University of Alabama Department of Physics and Astronomy PH 101 LeClair Summer Exam 1 Solutions
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1 University of Alabama Department of Pysics and Astronomy PH 101 LeClair Summer 2011 Exam 1 Solutions 1. A motorcycle is following a car tat is traveling at constant speed on a straigt igway. Initially, te car and te motorcycle are bot traveling at te same speed of 18 m/s, and te distance between tem is 92.0 m. After 2.50 s, te motorcycle starts to accelerate at a rate of 4.00 m/s 2. How long does it take from te moment te motorcycle starts to accelerate until it catces up to te car? Solution: Tis was also problem 5b from omework 2 (HW average: 85.9%, exam average 83.7%). Since te car and motorcycle are moving at te same velocity until te moment te motorcycle accelerates, tey will maintain a separation of x o =92 m until te moment te motorcycle accelerates. In tis case, te time of 2.50 s is not relevant for solving te problem - it is only telling you wen someone started measuring te motorcycle s and car s positions, and as suc represents an arbitrary coice of origin. It is not a ead start, like in previous problems similar to tis one, since bot veicles are in motion at te start of te problem. Let te motorcycle s position at te moment it accelerates be x = 0, and let time t = 0 be te moment wen te motorcycle begins accelerating. We will call te initial velocity of bot veicles v o, and te motorcycle s acceleration a m. Te motorcycle s position as a function of time is ten x m (t) = v o t at2 (1) wic correctly captures te initial condition x m (0)=0. Te car starts out a distance x o aead of te motorcycle, and continues at constant speed v o, so its position is x c (t) = x o + v o t (2) wic correctly captures te starting point x c (0)=x o. Te motorcycle catces te car wen teir positions are equal: x c = x m x o + v o t = v o t at2 x o = 1 2 at2 (3) (4) (5) 2xo t = 6.78 s (6) a
2 Note tat te initial velocity v o cancels out since bot objects start out wit te same initial velocity - only te relative velocity matters, wic is entirely determined by te motorcycle s acceleration. An equivalent question would be ow long does it take for te motorcycle to accelerate across te gap of distance x o? 2. A pilot flies orizontally at 1300 km/, at eigt =35 m above initially level ground. However, at time t=0, te pilot begins to fly over ground sloping upward at angle θ=4.3. If te pilot does not cange te airplane s eading, at wat time t does te plane strike te ground? Solution: (exam average 83.3%) Tis is a problem from a different textbook. Given: Te initial velocity and eigt of a plane flying toward an upward slope of angle θ. Find: How long before te plane its te slope? At time t=0, te plane is at te beginning of te slope, a eigt above level ground. Assuming te plane continues at te same orizontal speed, we wis to find te time at wic te plane its te slope. Given te plane s velocity and eigt and te slope s angle, we can relate te orizontal distance to intercept te ramp to te plane s eigt. Sketc: Assume a sperical plane (it doesn t matter). If te plane is at altitude, it will it te ramp after covering a orizontal distance d, were tan θ=/d. v θ d Relevant equations: We can relate te orizontal distance to intersect te ramp to te plane s altitude using te known slope of ground: tan θ = d (7) We can determine ow long te orizontal distance d will be covered given te plane s constant orizontal speed v: d = vt (8) Symbolic solution: Combining our equations above, te time t it takes for te plane to it te
