PHYS172_S24-Exam2-PRACTICE-SOLUTIONS-TWEAKS

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PHYS172 EXAM 2: PRACTICE [150 Points] SOLUTIONS & TWEAKS SPRING 2024 ALL PROBLEMS ARE WORTH 10 POINTS Problem 01 : A space craft, mass 100 𝑘𝑔 is in outer space at location < 1, 2, 5 > 𝑚 moving at an initial speed of 2 𝑚/? One set of thrusters are now turned on and exert a constant force < −100, 200, 0 > ? until the object reaches location < 2, 3, 5 > 𝑚 . At this point, these thrusters are turned off, and another set of thrusters are turned on which exert a constant force < 200 − 100, 0 > ? until the space craft reaches < 5, 4, 5 > 𝑚 . What is the speed of the spacecraft at the final location? A) 4.00 m/s B) 2.00 m/s C) 8.00 m/s D) 16.0 m/s E) None of the above HOW TO SOLVE: Energy Principle: ∆? ??? = 𝑊 ???? → ∆? ??? = 𝑊 ???? (No ∆? ??? ) → 1 2 𝑚𝑣 ? 2 − 1 2 𝑚𝑣 𝑖 2 = ? ⃗ ??? ∙ ∆?⃗ → 1 2 𝑚(𝑣 ? 2 − 𝑣 𝑖 2 ) = 〈? ? , ? ? , ? ? 〉 ∙ 〈? 2? − ? 1? , ? 2? − ? 1? , ? 2? − ? 1? 〉 → 1 2 𝑚(𝑣 ? 2 − 𝑣 𝑖 2 ) = ? ? (? 2? − ? 1? ) + ? ? (? 2? − ? 1? ) + ? ? (? 2? − ? 1? ) Everything is given except, 𝑣 𝑖 . Solve for 𝑣 𝑖 HOW TO TWEAK: • What is given and what is unknown may be changed e.g. instead of giving final speed and asking for initial speed, the opposite may be the case i.e. initial speed is given and you are asked final speed. • You can be told that the push is in the same (or opposite) direction as the motion, and the magnitudes of the displacement and forces are given, along with mass and initial(or final) speed, and you are asked final (or initial) speed. Here simply use: 1 2 𝑚𝑣 ? 2 − 1 2 𝑚𝑣 𝑖 2 = ±?𝑑 where + is used when the force is in same direction as displacement, and – is used when F is in opposite direction to displacement.

PHYS172 EXAM 2: PRACTICE [150 Points] SOLUTIONS & TWEAKS SPRING 2024 ALL PROBLEMS ARE WORTH 10 POINTS Problem 02 : Three protons, each of mass ? and charge +? , are initially held at the corners of a triangle that is ? on a side. They are then released from rest. Which of the following equations will you need to solve to determine the speed 𝑣 of each proton, when the protons are very far apart? A) 3 ( 1 2 𝑚 𝑣 2 ) = 3 ( 1 4𝜋 𝜀 0 ? 2 ? ) B) 3 ( 1 2 𝑚𝑣 2 ) = 2 ( 1 4𝜋𝜀 0 ? 2 ? ) C) ( 1 2 𝑚𝑣 2 ) = 3 ( 1 4𝜋𝜀 0 ? 2 ? ) D) 2 ( 1 2 𝑚𝑣 2 ) = 3 ( 1 4𝜋𝜀 0 ? 2 ? ) E) None of the above HOW TO SOLVE: In this situation, the only contributions to the system’s energy that change from the initial state to the final state are th e protons’ kinetic energy and their electric potential energy among each pair combination of protons. ∆? ??? = 𝑊 ???? = 0 ? → ∆? ??? = ∆? ??????? + ∆? 1,2 + ∆? 1,3 + ∆? 2,3 = 0 ? In the initial state Protons are initially at rest, so no kinetic energy: → ? 𝑖 = 0 In the final state Protons are extremely far part, so no potential energy because as , ? → ∞, ? ???? → 0 due to ?~ 1 ? Therefore, the protons only have PE in the initial state, and KE in the final state. So: ? ?,???? = ? 𝑖,??? Suppose the Kinetic Energy of each proton in the final state is ? ? 3? ? = ? 1,2 + ? 1,3 + ? 2,3 → 3? ? = 1 4𝜋𝜖 𝑜 ? 1,2 2 ? + 1 4𝜋𝜖 𝑜 ? 1,3 2 ? + 1 4𝜋𝜖 𝑜 ? 2,3 2 ? → 3 ( 1 2 𝑚𝑣 2 ) = 3 ( 1 4𝜋𝜖 𝑜 ? 2 ? ) HOW TO TWEAK: • The number of charges and their arrangement can be changed e.g. two charges, four charges, more charges. • The point is that for them to fly off at the same speed, they must be symmetrically arranged. • You must remember to take into account all possible pairs of interactions.

