Practice exam 1 PHYS1121_ Attempt review

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19/08/2021 Practice exam 1 PHYS1121: Attempt review https://moodle.telt.unsw.edu.au/mod/quiz/review.php?attempt=10071486&cmid=3806523 1/15 Dashboard  My courses  PHYS1121-PHYS1131-5216_00399  Quizzes  Practice exam 1 PHYS1121 Started on Wednesday, 18 August 2021, 3:33 PM State Finished Completed on Wednesday, 18 August 2021, 5:07 PM Time taken 1 hour 34 mins Marks 3.00/24.00 Grade 12.50 out of 100.00  Site Home  Announcements   User Guides School of Physics  COVID-19  Wellbeing & Health  Counselling PHYS1121-PHYS1131-PHYS1121-PHYS1131-Physics 1A-Higher Physics 1A - …  

19/08/2021 Practice exam 1 PHYS1121: Attempt review https://moodle.telt.unsw.edu.au/mod/quiz/review.php?attempt=10071486&cmid=3806523 2/15 Question 1 Partially correct Mark 2.00 out of 3.00 Question: A nerf gun is a toy that can shoot foam bullets. To use it, a bullet is loaded by pulling back on a spring. A button is then pressed that will release the foam bullet. The bullet has a mass of . When the nerf gun is fired horizontally from a height above the ground, the bullet lands on the ground horizontally from where it was fired. Assume that air resistance is negligible. Figure 1: A photo of a nerf gun and foam bullet. Part 1) How long does it take the bullet to reach the ground from when it leaves the nerf gun? 0.469 Your last answer was interpreted as follows: Part 2) What is the initial speed of the bullet as it leaves the nerf gun? 23.4 Your last answer was interpreted as follows: Part 3) What is the kinetic energy of the bullet as it leaves the nerf gun? 51.3 Your last answer was interpreted as follows: Your answer is partially correct. Feedback for Part 1) You have correctly answered this part. Feedback for Part 2) You have correctly answered this part. Feedback for Part 3) Kinetic energy is given by: .

19/08/2021 Practice exam 1 PHYS1121: Attempt review https://moodle.telt.unsw.edu.au/mod/quiz/review.php?attempt=10071486&cmid=3806523 3/15 Substitute into this equation, make sure you convert the mass of the bullet from to by dividing by 1000 when you substitute it in. Note: This answer should be correct to 3 significant figures. A correct answer is , which can be typed in as follows: 0.470 A correct answer is , which can be typed in as follows: 23.5 A correct answer is , which can be typed in as follows: 0.0516

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19/08/2021 Practice exam 1 PHYS1121: Attempt review https://moodle.telt.unsw.edu.au/mod/quiz/review.php?attempt=10071486&cmid=3806523 4/15 Question 2 Incorrect Mark 0.00 out of 3.00 Question: A mass is pushed (a quick push is given to start it moving, in the question consider the motion after this push has stopped) down a long plane inclined at an angle of The initial speed of the block down the plane is down the plane, after it has traveled 1.00 m the speed has decreased to . Figure 1: Diagrammatic representation of a block of mass sliding down an inclined plane with initial velocity . The plane is inclined at degrees to the horizontal. Part 1) What is the acceleration of the block up the plane (i.e. positive answer if acceleration is up the plane, negative answer if it is down the plane). 3.79 up the plane Your last answer was interpreted as follows: Part 2) What is the net force on the block as it slides down the inclined plane? 21.6 up the plane Your last answer was interpreted as follows: Part 3) What is the coefficient of kinetic friction between the block and the plane? 0.386 Your last answer was interpreted as follows: Incorrect answer. Note: Your answers should be correct to 3 significant figures. Feedback for Part 1) Use the kinematic equation, . Take down the plane as negative and up the plane as positive. Use , and . Rearrange and solve for . Feedback for Part 2) Use Newton's second law: The acceleration was found in the first part and the mass of the block is . Feedback for Part 3)

