3. In the figure below, the coefficient of friction between the center mass and the surface is

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1 Physics 04A Exa October 9, 05 Short-answer probles: Do any seven probles in your exa book. Start each proble on a new page and and clearly indicate the proble nuber for each. If you attept ore than seven probles, please indicate the one you do not wish to have graded. Six points each.. You want to launch a 500 g odel rocket straight up and hit a horizontally oving drone as it passes 5 above the launch point. The rocket engine provides onstant thrust of 8.0 N. The drone is flying at onstant velocity of 8.0 /s. The drone will be directly overhead in 5 seconds: how long should you wait to launch your rocket?. With odern pressure suits, a fighter pilot can anage 9 g s of acceleration without blacking out. (g = 9.8 /s.) How long would it take a fighter jet, at Mach (340 /s), to ake a 80 horizontal circular turn? 3. In the figure below, the coefficient of friction between the center ass and the surface is µ k = 0.. What is the acceleration of the asses? Assue the pulleys are frictionless Two asses, = 00 g and = 5 g, are attached to a string which is then draped over a frictionless pulley. Initially both asses are.0 fro the floor. How long until the heavier ass hits the floor? 5. A trained airline gorilla drops your 40-pound suitcase onto a baggage conveyor belt that is oving at.0 /s. The conveyor belt has oefficient of friction µ = 0.5. How far does your suitcase slide before its speed atches that of the conveyor belt? 6. Bubba s car has an air freshener hanging fro the rear-view irror. As he drives around a flat corner, he observes the air freshener swinging to the left, aking an angle of 0 to the vertical. What is Bubba s acceleration at that instant? 7. State Newton s three laws of otion.

2 8. Figure shows velocity versus tie for Bubba ( = 80 kg) in an elevator. What is Bubba s weight at tie (a) s, (b) 4 s, and (c) 7 s? 5 4 velocity (/s tie (s) Figure : Velocity / Tie graph

3 Physics 04A Key Exa October 9, 05 Short-answer probles: Do any seven probles in your exa book. Start each proble on a new page and and clearly indicate the proble nuber for each. If you attept ore than seven probles, please indicate the one you do not wish to have graded. Six points each.. Answer: Use Newton s nd law to calculate acceleration of the rocket, then use constantacceleration equations to find how long it takes the rocket to get to the desired altitude. Launch tie will be 5 seconds inus however uch tie it takes to reach altitude. a = F g = 6. /s y = y i + v iy t + a t y = a y t = t = a =.84 s That tie is the tie to reach altitude, subtract fro 5 to see that you need to launch in.6 s.. Answer: Use centripetal acceleration equation, solve for the turn radius r: = v r = r = v Now find the distance for a half-circle turn with this radius: S = πr = πv Finally divide distance by speed to get tie: t = S v = πv v = πv =. s A better answer would be to consider that the direction of the acceleration is not just horizontal in this case: there s always g downwards. The vector su of the downwards g and the horizontal will be the liiting 9g, so = 8.94g and t =. s... but that s within the precision stated in the proble, so I gave full credit for. s. 3. Answer: Start with a free-body diagra: Now either write an equation for each ass using F = a, or use the quick-and-dirty ethod of pretending it s a linear syste: F g 3µg g a = = = g 3µ = 0.65 /s Answer: Find the acceleration first, using free-body diagras and F = a: Now use kineatics equations: a = g + y = y i + v i t at 0 = y i g( ) y i ( + ) ( + ) t = t = =.7 s g( )

4 T Fn = 3g T g T Ff = 3μg 3 3g T g 5. Answer: The acceleration of the suitcase will be a = µg (you can show this with a free-body diagra and F f = µf N ) so find the distance until v =.0 /s. v f = v i + a x = x = v f µg = Answer: The tension in the string has horizontal and vertical coponents: the vertical coponent F v should equal the force of gravity and the horizontal coponent F h provides centripetal force. tan θ = F h F v = g = = g tan θ = 3.6 /s The direction of acceleration is to the right. 7. Answer: (a) Law of inertia: An object at rest reains at rest, an object oving continues oving at constant velocity, unless acted upon by an external force. (b) F = a (c) Forces coe in pairs.

5 8. Answer: The noral force on Bubba (AKA weight ) is enough to oppose the force of gravity (g) with enough left over (or left without) to cause the observed acceleration. Bubba s acceleration will be the slope of the velocity graph. (a) The slope is /s, so F N = (g + /s ) = 864 N. (b) The slope is zero, so F N = g = 784 /s. This is Bubba s noral weight. (c) The slope is /s, so F N = (g /s = 64 N. 3