PHYS172 EXAM 2: PRACTICE [150 Points] SOLUTIONS & TWEAKS SPRING 2024 ALL PROBLEMS ARE WORTH 10 POINTS Problem 03 : Starting from rest, a woman lifts a barbell of mass ? with a constant force of ? Newtons through a distance of ? meters, at which point she is still lifting, and the barbell has acquired a speed of 𝒗 m/s. Let ? ????? stand for the internal energy of the woman. For the system consisting of the woman + barbell + Earth , which equation below is the correct application of the energy principle from the initial state with the barbell at rest to the final state with the barbell having moved upward a distance ? and has a speed 𝒗 ? A) ∆? ????? + ( 1 2 𝑚𝑣 2 − 0) + (𝑚𝑔ℎ − 0) = ?ℎ B) ∆ ? ????? + ( 1 2 𝑚 𝑣 2 − 0) + ( 𝑚𝑔ℎ − 0) = 0 C) ∆? ????? + (0 − 1 2 𝑚𝑣 2 ) + (0 − 𝑚𝑔ℎ) = 0 D) ∆? ????? + (0 − 1 2 𝑚𝑣 2 ) + (0 − 𝑚𝑔ℎ) = ?ℎ E) None of the above HOW TO SOLVE: System : Woman + Barbell + Earth Surroundings: Nothing ∆? ??? = 𝑊 ???? → ∆? ??? + ∆? ??? + ∆? 𝑖??,??? = 0 → (? ? − ? 𝑖 ) + (? ? − ? 𝑖 ) + ∆? ????? + ∆? ??????? + ∆? ????ℎ = 0 In the above equation: ? ? = 1 2 𝑚𝑣 2 ? 𝑖 = 0 ? ? = 𝑚𝑔ℎ ? 𝑖 = 0 ∆? ??????? = 0 ∆? ????ℎ = 0 Substituting we get: ∆? ????? + ( 1 2 𝑚𝑣 2 − 0) + (𝑚𝑔ℎ − 0) = 0 HOW TO TWEAK: • What if the system and surrounding choices were different i.e. woman + barbell in system, earth in surroundings, or barbell in system, earth + woman in surroundings?

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PHYS172 EXAM 2: PRACTICE [150 Points] SOLUTIONS & TWEAKS SPRING 2024 ALL PROBLEMS ARE WORTH 10 POINTS Problem 04 : A block of mass M A and specific heat C A has an initial temperature of T A . It is brought into thermal contact with a second block of mass M B and specific heat C B at temperature T B . Which equation should you solve to find the final equilibrium temperature T ? A) ? ? 𝐶 ? ( ? − ? ? ) + ? ? 𝐶 ? ( ? − ? ? ) = 0 B) ? ? 𝐶 ? (? − ? ? ) + ? ? 𝐶 ? (? ? − ?) = 0 C) ? ? 𝐶 ? (? − ? ? ) + ? ? 𝐶 ? (? ? − ?) = 0 D) ? ? 𝐶 ? (? − ? ? ) + ? ? 𝐶 ? (? − ? ? ) = 0 E) None of the above HOW TO SOLVE: System : Both Blocks Surroundings: Nothing ∆? ??? = 𝑊 ???? + ? → ∆? ??? = 0 (No heat exchanged with surroundings) → ∆? 𝑖??,? + ∆? 𝑖??,? = 0 Use the fact that ∆? 𝑖?? = ?𝐶∆? : Change in internal energy: → ? ? 𝐶 ? (∆?) ? + ? ? 𝐶 ? (∆?) ? = 0 → ? ? 𝐶 ? (? − ? ? ) + ? ? 𝐶 ? (? − ? ? ) = 0 (because ∆? = ? ? − ? 𝑖 for each block. HOW TO TWEAK: • You could be asked to apply this equation to calculate actual final temperature of a system when everything else is given. • You could be asked to manipulate the equation to find the ratio of the specific heats, when everything else is given. • You could be asked to manipulate the equation to find the ratio of the masses, when everything else is given. • You could be asked to manipulate the equation to find the ratio of the changes in temperatures, when everything else is given.