19/08/2021 Practice exam 1 PHYS1121: Attempt review https://moodle.telt.unsw.edu.au/mod/quiz/review.php?attempt=10071486&cmid=3806523 5/15 The forces acting on the block parallel to the plane are the weight force (down the plane) and the frictional force (up the plane). Newton's second law tells us that the net force is the sum of these so: This can be re-arranged to give : Evaluate this. Feedback for Part 3) The forces acting on the block parallel to the plane are the weight force (down the plane) and the frictional force (up the plane). Newton's second law tells us that the net force is the sum of these so: This can be re-arranged to give : Evaluate this. A correct answer is , which can be typed in as follows: 2.36 A correct answer is , which can be typed in as follows: 4.84 A correct answer is , which can be typed in as follows: 0.644

19/08/2021 Practice exam 1 PHYS1121: Attempt review https://moodle.telt.unsw.edu.au/mod/quiz/review.php?attempt=10071486&cmid=3806523 6/15 Question 3 Incorrect Mark 0.00 out of 3.00 Question: Two masses and move towards each other on a frictionless track. has speed to the right while has a speed to the left ( Figure 1 ). They collide. The collision lasts s. After the collision moves to the right with a speed Figure 1: Diagrammatic representation of two masses, and moving towards each other on a frictionless track at speeds of to the right and to the left, respectively. Part 1) Before the collision what is the velocity of the centre of mass of the system? 0.28 Your last answer was interpreted as follows: Note: Give a negative answer if it is to the left. Part 2) What is the final velocity of 1.51 Your last answer was interpreted as follows: Note: Give a negative answer if it is to the left. Part 3) What is the average force on during the collision? 2.14 Your last answer was interpreted as follows: Note: Give a negative answer if it is to the left. Incorrect answer. Note: Your answers should be correct to 3 significant figures. Feedback for Part 1) The velocity of the centre of mass is given by: Substitute into this equation, except is negative as it is to the left. Feedback for Part 2) Momentum is conserved so: This can be rearranged to find In this case, is negative as it is to the left. Feedback for Part 3) The impulse is equal to the change in momentum so we can write:

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19/08/2021 Practice exam 1 PHYS1121: Attempt review https://moodle.telt.unsw.edu.au/mod/quiz/review.php?attempt=10071486&cmid=3806523 7/15 This can be rearranged to give the average force: Where is negative as it is to the left. Evaluate this. A correct answer is , which can be typed in as follows: - 0.0464 A correct answer is , which can be typed in as follows: - 1.95 A correct answer is , which can be typed in as follows: 327

19/08/2021 Practice exam 1 PHYS1121: Attempt review https://moodle.telt.unsw.edu.au/mod/quiz/review.php?attempt=10071486&cmid=3806523 8/15 Question 4 Incorrect Mark 0.00 out of 3.00 Question: In an experiment very similar to the one you saw/conducted, a hanging mass with g is released from rest (copy this link into your browser if you want to see a video of this: https://youtu.be/PD7RVOnL0CY). It is connected via a couple of very light pulleys to a very light rod with two masses, each with mass and a distance from the pivot point. The radius of the pulley supporting the masses is . These are shown in the diagram below. Figure 1: is a photo of the apparatus used in the lab, is a schematic diagram looking from above showing and , is a schematic diagram looking side on showing and . Part 1) Calculate the moment of inertia of the two masses on the rod. 0.424 Your last answer was interpreted as follows: Part 2) Using any variables you require from the list: write an expression for the magnitude of the tension in the string connecting the hanging mass to the device. In this list is the linear acceleration of the hanging mass in ms and is the acceleration due to gravity in ms . The other variables are shown in the diagram, and are in g (grams) and and are in cm. m(g-a) N Your last answer was interpreted as follows: The variables found in your answer were: Part 3) When the hanging mass is released it is observed to fall with an angular acceleration of . What is the acceleration of the hanging mass? 0 Your last answer was interpreted as follows: Incorrect answer. Feedback for Part 1) The moment of inertia of a point mass is given by . Both point masses contribute equally so the total moment of inertia is given by . Evaluate this. Remember to convert from g to kg and from cm to m. Feedback for Part 2) You have incorrectly answered this part. Consider the forces acting on the hanging mass, there is the weight force, pulling it down and the tension, , pulling it up. Newton's second law tells us that the net force is equal to so:

