A B = AB cos θ = 100. = 6t. a(t) = d2 r(t) a(t = 2) = 12 ĵ

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1 1. A ball is thrown vertically upward from the Earth s surface and falls back to Earth. Which of the graphs below best symbolizes its speed v(t) as a function of time, neglecting air resistance: The answer is e), the inverted triangle shape. The speed is 0 at the middle of the flight when the ball has reached its maximum height. 2. Two vectors have magnitudes 10 and 11, and their scalar product is 100. What is the magnitude of the vector sum of these two vectors? a) 6.6 b) * 4.6 c) 8.3 d) 9.8 e) none of these Double use of the scaler product. First Second, using C = A + B A B = AB cos θ = 100 C C = C 2 = ( A + B) ( A + B) = A 2 + B 2 + 2AB cos θ where θ is the angle between A and B. From the second equation above we have C 2 = = 21 = C = 21 = A particle moves along a path given by r(t) = 3t î t3 ĵ. What is the acceleration of the particle at t = 2.0 s (all quantities in MKS units, answer in m/s 2 )? a) * 12 ĵ b) 9 ĵ c) 6 ĵ d) 3 î 6 ĵ e) 3 î 12 ĵ 4. There is no net force on an object when a(t) = d2 r(t) = 6t dt ĵ 2 a(t = 2) = 12 ĵ a) it is moving in circle b) it is accelerating towards the Earth c) it is in free fall d) it is weightless e) * it is moving in a straight line at constant speed Motion is a straight line at constant speed means zero acceleration which means zero force. On the other hand, an object could appear to be weightless as in an orbiting space shuttle. That object still has a centripetal force acting on it to keep it in orbit. 5. A 1200 kg mass car is driving on a horizontal circle of radius 62 m. What is the minimum coefficient of static friction required to keep the car from slipping when its speed is 15 m/s? a) 0.15 b) 0.23 c) 0.35 d) * 0.37 e) 0.41 Centripetal force is equal to the static friction force. F c = mv2 r = µ s N = µ s mg = µ s = v2 rg =

2 6. A centripetal force does no work on a particle moving in a horizontal circle at constant speed because a) the motion is not along a straight line b) there is no acceleration c) * the force is perpendicular to the motion d) the force is parallel to the motion e) trick question: a centripetal force always does some work on such a particle A force which is always perpendicular to a motion does no work. 7. A ball of mass m is dropped off a building of height h. After the ball has dropped a distance d where d h, what is the kinetic energy of the ball (neglect air resistance)? a) mgh b) mgd c) * mgd d) mg(h d) e) 0.0 Conservation of total mechanical energy, where the initial kinetic energy is 0. mgh = mg(h d) mv2 = 1 2 mv2 = mgd 8. Two objects whose masses are in the ratio 4:1 are given the same impulse. If both masses were initially at rest, what is the ratio of their speeds after the impulse is finished? a) 2:1 b) 1:2 c) 1:1 d) * 1:4 e) 4:1 Impulse is the change in momentum. If the masses are initially at rest, then the change in momentum is the final momentum mv. So we have m 1 v 1 = m 2 v 2. If we say m 2 /m 1 = 4, then m 2 m 1 = 4 = v 2 v 1 = What fraction of c must a clock be moving such that a stationary observer perceives the clock to be running at half the correct rate? a) 0.50 b) * 0.87 c) 0.83 d) 0.91 e) 0.75 For a clock to appear to running at half the correct rate the Lorentz factor γ = v 2 /c = 2 = v 3 2 c = 2 = Two people are on a merry-go-round moving at 2 rad/s. One of the persons jumps off, in the radial direction. What happens to the angular speed of the merry-go-round? a) nothing b) * it increases c) it decreases d) it increases or decreases depending upon the ratio of the two masses e) it depends on whether one is in the Northern or the Southern hemisphere Conservation of angular momentum. 2