PHYS172 EXAM 2: PRACTICE [150 Points] SOLUTIONS & TWEAKS SPRING 2024 ALL PROBLEMS ARE WORTH 10 POINTS Problems 05-07: A spring of stiffness, ? ? and un-stretched length, ? 𝟎 is placed vertically on a horizontal surface. A block of mass ? is initially held at rest pressing down on the top of the spring, such that the compressed length of the spring is ? . The block is in contact with, but not attached to the spring, so when you release the block, the spring pushes it up all the way to the top of the top of the un-stretched length of the spring, at which point it is launched into the air.

PHYS172 EXAM 2: PRACTICE [150 Points] SOLUTIONS & TWEAKS SPRING 2024 ALL PROBLEMS ARE WORTH 10 POINTS Problem 05: Three students present partial claim, evidence, and reasoning arguments about how to find the speed vo at which the block is launched into the air at the top of the spring. Which is correct? Amy: Claim : I chose the block as the system and Earth and spring as the surroundings. Evidence : Based on the situation described in the problem, I was told only the block is interacting with the spring and Earth. Reasoning : Therefore, I would have a change in kinetic energy from the block, work from Earth, and work from the spring. Beth: Claim : I chose the block and spring as the system and Earth as the surroundings. Evidence : Based on the situation described in the problem, I was told only the block and spring interact with the Earth. Reasoning : Therefore, I would have a change in kinetic energy from the block, a change in potential energy of the spring, and work from Earth Cathy: Claim : I chose the block, spring, and Earth in the system, and nothing in the surroundings. Evidence : Based on the situation described in the problem, I was told only about the block, spring, and Earth. Reasoning : Therefore, I would have a change in kinetic energy from the block, work from the spring, and a change in potential energy from Earth. A) Amy is correct, but Beth and Cathy are wrong. B) Beth is correct, but Amy and Cathy are wrong. C) Cathy is correct, but Amy and Beth are wrong. D) Both Amy and Beth are correct, Cathy is wrong. E) None of them is correct. HOW TO SOLVE: Examine each of these statements. • If an object is in the surroundings , then the work done by it must be on the right-hand side . E.g. In Amy’s case Earth and Spring are on RHS. In Beth’s case Earth is on RHS. In Cathy’s case, nothing should be on RHS, but that’s not the case so Cathy’s equation is NOT consistent with her surroundings. • If an object is in the system , then the potential energy due to the object should be on the left-hand side . E.g. In Amy’s case only the block is in the system, so the only term on left hand side should be change in KE. In Beth’s case, block and spring are in system, so KE of block and PE of spring are on LHS. In Cathy’s case change in KE of block, change in PE of spring and change in PE of earth-mass should be also on LHS, the last of which is not on the left hand side.