19/08/2021 Practice exam 1 PHYS1121: Attempt review https://moodle.telt.unsw.edu.au/mod/quiz/review.php?attempt=10071486&cmid=3806523 9/15 The mass needs to be converted from g to kg by dividing this by 1000. Feedback for Part 3) The angular acceleration and linear acceleration are related through . Multiply the angular acceleration by , remembering to convert from cm to m first. A correct answer is , which can be typed in as follows: 0.0166 A correct answer is , which can be typed in as follows: ( M * g - M * a ) / 1000 A correct answer is , which can be typed in as follows: 0.476

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19/08/2021 Practice exam 1 PHYS1121: Attempt review https://moodle.telt.unsw.edu.au/mod/quiz/review.php?attempt=10071486&cmid=3806523 10/15 Question 5 Partially correct Mark 1.00 out of 3.00 Question: There are moles of an ideal monatomic gas that undergo a temperature change from to . Part 1) What is the change in internal energy of the gas during this temperature change? 13590 Your last answer was interpreted as follows: Part 2) If the process takes place at a constant volume, what is the value of the molar specific heat at constant volume, ? 3/2*R Your last answer was interpreted as follows: The variables found in your answer were: Part 3) What is the ratio of the specific heats for constant volume and pressure, ? 0 Your last answer was interpreted as follows: Your answer is partially correct. Feedback for Part 1) You have correctly answered this part. Feedback for Part 2) The specific heat for constant volume, , is given by: where is the degrees of freedom ( for a monatomic gas), and is the gas constant . Feedback for Part 3) The ratio of specific heats, , is given by: where is the gas constant and you know from Part 2). A correct answer is , which can be typed in as follows: 13590 A correct answer is , which can be typed in as follows: 12.47 A correct answer is , which can be typed in as follows: 1.667

19/08/2021 Practice exam 1 PHYS1121: Attempt review https://moodle.telt.unsw.edu.au/mod/quiz/review.php?attempt=10071486&cmid=3806523 11/15 Question 6 Incorrect Mark 0.00 out of 3.00 Question: A certain ideal gas is at and has an atomic mass of (atomic mass unit). Part 1) What is the average kinetic energy of a molecule of the gas? 7.80 Your last answer was interpreted as follows: Part 2) What is the root-mean-square, or rms speed, of the gas particles? 1.37*10^14 Your last answer was interpreted as follows: Part 3) How many molecules are in the gas if the gas pressure is and the volume of the gas is . 0 Your last answer was interpreted as follows: Incorrect answer. Feedback for Part 1) The average kinetic energy of the gas is given by: where is the Boltzmann constant, and is the temperature of the gas, which you've been given in the question. Feedback for Part 2) You can use the equation for the kinetic energy of a particle Use the value from Part 1 and solve for . Remember to convert mass into SI units. Feedback for Part 3) The number of molecules in a gas is given by where is the gas pressure, is the gas volume, , and is the temperature. Making sure you are working in SI units, rearrange for and solve. A correct answer is , which can be typed in as follows: 1.201e-20 A correct answer is , which can be typed in as follows: 394.8 A correct answer is , which can be typed in as follows: 1.526E24

19/08/2021 Practice exam 1 PHYS1121: Attempt review https://moodle.telt.unsw.edu.au/mod/quiz/review.php?attempt=10071486&cmid=3806523 12/15 Question 7 Incorrect Mark 0.00 out of 3.00 Question: Sound waves travel through a didgeridoo. You may approximate the shape of the didgeridoo to be cylindrical. The intensity of the sound waves at the end of the didgeridoo is and the power of the sound wave produced by the musician is . Assume all power is directed through the didgeridoo. Part 1) What is the cross sectional area of the didgeridoo? 0 Your last answer was interpreted as follows: Part 2) The displacement amplitude of the sound waves at the end of the tube can be described by: where is the the position of a small element relative to its equilibrium position in mm, is in metres, and is in seconds. What is the density of the medium that the sound wave is propagating through? 9 Your last answer was interpreted as follows: Part 3) What sound level, in dB, is the sound wave at the end of the didgeridoo? 0 dB Your last answer was interpreted as follows: Incorrect answer. Feedback for Part 1) The cross sectional area can be calculated using: where is the intensity of the wave (given in the question), is the power of the source (also given), and is the area. Feedback for Part 2) The displacement amplitude can be found from the following relation: where is the intensity of the sound wave, is the density of the medium that the sound wave propagates through, is the velocity of the sound wave, is the angular frequency, is the wave number, and is the amplitude of the displacement function. The displacement function given in the question is in the form: which you can use to find the values of , , and . Feedback for Part 3) The sound level in decibels is given by: where is the intensity given in the question, and is the reference intensity. A correct answer is , which can be typed in as follows: 14