3 slope is t = d v = v tan θ (9) Numeric solution: Using te numbers given, and converting units, t = v tan θ = 35 m 1300 km/ (1000 m/km) (1 /3600 s) (tan s (10) ) 3. A block of mass m = 5.00 kg is pulled along a orizontal frictionless floor by a cord tat exerts a force of magnitude F = 12.0 N at an angle of 65 wit respect to orizontal. (a) Wat is te magnitude of te block s acceleration? (b) Te force magnitude F is slowly increased. Wat is its value just before te block is lifted off te floor? Solution: (exam average 51.9%) Tis is a problem from a different textbook. Given: A block pulled along a frictionless floor by a force making an angle θ wit te orizontal. Find: Te block s acceleration, te maximum force before te block leaves te floor, and te block s acceleration at tat point. Sketc: Let te x and y axes be orizontal and vertical, respectively. We ave only te block s weigt, te normal force, and te applied force. n θ F y mg x Relevant equations: We need only Newton s second law and geometry. Symbolic solution: Along te vertical direction, a force balance must give zero for te block to remain on te floor. Tis immediately yields te normal force. Fy = F n mg + F sin θ = 0 = F n = mg F sin θ (11)
4 A orizontal force balance gives us te acceleration: X Fx = F cos θ = max = ax = F cos θ m (12) If te magnitude of te force is increased, te block will leave te floor as te normal force becomes zero: Fn = mg F sin θ = 0 = F= mg sin θ (13) At tat point, its acceleration will be ax = F g cos θ = m tan θ (14) Numeric solution: Using te numbers given, te initial acceleration is ax = F 12.0 N cos θ = cos m/s2 m 5.00 kg (15) At te point te block is about to leave te floor, te required force is (5.00 kg) 9.81 m/s2 mg 54 N F= = sin θ sin 65 (16) 4. Consider te figure below. Let = 1 m, θ = 15, and m = 0.5 kg. Tere is a coefficient of kinetic friction µk = 0.2 between te mass and te inclined plane, and te mass m starts out at te very top of te incline wit a velocity of vi = 0.1 m/s. Wat is te speed of te mass at te bottom of te ramp? Solution: (exam average 84.7%) Tis is a problem from te PH105 textbook. First focus on te block and incline, and draw te free body diagram for tose two. Let te x axis point down te incline, and te y axis up, normal (perpendicular) to te incline. In te y direction,
5 we ave te normal force, and part of te weigt of te block: ΣF y = F n mg cos θ = ma y = 0 F n = mg cos θ (17) In te x direction, we ave te oter component of te weigt, and opposing tat we ave friction: ΣF x = mg sin θ f k = mg sin θ µ k F n = mg sin θ µ k mg cos θ = ma x (18) a x = g (sin θ µ k cos θ) (19) Given te acceleration a x, te initial velocity v i = 0.1 m/s, and te lengt of te ramp being / sin θ, we can readily calculate te speed at te bottom of te ramp, wic we ll call v f : ( v 2 f = v2 i + 2a x x = v 2 i + 2g (sin θ µ k cos θ) v f = ) ( = v 2i + 2g 1 µ ) k tan θ (20) sin θ ( v 2i + 2g 1 µ ) k 2.23 m/s (21) tan θ 5. A baseball leaves te bat at 30.0 above te orizontal and is caugt by an outfielder 375 ft from ome plate at te same eigt from wic it left. Wat is te initial speed of te ball? Solution: (exam average 89.2%) We know te launc angle and range of te projectile over level ground. For tis situation, we ve already derived te range equation, so we may as well use it: R = v2 o sin 2θ g (22) were θ = 30.0 is te launc angle, v o te launc velocity, and R = 375 ft 114 m is te range. Solving for v o, v o = gr 36 m/s 118 ft/s 81 mp (23) sin 2θ 6. An advertisement claims tat a particular automobile can stop on a dime. Wat net force would actually be necessary to stop a 850 kg automobile traveling initially at 45.0 km/ in a distance equal to te diameter of a dime, wic is 1.8 cm. Hint: watc te units! Solution: (exam average 76.4%) We know te initial velocity v o = 45 km/ = 12.5 m/s, te final velocity v f = 0, and te distance over wic te car accelerated, x = m. Tis is enoug to
6 get us te net acceleration, wic is ten enoug to get us te net force using Newton s second law. First te acceleration: v 2 f = v2 o + 2a x a = v2 o 2 x (24) (25) Te minus sign reminds us tat te acceleration is in te direction opposite x. If tis is te net acceleration, te net force causing it must follow F=ma, so te net force is in magnitude (i.e., we don t care about te sign) F net = ma = mv2 o 2 x N 415 tons (26) A scary, unsurvivable amount of force, equivalent to pulling about 440 g s (were 50 g s generally means serious injury or deat). One could also solve tis wit te work-energy teorem: te force F stopping te veicle troug a displacement x does work W = F x, and tis must be te same as te car s cange in kinetic energy, wic is 1 2 mv2 o. Equating and solving for F, we arrive at te same result.
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