Problem_Set_9

Anthony Putrino 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; if the object’s kinetic energy momentum is doubled then it is changed by a factor of two. 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 system of particles is known to have zero momentum then it does not follow that the kinetic energy of the system is also zero, this is because the particles can have zero momentum but if these particles are actively moving then they can be considered to have kinetic energy. 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. Find momentum by using the component method of vector addition. P total = P cat +P dog = m cat v cat +m dog v dog =(5.30kg)(3.04m/s)+(26.2kg)(2.70m/s) = (16.1kg*m/s) x+(70.7kg*m/s)y Find the velocity by dividing the momentum by mass. Vowner = P total /m 0 =(16.1kg*m/s)x + (70.7kg*m/s)y/74kg = (0.218m/s) 2 + (0.955m/s) 2 ) Find the direction and magnitude. Angle = tan -1 (.955/.218) = approx 7 degrees north east Vowner = sqrt((0.218m/s) 2 +(0.955m/s) 2 ) = 0.980m/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?

Solve for mass using Impulse equation. m = I/ ∆ v = (-9.3kg*m/s)/-24m/s - 4.2m/s = .33kg 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. Solve for v 2 by setting P=0. .v 2 = -m 1 v 1 /m 2 +m mouse = (-.35kg)(-.88m/s)/(.25) + (.042kg) = 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? Solve using conversation of momentum with all combined mass. V a+s+g = m bullet v bullet /m arm + m shoulder + m bullet = -(.038kg)(385m/s)/(15)+(3.3kg) = -.80m/s 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 ms 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? Solution: (a) Solve using Newton’s Second Law. v f = Ϝ ∆t /m c =(2800N)(6.5m/s*1s/1000ms)/(3.3kg) = 5.5m/s (b) Solve for v f by p i = p f . m c v c = (m c +m b )v c+b → v c+b = (m c /m c +m b )v f =(3.3kg/3.3kg+1.5kg)*(5.5m/s) = 3.8m/s