3 11. The space shuttle is in circular orbit around the Earth at a radius 3R E. If it moves to a circular orbit of radius 4R E, what is the ratio of the total energy between the second and the first orbits? a) * 3:4 b) 4:3 c) 1:1 d) 9:16 e) 16:9 For a particle of mass m in orbit about the Earth at radius R, the total energy is (see page 376) E(R) = 1 GM E m 2 R So for a second radius R 2 = 4R E and a first radius R 1 = 3R E we have E(R 2 ) = 1 GM E m, E(R 1 ) = 1 GM E m = E(R 2) 2 4R E 2 3R E E(R 1 ) = A chisel has a point with an area of 0.6 cm 2. It is struck with a hammer having a force of 42 N. What pressure does the chisel exert at its point on a surface (answer in N/m 2 ). a) 25.2 b) 30,000 c) 300,000 d) * 700,000 e) 650,000 Pressure is force per unit area. We must convert the area to MKS units P = 42 = 700, 000 (0.6)(10 2 N/m2 ) A student wants to set up a standing wave on a wire 1.8 m long which is fixed at both ends. The speed of the waves in the wire is 540 m/s. What is the minimum frequency in Hz? a) * 150 b) 50 c) 100 d) 125 e) 175 The lowest frequency will have the longest wavelength. The longest wavelength is twice the length of the wire, meaning 3.6 m. We get the frequency from the speed equation v = fλ = f = v λ = = 150 Hz 14. A ping pong ball has a density of gm/cm 3, and a radius of 1.9 cm. How much force is required in Newtons to hold the ping pong ball completely under the water? a) b) * c) d) e) 1.10 The force will be the buoyant force less the weight of the ping pong ball F = F B w = ρ W gv ρ P gv = (ρ W ρ P )gv where ρ W is the density of water, and ρ P is the density of the ping pong ball. Using the formula for the volume of a sphere, and converting the densities to the MKS units, we get B = ( )(9.8) 4 3 π(0.019)3 = N 3

4 15. Two ships pass close to each other on the water, each going in the same direction. According to Bernoulli s law, what kind of force might these ships experience (Note: this actually happened to the Titanic and the New York as these ocean liners were pulling out of Southhampton Harbor on the Titanic s first and only voyage.) a) There will be a repulsive force between the ships pushing the ships apart b) Each ship will experience a decelerating force in the direction it is going c) Each ship will experience an accelerating force in the direction it is going d) * Each ship will experience an attractive force pushing the ships together e) There will be no unusual force acting on either ship Explained in the text and in class as an example of Bernoulli s law. 16. The density of a certain ideal fluid is 850 kg/m 3. This fluid flows in a pipe of radius 0.01 m at a rate of 0.25 kg/s. What is the speed v of the fluid in m/s? a) 3.0 b) 0.94 c) 0.85 d) 1.3 e) 0.75 The mass flow is ρav, so v = 0.25/(850 π (0.01) 2 ) = A simple harmonic oscillator is displaced 5.0 cm from its equilibrium position at t = 0, and released. At t = 1.50 seconds, it is located at 2.0 cm from its equilibrium position. What is the angular frequency ω of this oscillator in rad/s? a) 0.50 b) 2.0 c) 4.0 d) π e) * π/4 From the initial conditions we have the SHM equation for this case x(t) = 0.05 cos(ωt) where x is given in meters, t in seconds, and ω is in radians/second. Substituting in t = 1.5 we have x(1.5) = 0.02 = 0.05 cos(1.5ω) = ω = cos 1 (0.4) = π where you must evaluate the inverse cosine in units of radians. 18. The Tacoma Narrows Bridge collapse, as shown on the video in class, is an example of a) electrical resonance b) magnetic resonance c) * mechanical (wind) resonance d) earthquake resonance e) Tsunami (tidal wave) resonance Continuous gusts of wind sent the bridge into higher and higher amplitude oscillations which caused its collapse. 4

5 Part II Worked Problems Solve each of the problems. Show clearly all your work and which equations you use. Partial credit will be given. All numerical answers must have units attached where appropriate. 1. A bungee-cord jumper with a mass of 61 kg is on a bridge 45 m above a river. The bungee cord has an unstretched length L = 25 m, and obeys Hooke s Law with a force constant k = 160 N/m. The jumper drops off the bridge and stops at a height h above the water as shown. What is the distance h as shown? (16 points) Energy is conserved in this problem. The jumper starts with zero kinetic energy. At the bottom of the fall the jumper has been stopped meaning zero kinetic energy again. The jumper s potential energy has decreased and that energy has gone into the stretching of the bungee-cord a distance d. Taking m as the mass of the jumper, then the energy conservation is written has mg45 = mgh kd2 We also know that the bridge height is 45 = L + d + h = 25 + d + h = d = 20 h Substituting this expression for d in the energy conservation equation we get mg(45 h) = 1 k(20 h)2 2 This leads to a quadratic equation which we solve for h The solution to two significant figures is h = 2.0 meters. 5

6 2. A piece of aluminum is suspended from a string and then completely immersed in a container of water as in the figure. The mass of the piece is 1.0 kg, and aluminum has a density of 2.7 gm/cm 3. What is the tension in the string before and after the aluminum is placed in the water? (16 points) The initial tension is just the weight of the aluminum mass w = mg = (1.0)(9.8) = 9.8 N The final tension is the weight of the aluminum less the buoyant force. The buoyant force is the weight of the displaced water. We need to determine the volume of the aluminum mass in order to compute the buoyant force of the water. We can get the volume by using the density and the mass ρ m V = V = m ρ Working in MKS units we have ρ = 2700 kg/m 3 so the volume is V = = m 3 From this volume we calculate the buoyant force as F B = ρ w gv = 1000(9.8)( ) = 3.63 N This leaves the second tension as = 6.17 Newtons. 6

