F s = k s. W s i. F ds. P W = dw t dt. K linear 1 2 mv2. U g = mgh. U s = 1 2 k s2. p mv

Transcription

1 v = v i + at F f = µf n p i = p f x = x i + v i t at2 v 2 = v 2 i + 2a x F = ma = dp dt = U x = r cos(θ) y = r sin(θ) r = x 2 + y 2 tan(θ) = y x F s = k s W sf s i F ds P W = dw t dt K linear 1 2 mv2 U g = mgh U s = 1 2 k s2 p mv J t2 t 1 F dt = p ( ) ( ) m1 m 2 2m2 v 1f = v 1i + v 2i m 1 + m 2 m 1 + m 2 ( ) 2m1 v 2f = v 1i + m 1 + m 2 r CM = 1 M r dm 1 M a r = v2 r = rω2 F c = m v2 r ( m2 m 1 m 1 + m 2 n m i r i i=1 ) v 2i 1

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3 Name: Score: / Physics 204A Exam 3 11/2/08 Short-answer problems: Do any seven problems and clearly mark the one you wish to omit by drawing a diagonal line through the answer space. Show your work for complete credit. Six points each. 1. A skier starts with a speed of 3.0 m/s then slides down a nearly-frictionless hill, losing 18 m of altitude before hitting a horizontal patch of rough snow with a coefficient of friction µ = 0.2. The rough snow extends for x = 21 m. How fast is the skier going when she reaches the end of this rough patch? 2. Dr. Ayars can, for short bursts of time, generate 500 W of power while riding his bicycle. His mass is 69 kg, and his carbon-fiber triathlon bike is an additional 8 kg. Assuming he can maintain this 500-W power output for the necessary time, how long does it take him to go from v = 0 to v = 12 m/s? (Note: Air drag is not negligible at 12 m/s (27 mph), but ignore drag for this problem.) 3

4 3. In physics lab, a spring-loaded cart launcher is used to launch a 750 gram cart. The initial spring compression was 12 mm and the spring constant of the spring is k = 510 N/m. What is the launch velocity of the cart? 4. Bubba (M = 80 kg) runs off a dock with a horizontal speed v i = 4.5 m/s and lands in a rowboat (m = 40 kg) that is initially at rest. What is the speed of the rowboat immediately after Bubba s arrival? 4

5 5. You re driving down the road at 60 mph when a giant custard beetle initially flying at 60 mph directly towards you hits your window. (He won t have the guts to do that again!) (a) Which experiences a greater force, your car or the beetle? (b) Which experiences a greater acceleration? (c) Which experiences a greater change in velocity? (d) Which experiences a greater change in kinetic energy? (e) Which experiences a greater change in momentum? Carefully explain each of your answers. 6. A line of identical train cars collide inelastically. (They couple.) Two carts are initially coupled and moving at speed v: the rest are stationary. The two moving cars hit a third, then those three hit a fourth, and so on. This continues until the speed of the collection is 1/4 the speed of the initial two carts. How many carts are in the final collection? 5

6 7. At the ice rink one day, you observe the following delightful scenario: Hockey player a has mass m a = 110 kg, and is moving due east at v a = 7.0 m/s. Physics professor b has mass m b = 65 kg, and is wobbling due north at v b = 2.5 m/s. After they hit, and the physics professor grabs the hockey player in a panic so he doesn t fall down, what is the new direction of motion θ of the hockey player? a θ Figure 1: Ice-rink collision for problem 7 b 8. A lab cart with mass M = 1.5 kg is rolling across the lab table with speed v = 0.65 m/s. Your goal is to stop it by shooting a single sticky lump of clay (m = kg) directly at it. How fast should that lump be moving? 6

7 Physics 204A Key Exam 3 11/2/08 Short-answer problems: Do any seven problems and clearly mark the one you wish to omit by drawing a diagonal line through the answer space. Show your work for complete credit. Six points each. 1. Answer: The initial energy plus the work done by friction equals the final energy. E i = 1 2 mv2 i + mgh W = F f x = µmgx 2. Answer: E f = 1 2 mv2 f 1 2 mv2 i + mgh µmgx = 1 2 v2 f v 2 f = v2 i + 2g(h µx) = v f = 16.7 m/s P = W t = t = W P = mv2 2P = 11.1 s 3. Answer: Spring potential energy becomes cart kinetic energy, but energy is conserved in this situation so 1 2 mv2 = 1 k 2 kx2 = v = m x2 =.31 m/s 4. Answer: This is a straightforward (pardon the pun) application of conservation of momentum. Bubba s initial momentum is The final momentum is Momentum is conserved, so p i = p f and p i = Mv i p i = (m + M)v f M Mv i = (m + M)v f = v f = v i m + M = 3.0 m/s 5. Answer: (a) They are equal. The force of the beetle on your car is according to Newton s 3 rd law, equal and opposite to the force of the car on the beetle. (b) Acceleration is force per mass, so since the forces are the same but the mass of the beetle is much less than the mass of your car, the beetle experiences a much greater acceleration! (c) The beetle has a much larger change in velocity, due to its larger acceleration. (A common-sense answer works here also: your car barely changed speed at all, but the beetle went from whatever speed he was beetling along at to highway speed.) 1

8 (d) Both the car and beetle have essentially zero change in kinetic energy, because neither of them changes speed. Although the beetle s velocity changes by 120 mph, it goes from +60 to 60 and 1 2 mv2 doesn t change. (e) They are exactly equal. This is a closed system, car and beetle, so momentum is conserved. The change in momentum of the beetle is equal and opposite to the change in momentum of the car. 6. Answer: Solve for N. p i = p f = 2mv i = Nmv f 2mv i = Nmv f = Nm N = 8 ( vi ) = 2 = N Answer: This is a conservation of momentum equation. Momentum is a vector, so the x and y components are both conserved. The initial x momentum is p x = m a v a The final x momentum will be the same. The initial y momentum is p y = m b v b The angle θ is just the inverse tangent of py p x. ( ) θ = tan 1 mb v b m a v a = Answer: Momentum is conserved, so you need for the clay to have momentum equal and opposite to the momentum of the cart. p clay = m clay v clay = m cart v cart = v clay = v cart m cart m clay = 97.5 m/s This is probably not a realistic goal. 2