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PHYS172 EXAM 2: PRACTICE [150 Points] SOLUTIONS & TWEAKS SPRING 2024 ALL PROBLEMS ARE WORTH 10 POINTS Problem 06: What is the maximum height, ℎ ??? achieved by the block before it starts to come down again? A) ℎ ??? = ? 0 2 2? B) ℎ ??? = ? 0 2 ? C) ℎ ??? = 2? 0 2 ? D) ℎ ??? = 3? 0 2 ? E) None of the above HOW TO SOLVE: System: Block + Earth Surroundings: Nothing Use Energy Principle starting with mass at top of ramp (initial) to highest point in flight (final) ∆? ??? = 𝑊 ???? → ∆? ??? + ∆? ??? + ∆? 𝑖?? = 0 → (? ? − ? 𝑖 ) + (? ? − ? 𝑖 ) + 0 = 0 In the above equation: ? ? = 0 [Because the block is at rest at the highest position] ? ? = 1 2 𝑚𝑣 0 2 ? ? = 𝑚𝑔ℎ ??? ? 𝑖 = 0 [Because top of ramp is the reference] Substituting we get: (0 − 1 2 𝑚𝑣 0 2 ) + (𝑚𝑔ℎ ??? − 0) = 0 → 1 2 𝑚𝑣 0 2 = 𝑚𝑔ℎ ??? Use identity that 1 − ( cos 𝜃) 2 = ( sin 𝜃) 2 → ℎ ??? = 𝑣 0 2 2𝑔 HOW TO TWEAK: Instead of being launched vertically upward, what if the block were launched off an incline plane ? • Basically, the same concept, but we must only consider the VERTICAL component of the velocity when it leaves the ramp. i.e. replace 𝑣 0 in the above problem with 𝑣 0 sin 𝜃 • You are asked to calculate the maximum height reached by the block? Use Energy Principle: ∆? ??? = 𝑊 ???? → ∆? ??? + ∆? ??? + ∆? 𝑖?? = ? + 𝑊 ???? System: Block, Earth Surroundings: Nothing (no need to have spring as it does not exert any force once block leaves it) (0 − 1 2 𝑚(𝑣 0 sin𝜃) 2 ) + (𝑚𝑔ℎ ??? − 0) + 0 = 0 + 0 → Solve for ℎ ???

PHYS172 EXAM 2: PRACTICE [150 Points] SOLUTIONS & TWEAKS SPRING 2024 ALL PROBLEMS ARE WORTH 10 POINTS Problem 07: What is the time ? ??? after it is launched, for the block to reach the highest point on its flight? A) ? ??? = ? 0 ? B) ? ??? = 2? 0 ? C) ? ??? = ? 0 2? D) ? ??? = ? 0 4? E) None of the above HOW TO SOLVE: Use momentum principle. ∆𝑝⃗ ??? = ? ⃗ ??? ∆? Consider only the y-component of the momentum principle because we know that gravity acts in y-direction: ∆𝑝⃗ ???,? = ? ⃗ ???,? ∆? → (0 − 𝑚𝑣 0 ) = (−𝑚𝑔)∆? → ∆? = ? 0 ? HOW TO TWEAK: Instead of being launched vertically upward, what if the block were launched off an inclined plane? • Basically, the same concept, but we must only consider the VERTICAL component of the velocity when it leaves the ramp. i.e. replace 𝑣 0 in the above problem with 𝑣 0 sin 𝜃 • You are asked to calculate the maximum height reached by the block? Use Mom. Principle: ∆𝑝⃗ ???,? = ? ⃗ ???,? ∆? ∆𝑝⃗ ???,? = ? ⃗ ???,? ∆? → (0 − 𝑚𝑣 0 sin 𝜃) = (−𝑚𝑔)∆? → ∆? = ? 0 sin 𝜃 ?

