude of the average force exerted on the ball by the bat. Show your work. The impulse that the bat delivers to the ball equals the average force of the bat on the ball multiplied by the time interval over which that force is applied. We already know the impulse since the change in momentum was calculated in part a and, according to the impulse-momentum theorem, the impulse is equal to the change in momentum. The magnitude of the average force is therefore 𝐹𝐹 𝑎𝑎𝑎𝑎𝑎𝑎 = 19 N-s 0.001 𝑠𝑠 = 19000 N

Physics 114, Final Exam 32 g = 9.8 m/s 2 k B = R/N A = 1.38 x 10 -23 J/K 1 in. = 2.54 cm R = 8.31 J/mol-K 1 ft. = 0.305 m N A = 6.02 x 10 23 mol -1 1 mi. = 1609 m 1 atm = 1.01 x 10 5 Pa �𝑆𝑆 ⃗ � = 𝑘𝑘 | 𝑥𝑥 ⃗ | r C circle π 2 =