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19/08/2021 Practice exam 1 PHYS1121: Attempt review https://moodle.telt.unsw.edu.au/mod/quiz/review.php?attempt=10071486&cmid=3806523 13/15 A correct answer is , which can be typed in as follows: 2.9 A correct answer is , which can be typed in as follows: 141.6

19/08/2021 Practice exam 1 PHYS1121: Attempt review https://moodle.telt.unsw.edu.au/mod/quiz/review.php?attempt=10071486&cmid=3806523 14/15 Question 8 Incorrect Mark 0.00 out of 3.00 Question: Figure 1: Sebastian Vettel's Ferrari SF90 Formula One racing car. Formula One cars travel at very high speeds. The engine creates a sound at multiple frequencies, though there is one particular frequency that dominates the other frequencies, and that is the frequency that we hear. As the cars travel towards and away from you (such as when you are watching the cars from the side of the track), the frequency you hear changes. This is due to the Doppler effect. Part 1) You are watching your favourite Formula One race, and because you are a keen physics student, you measure the frequency of the sound from the car. You observe that the car travels at a constant speed while you make the measurements. When the car travels towards you, from your position at the side of the track, you measure the frequency to be . As it travels away from you, you measure the frequency to be . What is the velocity of the car? Take the speed of sound in air to be . 0 Your last answer was interpreted as follows: Part 2) If the car were not moving, what would be the frequency you would hear from the engine? 0 Your last answer was interpreted as follows: Part 3) The driver in the car described in Part 1) is behind another car that is travelling at a speed . If the driver in Part 1) could hear the sound from the car's engine in front, what frequency would they hear as they drive? Assume that both engines, when at rest, produce sound at the same frequency. 9 Your last answer was interpreted as follows: Incorrect answer. Feedback for Part 1) The general Doppler effect equation is given by: where is the frequency detected by the observer, is the frequency produced by the source, is the speed of the wave in the medium (in this case, air, 343 m/s), is the frequency that the observer is travelling, and is the velocity of the source. Recall that the symbols are set according to the direction that the observer and source are moving.

19/08/2021 Practice exam 1 PHYS1121: Attempt review https://moodle.telt.unsw.edu.au/mod/quiz/review.php?attempt=10071486&cmid=3806523 15/15 In this part, we do not know what , the frequency of the engine, is. However, we do know that since we, the observer, are not moving. We are trying to solve for the speed of the car, . When the car is approaching us, the frequency we detect (given in the question) will be given by: When the car is receding, the frequency we detect (given in the question) will be given by: We can combine these two equations by allowing one to equal , and rearrange and solve to find in km/hr. Feedback for Part 2) Using our answer to Part 1), we can use the Doppler equation to solve for , recalling that when the car is approaching us, the frequency we detect (given in the question) will be given by: and when the car is receding, the frequency we detect (given in the question) will be given by: Choose one of these to substitute into and solve for . Feedback for Part 3) In this case, the driver is the observer, and their velocity is nonzero. The source is the other car. Our driver is approaching the source, and the source is moving away from the observer. Therefore, the equation to describe this is: where is what we calculated in Part 2), is what we calculated in Part 1), is the speed of the other driver, and is what we are trying to solve for. A correct answer is , which can be typed in as follows: 353.56 A correct answer is , which can be typed in as follows: 734.43 A correct answer is , which can be typed in as follows: 762.56

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