Ps 113 - Quiz 2

12 flashcards · Physics · by Nicholas Enoul

  1. During a circus act, an elderly performer thrills the crowd by catching a cannon ball shot at him. The cannon ball has a mass of 67.0 kg and its horizontal component of velocity is 9.00 m/s when the 65.0 kg performer catches it. If the performer is on nearly frictionless roller skates, what is his recoil velocity in meters per second? #8.10 https://www.youtube.com/watch?v=cR92AwexbBg

    Answer: 4.57 m/s (an attempt at an explanation) This may seem daunting, but it is a super easy plug n' play question with one funky aspect. Here's the steps: We will have to use the following formula: \\ With this, we have our variables: M1 = 67 V1i = 9 M2 = 65 V2i = 0 <-- This is because we're assuming the performer is stationary! V2f and V1f = What we are trying to find. \\ With this first part done, we can now plug stuff in, which should look like: (67)(9) + (65)(0) = (67)V1f + (65)V2f \\ Th…

  2. During a circus act, an elderly performer thrills the crowd by catching a cannon ball shot at him. The cannon ball has a mass of 54.0 kg and its horizontal component of velocity is 7.50 m/s when the 65.0 kg performer catches it. If the performer is on nearly frictionless roller skates, what is his recoil velocity in meters per second? #8.10 https://www.youtube.com/watch?v=cR92AwexbBg

    Answer: 3.4 m/s (an attempt at an explanation) This may seem daunting, but it is a super easy plug n' play question with one funky aspect. Here are the steps: We will have to use the following formula: \\With this, we have our variables: M1 = 54 V1i = 7.5 M2 = 65 V2i = 0 <-- This is because we're assuming the performer is stationary! V2f and V1f = What we are trying to find. \\With this first part done, we can now plug stuff in, which should look like: (54)(7.5) + (65)(0) = (54)V1f + (65)V2f \\…

  3. Train cars are coupled together by being bumped into one another. Suppose two loaded train cars are moving toward one another, the first having a mass of 120,000 kg and a velocity of 0.300 m/s, and the second having a mass of 95,000 kg and a velocity of −0.120 m/s. (The minus indicates direction of motion.) What is their final velocity (in m/s)? #8.5 https://www.youtube.com/watch?v=UdN6nBXdpy0

    Answer: 0.1144 m/s (an attempt at an explanation) This may seem daunting, but it is a super easy plug n' play question with one funky aspect. Here's the steps: We will have to use the following formula: \\With this, we have our variables: M1 = 120,000 V1i = 0.3 M2 = 95,000 V2i = -0.12 V2f and V1f = What we are trying to find. \\With this first part done, we can now plug stuff in, which should look like: (120,000)(0.3) + (95,000)(-.12) = (120,000)V1f + (95,000)V2f \\ As the train cars are couple…

  4. Train cars are coupled together by being bumped into one another. Suppose two loaded train cars are moving toward one another, the first having a mass of 180,000 kg and a velocity of 0.300 m/s, and the second having a mass of 95,000 kg and a velocity of −0.120 m/s. (The minus indicates direction of motion.) What is their final velocity (in m/s)? #8.5 https://www.youtube.com/watch?v=UdN6nBXdpy0

    Answer: 0.155 m/s (an attempt at an explanation) This may seem daunting, but it is a super easy plug n' play question with one funky aspect. Here's the steps: We will have to use the following formula: \\With this, we have our variables: M1 = 180,000 V1i = 0.3 M2 = 95,000 V2i = -0.12 V2f and V1f = What we are trying to find. \\With this first part done, we can now plug stuff in, which should look like: (180,000)(0.3) + (95,000)(-.12) = (120,000)V1f + (95,000)V2f \\ As the train cars are coupled…

  5. Space probes may be separated from their launchers by exploding bolts. (They bolt away from one another.) Suppose a 6000 kg satellite uses this method to separate from the 1600 kg remains of its launcher, and that 8000 J of kinetic energy is supplied to the two parts. What are their subsequent velocities (in m/s) using the frame of reference in which they were at rest before separation? (Assume the final velocity of the satellite is in the positive direction. Indicate the direction with the sig…

    Velocity of Satellite = 0.749 Velocity of Launcher = -2.81 (an attempt at an explanation) This seems stupidly daunting, but let's break it down step by step. The first thing to do is to use the formula: which then transforms to: \\ As the initial velocity is zero, the whole left side is canceled. Leaving us with this: \\ The next step is to start labeling. This way, we can find out what we want. Mass and Velocity of Launcher = ML & VL Mass and Velocity of Satellite = VS & VS \\ With this, it sh…

  6. Space probes may be separated from their launchers by exploding bolts. (They bolt away from one another.) Suppose a 5200 kg satellite uses this method to separate from the 1200 kg remains of its launcher, and that 3000 J of kinetic energy is supplied to the two parts. What are their subsequent velocities (in m/s) using the frame of reference in which they were at rest before separation? (Assume the final velocity of the satellite is in the positive direction. Indicate the direction with the sig…

    Velocity of Satellite = 0.466 Velocity of Launcher = -2.015 (an attempt at an explanation) This seems stupidly daunting, but let's break it down step by step. The first thing to do is to use the formula: which then transforms to: \\ As the initial velocity is zero, the whole left side is canceled. Leaving us with this: \\ The next step is to start labeling. This way, we can find out what we want. Mass and Velocity of Launcher = ML & VL Mass and Velocity of Satellite = MS & VS \\ With this, it s…