PHYS172 EXAM 2: PRACTICE [150 Points] SOLUTIONS & TWEAKS SPRING 2024 ALL PROBLEMS ARE WORTH 10 POINTS Problem 08-09: Your friend, mass ? jumps up from a crouched position to a vertical standing position, such that their center of mass moves vertically upward through height ? . During this process, the point of contact between your friend’s feet and the ground is stationary. Your friend experiences a normal force ? ? acting upward at the point of contact between the feet and the ground. Your friend’s change in internal energy in the process is ∆? ??? and their change in kinetic energy due to their torso moving relative to the center of mass is ∆? ??? Problem 08: Which of the following equations represents the energy principle of the point particle system as applied to your friend from the initial to the final positions shown above? A) ∆? ????? = ? ? ℎ B) ∆? ????? + ∆? 𝑖?? + ∆? ??? = −?𝑔ℎ C) ∆? ????? = ( ? ? − ?𝑔)ℎ D) ∆? ????? + ∆? 𝑖?? + ∆? ??? = (? ? − ?𝑔)ℎ E) None of the above HOW TO SOLVE: System: Person Surroundings: Earth EP Point Particle System: ∆? ????? = ? ⃗ ??? ∙ ∆?⃗ ?? In this case: ? ⃗ ??? = ? ⃗ ? + ? ⃗ ???? = 〈0, ? ? , 0〉 + 〈0, −?𝑔, 0〉 = 〈0, (? ? − ?𝑔), 0〉 and ∆?⃗ ?? = 〈0, ℎ, 0〉 So: EP of Point Particle system becomes: ∆? ????? = 〈0, (? ? − ?𝑔), 0〉 ∙ 〈0, ℎ, 0〉 → ∆? ????? = (? ? − ?𝑔)ℎ Problem 09: Which of the following equations represents the energy principle of the extended system as applied to your friend from the initial to the final positions shown above? A) ∆? ????? = ? ? ℎ B) ∆? ????? + ∆? 𝑖?? + ∆? ??? = −?𝑔ℎ C) ∆? ????? = (? ? − ?𝑔)ℎ D) ∆? ????? + ∆? 𝑖?? + ∆? ??? = (? ? − ?𝑔)ℎ E) None of the above HOW TO SOLVE: System: Person Surroundings: Earth EP Extended System: ∆? ????? + ∆? ??? + ∆? 𝑖?? = ∑ (? ⃗ 𝑖 ∙ ∆?⃗ 𝑖 ) ? 𝑖=1 Right hand Side has two forces: ∑ (? ⃗ 𝑖 ∙ ∆?⃗ 𝑖 ) ? 𝑖=1 = ? ⃗ ? ∙ ∆?⃗ ? + ? ⃗ ???? ∙ ∆?⃗ ???? = ? ⃗ ? ∙ 〈0,0,0〉 + 〈0, −?𝑔, 0〉 ∙ 〈0, ℎ, 0〉 = −?𝑔ℎ This is because.. (i) Point of application of normal force does not move, so ∆?⃗ ? = 〈0,0,0〉 (ii)Point of application of the gravitational force acts through CM, so ∆?⃗ ???? = ∆?⃗ ?? = 〈0, ℎ, 0〉 So, substituting back : ∆? ????? + ∆? ??? + ∆? 𝑖?? = −?𝑔ℎ HOW TO TWEAK: • What is the object were pushing off horizontally, instead of vertically. (See example in lecture) • Same idea, but we will NOT have force of gravity acting.

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PHYS172 EXAM 2: PRACTICE [150 Points] SOLUTIONS & TWEAKS SPRING 2024 ALL PROBLEMS ARE WORTH 10 POINTS Problem 10-11: You, mass ? hang by your hands from a tree limb that is a height ? above the ground, with your center of mass a height ? above the ground, and your feet a height ? above the ground. You then let yourself fall. You absorb the shock by bending your knees, ending up momentarily at rest in a crouched position with your center of mass a height 𝒃 above the ground. Problem 10: What is your kinetic energy at the instant your feet first touch the ground? A) ?𝑔ℎ B) ?𝑔(ℎ − 𝑑) C) ?𝑔𝑑 D) ?𝑔(ℎ − 𝑑 − 𝑏) E) ?𝑔(ℎ − 𝑏) HOW TO SOLVE: Point Particle System from initial state (hanging) to intermediate state (just before it hits ground) [not shown in fig.] ∆? ????? = ? ⃗ ??? ∙ ∆?⃗ ?? Here : ∆? ????? = ? ? − ? 𝑖 = ? ? − 0 = ? ? (b/c initially at rest) ? ⃗ ??? = 〈0, −?𝑔, 0〉 ∆?⃗ ?? = ?⃗ ? − ?⃗ 𝑖 = 〈0, ℎ − 𝑑, 0〉 − 〈0, ℎ, 0〉 = 〈0, −𝑑, 0〉 [Note: ?⃗ ? = 〈0, ℎ − 𝑑, 0〉 b/c when his feet hit the ground, his CM has moved distance 𝑑 downward from its initial height ℎ , so final height of CM = ℎ − 𝑑 ] Plugging into Point Particle equation: ? ? = 〈0, −?𝑔, 0〉 ∙ 〈0, −𝑑, 0〉 = ?𝑔𝑑 HOW TO TWEAK: Instead of a person it could be some other object dropping that gets squished when it hits the ground. The way you find the KE just before it hits the ground will be identical. Basically, it is based on how far the CM of the object has fallen. Problem 11: What is the change in internal energy in your body from the initial to final position? A) ?𝑔 (ℎ − 𝑏 ) B) −?𝑔(ℎ − 𝑏) C) −?𝑔(ℎ − 𝑑 − 𝑏) D) ?𝑔(ℎ − 𝑑 − 𝑏) E) ?𝑔(ℎ − 𝑑) HOW TO SOLVE: Extended System from initial state (hanging) and final state (crouched position) [shown in fig.] ∆? ????? + ∆? ??? + ∆? 𝑖?? = (? ⃗ 1 ∙ ∆?⃗ 1 ) + (? ⃗ 2 ∙ ∆?⃗ 2 ) + ⋯ Here : ∆? ????? = ? ? − ? 𝑖 = 0 − 0 = 0 (b/c initially at rest which hanging, finally also at rest after crouching) ∆? ??? = 0 ? ⃗ 1 = 〈0, −?𝑔, 0〉 ∆?⃗ ?? = ?⃗ ? − ?⃗ 𝑖 = 〈0, 𝑏, 0〉 − 〈0, ℎ, 0〉 = 〈0, 𝑏 − ℎ, 0〉 ? ⃗ 1 = 〈0, ? ???? , 0〉 ∆?⃗ 2 = 〈0,0,0〉 [Feet experience force from ground, but do not move] Plugging into Extended System equation0 +0 + ∆? 𝑖?? = −?𝑔(𝑏 − ℎ) + 0 → ∆? 𝑖?? = ?𝑔(ℎ − 𝑏) HOW TO TWEAK: Instead of a person dropping it could be some other object dropping that gets squished when it hits the ground. The way you find the ∆? 𝑖?? is just as we did above. Very Important to keep in mind here is that NOT to use ∆? ????? from Prob. 10 in Prob. 11. This is because the final states for Prob. 10 and Prob. 11 are different. In Prob. 10, final state is just before it hits ground. In Prob. 11 final state is when it has come to rest on the ground.

