final 01 TAYLOR, JEFFREY W Due: May , 1:00 pm 1 Version number encoded for clicker entry: V1:1, V2:1, V3:5, V4:1, V5:5.

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1 final 0 TAYLOR, JEFFREY W Due: May 0 006, :00 pm Version number encoded for clicker entry: V:, V:, V:5, V4:, V5:5. Question Part of. 0 points. A heavy liquid with a density g/cm is poured into a U-tube as shown in the lefthand figure below. The left-hand arm of the tube has a cross-sectional area of 9.79 cm, and the right-hand arm has a cross-sectional area of 5.69 cm. A quantity of 9. g of a light liquid with a density. g/cm is then poured into the right-hand arm as shown in the right-hand figure below cm 5.69 cm 9.79 cm 5.69 cm ρ h = g/cm. Using the definition of density L = ρ l = m l V l = m l A L m l A ρ l = cm. After the light liquid has been added to the right side of the tube, a volume A h of heavy liquid is displaced to the left side, raising the heavy liquid on the left side by a height of h with a displaced volume of A h. Since the volume of heavy liquid is not changed, we have A h = A h. At the level of the heavy-light liquid interface in the right side, the absolute pressure is h h L P = P atm + ρ l g L, and at the same level in the left tube, P = P atm + ρ h g h + h ]. Equating these two values, we obtain heavy liquid g/cm light liquid. g/cm If the density of the heavy liquid is g/cm, by what height h does the heavy liquid rise in the left arm? cm cm cm cm cm correct cm ρ l g L = ρ h g h + h ] = ρ h g h + A ] h A = ρ h g h + A ]. A Solving for h, we have h = ρ l L ρ h + A ] A = (. g/cm ) ( cm) ( g/cm ) + (9.79 ] cm ) (5.69 cm ) = cm. Let : m l = 9. g, A = 9.79 cm, A = 5.69 cm, ρ l =. g/cm, and Question Part of. 0 points. A ski jumper travels down a slope and leaves the ski track moving in the horizontal direction with a speed of 6 m/s as in the

2 final 0 TAYLOR, JEFFREY W Due: May 0 006, :00 pm figure. The landing incline below her falls off with a slope of θ = 4. The acceleration of gravity is 9.8 m/s. 6 m/s Excluding t from these equations gives d = v 0 sin θ g cos θ = () (6 m/s) sin 4 (9.8 m/s ) cos 4 = m. y d Question Part of. 0 points. x v f φ 4 Calculate the distance d she travels along the incline before landing m m correct m m m m It is convenient to select the origin (x = y = 0) at the beginning of the jump. Since v x0 = 6 m/s and v y0 = 0 m/s in this case, we have x = v x0 t y = v y0 t g t = g t. The distance d she travels along the incline before landing is related to her x and y coordinates by x = d cos θ y = d sin θ. Substituting these expressions for x and y into the two equations above, we obtain d cos θ = v x t d sin θ = g t. Determine how long the ski jumper is airborne s s correct s s s s Excluding d rather than t from the system above, we obtain t = v 0 tan θ g = () (6 m/s) tan(4 ) 9.8 m/s = s. Question 4 Part of. 0 points. What is the magnitude of the relative angle φ with which the ski jumper hits the slope? correct v y = g t = (9.8 m/s ) (4.654 s) = m/s, and v x = 6 m/s, given.

3 final 0 TAYLOR, JEFFREY W Due: May 0 006, :00 pm The direction φ t of the velocity vector (relative to the positive x axis) at impact is ( ) vy φ t = arctan v ( x ) m/s = arctan 6 m/s = Therefore the relative angle of impact φ on the slope is φ = φ t θ = (60.09 ) (4 ) = Question 5 Part of. 0 points. A runner is jogging at a steady v r = 7 km/hr. When the runner is L = 7.5 km from the finish line a bird begins flying from the runner to the finish line at v b = 5 km/hr (5 times as fast as the runner). When the bird reaches the finish line, it turns around and flies back to the runner. Even though the bird is a dodo, we will assume that it occupies only one point in space, i.e., a zero length bird. How far does the bird travel?. km..5 km correct..9 km 4.. km km km v b v r L finish line Let, dodo birds fly, and d r be the distance the runner travels. d b be the distance the bird travels. v r be the speed of the runner. v b be the speed of the bird. L = d r be the original distance to the finish line. L be the distance to the finish line after the first encounter.. L i be the distance to the finish line after the i th encounter. dr db L finish line Since the bird travels 5 times as fast as the runner at the first meeting between the bird and runner, d b = 5 d r. () The sum of the bird s and runner s distances is 5 times L. d b + d r = L. () Therefore, substituting for d b from Eq. () d r + 5 d r = L d r = 6 L = (7.5 km) =.5 km. () 6 Thus the distance the bird flies is d b = 5 d r = 0 6 L = 0 6 (7.5 km) =.5 km, (4) and the distance for the runner to travel after this first encounter is L = 4 6 L = 4 (7.5 km) = 5 km. 6 Question 6 Part of. 0 points. After this first encounter, the bird then turns around and flies from the runner back to the finish line, turns around again and flies back to the runner. The bird repeats the back and forth trips until the runner reaches the finish line.