  7. A 60.0 kg skier with an initial speed of 13.0 m/s coasts up a 2.50 m high rise as shown in the following figure. Question: Find her final speed at the top (in m/s), given that the coefficient of friction between her skis and the snow is 0.0800. (Hint: Find the distance traveled up the incline assuming a straight-line path as shown in the figure.) #7.6 NOTE: I needed to use brute-force memorization for parts of this question. They will be on the answers card if you want to know them. Video Guide…

    Answer: 10.7 (an attempt at an explanation) Please note that I needed to use brute-force memorization for two particular things: \\ It is highly unlikely that, at the very least, the first one would be on the equations sheet, as we were meant to use the base Wnc = FDcos(theta), but by God I could not explain how or why this gets from that to -muse bullcrap. PLEASE DO EXTRA RESEARCH if you want to guarantee a 100%. Now, let's actually start on the first thing we need to do. We will need to use t…

  8. A 65.0 kg skier with an initial speed of 12.0 m/s coasts up a 2.50 m high rise as shown in the following figure. Question: Find her final speed at the top (in m/s), given that the coefficient of friction between her skis and the snow is 0.0800. (Hint: Find the distance traveled up the incline assuming a straight-line path as shown in the figure.) #7.6 NOTE: I needed to use brute-force memorization for parts of this question. They will be on the answers card if you want to know them. Video Guide…

    Answer: 9.46 Margin of .2 (an attempt at an explanation) Please note that I needed to use brute-force memorization for two particular things: \\ It is highly unlikely that, at the very least, the first one would be on the equations sheet, as we were meant to use the base Wnc = FDcos(theta), but by God I could not explain how or why this gets from that to -muse bullcrap. PLEASE DO EXTRA RESEARCH if you want to guarantee a 100%. Now, let's actually start on the first thing we need to do. We will…

  9. (a) How long (in s) will it take an 875 kg car with a useful power output of 42.0 hp (1 hp = 746 W) to reach a speed of 13.0 m/s, neglecting friction? (Assume the car starts from rest.) (b) How long (in s) will this acceleration take if the car also climbs a 2.80 m-high hill in the process?

    (a) 2.36 (b) 3.13 (an attempt at an explanation) --- Solving for (a) --- \\The first formula we would want to use is this one: \\ The good news is that we have P and we have the t to solve. Bad news: we don't know W yet. It is super easy though, use this formula: \\ This looks bad, but the great news is that Ki (initial kinetic energy) is completely gone as we are starting from rest. This leaves the following formula: \\ With this, we can solve for W M = 875 V = 13 (1/2) *875*13^2 --> W = 73937…

  10. (a) How long (in s) will it take an 845 kg car with a useful power output of 44.0 hp (1 hp = 746 W) to reach a speed of 17.0 m/s, neglecting friction? (Assume the car starts from rest.) (b) How long (in s) will this acceleration take if the car also climbs a 2.60 m-high hill in the process? #7.9

    (a) 3.72 (b) 4.38 (an attempt at an explanation) --- Solving for (a) --- \\The first formula we would want to use is this one: \\ The good news is that we have P and we have the t to solve. Bad news: we don't know W yet. It is super easy though, use this formula: \\ This looks bad, but the great news is that Ki (initial kinetic energy) is completely gone as we are starting from rest. This leaves the following formula: \\ With this, we can solve for W M = 845 V = 17 (1/2) *845*17^2 --> W = 12210…

  11. (a) How high a hill can a car coast up (engine disengaged) if friction is negligible and its initial speed is 84.0 km/h? (b) If, in actuality, a 750 kg car with an initial speed of 84.0 km/h is observed to coast up a hill to a height 13.0 m above its starting point, how much thermal energy was generated by friction? (c) What is the average force of friction if the hill has a slope 2.5° above the horizontal? NOTE: I am not sure about this, but you may need to brute-force memorize the formula for…

    (a) 27.8 (b) 108616.67 or 1.09e+05 (c) 364 newtons (an attempt at an explanation) First thing to do: Convert the units we have to MKS. 84 km/h * 1000m/km * 1hr/3600 seconds 84 km/hr --> ~23.33 m/s --- Solving for (a) --- We will first need the following formulas or (a): \\ Please note that the final Kinetic force (Kf) and PEi are both 0. Why? Idk, I ain't paid to do this! However, with all of this info, we can make the following formula: \\ With this, we can now solve for Y, and fun fact, we do…

  12. (a) How high a hill can a car coast up (engine disengaged) if friction is negligible and its initial speed is 108.0 km/h? (b) If, in actuality, a 750 kg car with an initial speed of 108 km/h is observed to coast up a hill to a height 24.0 m above its starting point, how much thermal energy was generated by friction? (c) What is the average force of friction if the hill has a slope 2.5° above the horizontal? NOTE: I am not sure about this, but you may need to brute-force memorize the formula for…

    (a) 45.9 (b) 161100 or 1.61e+05 (c) 293 newtons (an attempt at an explanation) First thing to do: Convert the units we have to MKS. 108 km/h * 1000m/km * 1hr/3600 seconds 108 km/hr --> 30 m/s --- Solving for (a) --- We will first need the following formulas or (a): \\ Please note that the final Kinetic force (Kf) and PEi are both 0. Why? Idk, I ain't paid to do this! However, with all of this info, we can make the following formula: \\ With this, we can now solve for Y, and fun fact, we don't n…