PHYS172 EXAM 2: PRACTICE [150 Points] SOLUTIONS & TWEAKS SPRING 2024 ALL PROBLEMS ARE WORTH 10 POINTS Problem 12 : A rod of length ? and negligible mass is attached to a uniform disk of mass ? and radius 𝑹 . A string is wrapped around the disk, and you pull on the string with a constant magnitude force ? . Two small balls each of mass m slide along the rod with negligible friction. The apparatus starts from rest, and when the center of the disk has moved a distance ? , a length of string s has come off the disk, and the balls have collided with the ends of the rod and stuck there. The apparatus slides on a nearly frictionless table. What is the angular speed , 𝝎 of the disk at this instant? Note: The moment of inertia of a disk is ? = 1 2 ?? 2 . A) 𝜔 = √ 2𝐹? 2?? 2 +?𝑅 2 B) 𝜔 = √ 2𝐹? ?? 2 +?𝑅 2 C) 𝜔 = √ 𝐹? ?? 2 +?𝑅 2 D) 𝜔 = √ 2𝐹? 4?? 2 +?𝑅 2 E) 𝜔 = √ 4𝐹? ? ? 2 +? 𝑅 2 HOW TO SOLVE: This is one of the vast set of problems which follow a similar procedure. Step 1: EP for Point Particle System: ∆? ????? = ? ⃗ ??? ∙ ∆?⃗ ?? → ∆? ????? = ?𝑑 Step 2: EP for Extended System: ∆? ????? + ∆? ??? + ∆? 𝑖?? = (? ⃗ 1 ∙ ∆?⃗ 1 ) + (? ⃗ 2 ∙ ∆?⃗ 2 ) + ⋯ → ∆? ????? + ( 1 2 ?𝜔 2 − 0) + 0 = ? ⃗ ∙ ∆?⃗ = ?(𝑑 + ?) (From figure above 𝑑 + ? is magnitude of displacement of point of application of force ) Get ∆? ????? from Step 1 above, and substitute → ?𝑑 + 1 2 ?𝜔 2 = ?(𝑑 + ?) → 𝜔 = √ 2𝐹? 𝐼 Now, substitute for ? = 1 2 ?? 2 + 2 (𝑚 ( ? 2 ) 2 ) Reason: Mom of Inertia = Disk + 2 Point Masses (distance ?/2 from center) So: 𝜔 = √ 2𝐹? ( 1 2 ?𝑅 2 + 𝑚𝐿 2 2 ) → 𝜔 = √ 4𝐹? ?𝑅 2 +?? 2 HOW TO TWEAK: • Object might have different moment of inertia. • Object might be moving vertically and also rotating (such as a Yo-Yo). See next problem.

