problem set 9

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Physics

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Feb 20, 2024

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Problem Set #9- Conceptual Questions 2, 4 and problems and Exercises 4, 16, 24, 27, 33, 44 Conceptual Questions 2. By what factor does an object’s kinetic energy change if its speed is doubled? By what factor does its momentum change? If an object’s kinetic energy speed is doubled then it is changed by a factor of four; the momentum of the object also increases by a factor of 2. 4. A system of particles is known to have zero momentum. Does it follow that the kinetic energy of the system is also zero? Explain. If a collection of particles is determined to possess zero momentum, it does not necessarily imply that the system's kinetic energy is also zero. This is because the particles can exhibit zero momentum while still having kinetic energy if they are in motion. Problems 4, 16, 24, 27, 33, 44 4. A 26.2-kg dog is running northward at 2.70 m/s, while a 5.30-kg cat is running eastward at 3.04 m/s. Their 74.0-kg owner has the same momentum as the two pets taken together. Find the direction and magnitude of the owner’s velocity. (26.2 kg) x (2.70 m/s) + (5.30 kg) x (3.04 m/s) = (74.0 kg) x (velocity of owner) 86.852 kg x m/s = (74.0 kg) x (velocity of owner) velocity of owner = 1.173 m/s tan(theta) = (2.70 m/s) / (3.04 m/s) theta = 41.37 degrees So, theta is approximately 41.4 degrees. This means that the vector sum of the velocities of the dog and cat points 41.4 degrees north of east. Therefore, the direction of the owner's velocity is 41.4 degrees north of east, and the magnitude of the owner's velocity is 1.173 m/s.

16. When spiking a volleyball, a player changes the velocity of the ball from 4.2 m/s to −24 m/s; along a certain direction. If the impulse delivered to the ball by the player is −9.3 kg m/s, what is the mass of the volleyball? -9.3 kg x m/s = (mass of the volleyball) x (-24 m/s) - (mass of the volleyball) x (4.2 m/s) Now, factor out the mass of the volleyball from both terms: -9.3 kg x m/s = (mass of the volleyball) x (-24 m/s - 4.2 m/s) -9.3 kg x m/s = (mass of the volleyball) x (-28.2 m/s) mass of the volleyball = -9.3 kg x m/s / (-28.2 m/s) mass of the volleyball = 0.33 kg 24. A 0.042-kg pet lab mouse sits on a 0.35 kg air-track cart. The cart is at rest, as is a second cart with a mass of 0.25 kg. The lab mouse now jumps to the second cart. After the jump, the 0.35-kg cart has a speed of υ1=0.88m/s. given: A lab mouse with a mass of 0.042 kg is sitting on a 0.35 kg air-track cart. The initial state has both the 0.35 kg cart and a second cart with a mass of 0.25 kg at rest. The lab mouse jumps from the first cart to the second cart. After the jump, the first cart (0.35 kg) has a speed of υ1 = 0.88 m/s. (-.35kg) (-.88m/s) / (.25) + (.042kg) = 1.1m/s So, the speed of the second cart after the lab mouse jumps onto it is approximately 1.1m/s.

27. The recoil of a shotgun can be significant. Suppose a 3.3-kg shotgun is held tightly by an arm and shoulder with a combined mass of 15.0 kg. When the gun fires a projectile with a mass of 0.038 kg and a speed of 385 m/s, what is the recoil velocity of the shotgun and arm–shoulder combination? Final Momentum = (Mass of Shotgun + Mass of Arm-Shoulder) x V + (Mass of Projectile) x (Projectile Velocity) Final Momentum = (3.3 kg + 15.0 kg) x V + (0.038 kg) x (385 m/s) Final Momentum = 18.3 kg * V + 14.63 kg m/s The law of conservation of momentum states that the initial momentum is equal to the final momentum, so: 0 = 18.3 kg x V + 14.63 kg m/s Now, solve for V: 18.3 kg x V = -14.63 kg m/s V = (-14.63 kg m/s) / 18.3 kg V = -0.80 m/s The recoil velocity of the shotgun and arm-shoulder combination is approximately -0.80 m/s. The negative sign indicates that the combination moves in the opposite direction of the projectile. 33. The human head can be considered as a 3.3-kg cranium protecting a 1.5-kg brain, with a small amount of cerebrospinal fluid that allows the brain to move a little bit inside the cranium. Suppose a cranium at rest is subjected to a force of 2800 N for 6.5 m/s in the forward direction. (a) What is the final speed of the cranium? (b) The back of the cranium then collides with the back of the brain, which is still at rest, and the two move together. What is their final speed? (c) The cranium now hits an external object and suddenly comes to rest, but the brain continues to move forward. If the front of the brain interacts with the front of the cranium over a period of 15 m/s before coming to rest, what average force is exerted on the brain by the cranium? Given: F = force (2800 N) m = mass (3.3 kg, mass of the cranium) (a) we would use newtons 2 nd law to find the final speed of the cranium when subjected to a force of 2800 N for 6.5 s (2800N) (6.5m/s x 1s/1000ms) / (3.3kg) = 5.5m/s (b) (3.3kg/3.3kg + 1.5kg) x (5.5m/s) = 3.8m/s (c)

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-(1.5kg) (3.8m/s)/(15m/s) x 1s/1000m/s = -380N, the average force exerted on the brain is -380N 44. Find the x coordinate of the center of mass of the bricks shown. m1x1+m2x2+m3x3/m1+m2+m3 = m((L/2) +L+(5L/4))/3m = 1/3(11L/4) =11/12L

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