7 3. A string has a length L and a mass M, and is under a tension T. The string is vibrating at its fundamental frequency. The fundamental (lowest) frequency in a vibrating string depends upon the length of the string and the wavelength of the standing wave. The longest wavelength will give the lowest frequency consistent with the speed of the wave v = fλ The longest wavelength is 2L. The speed of the wave is given by v = T T L µ = M Hence the lowest frequency for a give L and M is determined by T L M = f(2l) = f = 1 T 2 LM a) If the length of the string is doubled, and everything else remains the same as stated in the problem, by what factor does the fundamental frequency change? (5 points) Doubling L while keeping T and M constant will reduce the fundamental frequency by a factor 1/ 2. b) If the mass of the string is doubled, and everything else is the same as first stated in the problem, by what factor does the fundamental frequency change? (5 points) Doubling the mass M while keeping L and T the same has the same effect as doubling L. The fundamental frequency is reduced by a factor 1/ 2. c) If the tension in the string is doubled, and everything else is the same as first stated in the problem, by what factor does the fundamental frequency change? (5 points) Doubling T while keeping L and M the same will increase the fundamental frequency by 2. 7

8 4. A bullet of mass 10.0 grams traveling horizontally at 300 m/s is fired into a 2.50 kg block. The bullet stops inside the block, and the block begins to move horizontally. The block was initially on the edge of a frictionless table at a height of 1.00 m above the surface of a floor. a) What was the initial momentum of the bullet? (3 points) p = mv = 0.010(300) = 3.0 kg-m/s b) What was the final momentum of the block+bullet system just after the bullet stopped inside the block? (3 points) Same as initial momentum because only internal forces are acting. c) How much time did it take the block to fall to the floor? (3 points) We compute the time of fall by knowing the initial height, and that the block had no initial vertical speed y(t) = 0 = y gt2 = t = 2y0 2 g = 9.8 = seconds d) How far horizontally did the block travel before it hit the floor (3 points) For this we first need to know the initial horizontal speed of the block which we get from momentum conservation (m + M)V = mv = V = m M + m v = 0.01 (300) = m/s From the time of fall seconds, we can compute the horizontal distance traveled as (0.452)(1.195) = 0.45 m. e) What was the final kinetic energy of the block? (3 points) The final kinetic energy of the block is the initial kinetic energy plus the change in potential energy. Energy is conserved. In this case h = 1.00 m. K f = K i + Mgh = 1 2 MV 2 + Mgh 8

9 5. A large storage tank with an open top is filled to a height h 0. The tank is punctured at a height h above the bottom of the tank as shown. Find an expression for how far in horizontal distance from the tank the stream of water lands on the floor. (16 points) Problem taken directly from Chapter 15, page 31, and was done in the last class. We use Bernoulli s equation to get the exit speed of the water at the puncture hole. The speed at the top surface is considered to be zero because the tank area is large. The pressure at the top surface and at the exit is atmospheric pressure. Bernoulli s equation becomes P a ρgh 0 = P a ρv2 + ρgh v 2 = 2g(h h 0 ) Now to get the time of fall, we use the same method as in the problem 4. The horizontal distance traveled becomes y(t) = 0 = h 1 2 gt2 = t = 2h g x(t) = vt = 2g(h h0 )2h g = 2 h(h h 0 ) 9

10 6. A student stands in the center of a rotating platform which has frictionless bearings. He has a 2.0 kg object in each hand, held out horizontally 1.0 m from the axis of rotation of the platform. The platform, and the student, are initially rotating at 12 revolutions/minute. The moment of inertia of the student and the platform is 2.0 kg-m 2, and stays constant. a) What is the initial angular velocity, in radians/second, of the system? (3 points) ω 1 = 12(2π) 60 = 1.26 rad/sec b) The student brings the two object each to a distance of 0.2 m from the axis of rotation. What is the new angular velocity? (4 points) We use conservation of angular momentum L 1 = L 2 = I 1 ω 1 = I 2 ω 2 ω 2 = I 1 I 2 ω 1 The total moment of inertia is the moment of inertia of the student plus the moment of inertia of the platform plus the moment of inertia of the two objects. Initially we have Finally we have I 1 = (2.0)(1.0) 2 = 6.0 kg-m 2 I 2 = (2.0)(0.2) 2 = 2.16 kg-m 2 ω 2 = = 3.50 rad/sec 2.16 c) What is the change in rotational kinetic energy of this system? (3 points) K 1 = 1 2 I 1ω 2 1 = (0.5)(6.0)(1.26) 2 = 4.76 J K 2 = 1 2 I 2ω 2 2 = (0.5)(2.16)(3.50) 2 = J So the change is kinetic energy is J d) What causes the change in kinetic energy? (4 points) The student does work to bring the objects closer in. 10