PHYS172 EXAM 2: PRACTICE [150 Points] SOLUTIONS & TWEAKS SPRING 2024 ALL PROBLEMS ARE WORTH 10 POINTS Problem 13-14 Two identical masses on a frictionless surface are connected to each other with a spring. The initial and final configurations of the masses are shown in the figure. Initially, the spring is at its equilibrium length. The rightmost mass is pulled with a force ? = 100 ? to the right. Problem 13 : What is the change in translational kinetic energy of the system from the initial to the final position? A) 15.0 J B) 8.0 J C) 7.0 J D) 1.0 J E) None of the above HOW TO SOLVE: This is one of the vast set of problems which follow a similar procedure. Step 1: EP for Point Particle System: ∆? ????? = ? ⃗ ??? ∙ ∆?⃗ ?? → ∆? ????? = 〈100,0,0〉 ∙ 〈0.10 − 0.03,0,0〉 → ∆? ????? = 7 ? (See figure above to show how ∙ ∆?⃗ ?? = 〈0.07,0,0〉 ) HOW TO TWEAK: -- Similar problem, but no spring instead there is a rope connecting the two. -- Similar problem, but it is moving vertically rather than horizontally. Problem 14 : What is the change in internal energy of the system from the initial to the final position? A) 15.0 J B) 8.0 J C) 7.0 J D) 1.0 J E) None of the above HOW TO SOLVE: Step 2: EP for Extended System: ∆? ????? + ∆? ??? + ∆? 𝑖?? = ? ⃗ ∙ ∆?⃗ 𝐹 → ∆? ????? + 0 + ∆? 𝑖?? = ? ⃗ ∙ ∆?⃗ 𝐹 (See figure above to show how ∙ ∆?⃗ 𝐹 = 〈0.08,0,0〉 This is displacement of point of application of force ) Get ∆? ????? from Step 1 above, and substitute → 7? + 0 + ∆? 𝑖?? = 〈100,0,0〉 ∙ 〈0.08,0,0〉 → ∆? 𝑖?? = 1.0? HOW TO TWEAK: See above. ∆?⃗ ?? = 0.07𝑖Ƹ 0.10𝑚 0.03𝑚 ∆?⃗ 𝐹 = 0.08𝑖

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PHYS172 EXAM 2: PRACTICE [150 Points] SOLUTIONS & TWEAKS SPRING 2024 ALL PROBLEMS ARE WORTH 10 POINTS Problem 15 : Four objects of the same mass, ? are initially at rest and positioned as shown on an inclined plane of height ? . The “particle” is a flat, slippery object that can slide down the plane without friction. The other objects, all having the same radius, 𝑹 are a solid cylinder, a hoop and a solid sphere that roll down the plane without slipping. Note: ? ℎ??? = ?? 2 ? ???𝑖???? = 1 2 ?? 2 ? ??ℎ??? = 2 5 ?? 2 If they are all released at the same time, in what order (first to last) will the objects reach the finish line? A) particle, cylinder, hoop, sphere B) particle, sphere, cylinder, hoop C) particle, hoop, sphere, cylinder D) hoop, cylinder, sphere, particle E) sphere, cylinder, hoop, particle HOW TO SOLVE: The object with the smallest ? has the smallest ? ?????𝑖???? and therefore the largest ? ????? . → Because it has the greatest ? ????? it will achieve the greatest speed of center of mass 𝑣 ?? and reach the bottom first. In our list of objects: • The particle has the smallest (zero) moment of inertia. • The sphere has the next smallest (pre-factor = 2/5 = 0.40) • The cylinder has the next smallest (pre-factor = ½ = 0.50) • The hoop has the largest (pre-factor = 1) Hence the order in which they arrive at the bottom: HOW TO TWEAK: Rather than sliding down an incline, all objects could be pulled with the same horizontal force along a flat surface. The results would be the same as above: Particle has largest translational KE, hoop has smallest translational KE.

PHYS172_S24-Exam2-PRACTICE-SOLUTIONS-TWEAKS

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