4 final 0 TAYLOR, JEFFREY W Due: May 0 006, :00 pm 4 How far does the bird travel from the beginning? (i.e., include the distance traveled to the first encounter). 6 km. 7.5 km correct. 9 km km 5. 4 km km Repeating this scenario a second time the distance for the runner to travel after the second encounter is L = 4 ( ) 4 6 L = L, 6 and the third time and the i th time L = 4 6 L = L i = 4 6 L i = ( ) 4 L, 6 ( ) i 4 L. (5) 6 Note: The distance the bird travels between the (i ) th and i th time is see Eq. (4)] d bi = 0 6 L ( 4 6) i (6) and summing over all terms d bi d b = d bi = 0 6 L ( ) ] i 4 (7) 6 i=0 i=0 = 0 6 L + 4 ( ( ) ) + + (8) 6 ( ) 4 ( ) 5 ( ) ] Or, by factoring from the second term on 6 { d b = 0 6 L ( ) + (9) ( ) ( ) 4 ( ) ]} By comparing Eq. (8) with (9), and generalizing (l = 4, and k = 6), the infinite series i=0 ( ) i l = + l k k then solving Eq. (0) for i=0 i=0 i=0 ( ) i l (0) k ( ) i l k ( ) i l = k k k l () i=0 k=6, l=4 ( ) i 4 = = 6. Therefore from Eq. (7)] d b = 0 6 L i=0 ( ) i 4 6 = 0 6 L 6 = 5 L = 5 (7.5 km) = 7.5 km. Elegant Alternative Solution: The bird will travel 5 times as far as the runner in the same time. Since the bird and jogger travel for the same length of time, the bird will travel d b = 5 L = 5 (7.5 km) = 7.5 km Question 7 Part of. 0 points. The two blocks are connected by a light string that passes over a frictionless pulley with a negligible mass. The 4 kg block lies on a rough horizontal surface with a constant coefficient of kinetic friction 0.. This block is connected to a spring with spring constant 4 N/m. The second block has a mass of 9 kg. The system is released from rest when the spring is unstretched, and the 9 kg block falls a distance h before it reaches the lowest point. The acceleration of gravity is 9.8 m/s. Note: When the 9 kg block is at the lowest point, its velocity is zero.

5 final 0 TAYLOR, JEFFREY W Due: May 0 006, :00 pm 5 4 N/m 4 kg µ = 0. h 4 kg 9 kg 9 kg Calculate the mechanical energy removed by friction durning the time when the 9 kg mass falls a distance h J correct J J J J J Basic Concepts: Work-Energy Theorem Spring Potential Energy Frictional Force according to the Work- Energy Theorem Solution: WA B ext = (K B K A ) + (U g B U g A ) + (U sp B U sp A ) + W A B dis. For the present case, the external work = 0, A corresponds to the initial state and B the state where m has descended by a distance s. The sum of the kinetic energy of m plus that of m at B is given by WA B ext K = K B = (U g A U g sp B ) + (UA U sp B ) W A B dis = m g s k s µ m g s. () Based on the Eq. at s = h, K B = 0, we have In turn, µ m g h = m g h k h. h = g m µ m ] k = (9.8 m/s ) (9 kg) (0.) (4 kg)] (4 N/m) = 8. m. h () We know that E initial = E final + E µ, where E µ is the mechanical energy removed by friction. In order to solve the second part of the problem we need to calculate the initial and final energies. Let y be the vertical position of m and y the vertical position of m, where say y = 0 is the initial vertical position of m. The total energy of the system is E = U + K, where U = U grav + U grav + U spring. Initially, U grav,in = m g y, U grav,in = 0, U springin = 0 (the spring is unstretched) and K in = 0. In the final situation the masses have a null velocity, and so we have once again K fin = 0. The potential energies are U grav = m g y, U grav,fin = m g (y h) and U spring = k h. Finally, letting y = 0 m and using Eq. for h, we get E µ = E initial E final = m g ( h) k h = m g h k h = g k = µ m g k m (m µ m ) (m µ m ) ] m µ m ] = (0.) (4 kg) (9.8 m/s ) (4 N/m) (9 kg) (0.) (4 kg)] = J. Question 8 Part of. 0 points. Given: The speed of sound in air is 4 m/s. An open vertical tube has water in it. A tuning fork vibrates over its mouth. As the water level is lowered in the tube, the seventh resonance is heard when the water level is 4 cm below the top of the tube.

6 final 0 TAYLOR, JEFFREY W Due: May 0 006, :00 pm 6 = 44 cm. Question 9 Part of. 0 points. 4 cm What is the frequency of the sound wave; i.e., the tuning fork? s s. 686 s s s s correct The frequency is What is the wave length of the sound wave?. 6 cm. 8 cm. 40 cm 4. 4 cm cm correct cm There are seven nodes (N = 7) in the air column. The number of quarter wavelengths λ 4 in the length of the pipe is J =. Since J = l, then λ = 4 l λ J = 4 l. 4 The number of quarter wavelengths is odd which is indicative of a node at one end and an anti-node at the other end of the tube. λ = 4 l N, where N = 7 4 l = (7) 4 (4 cm) = (7) f = v λ (4 m/s) = (44 cm) (0.0 m/cm) = Hz. Question 0 Part of. 0 points. The water continues to leak out the bottom of the tube. When the open vertical tube next resonates with the tuning fork, what is its length cm. 44 cm cm cm cm cm correct The next resonance will occur when the open vertical tube has a length of onehalf wavelength greater than its initial wave length. l = l + λ = (4 cm) + = 65 cm. (44 cm)

7 final 0 TAYLOR, JEFFREY W Due: May 0 006, :00 pm 7 Question Part of. 0 points. You are given f (x), a transverse wave that moves on a string that ends and is FIXED in place at x = 5 m. As the problem begins, the wave is moving to the right at v = m/s. Amplitude (centimeter) Distance (meter) What is the shape of the wave on the string after 5 s? Distance (meter) Distance (meter) v Distance (meter) Distance (meter) correct Distance (meter) Distance (meter) Consider the image of the wave reflected

8 final 0 TAYLOR, JEFFREY W Due: May 0 006, :00 pm 8 about the FIXED point x = 5 m in the following diagram. The image will be moving to the left at v = m/s (in the opposite direction from the real wave). The initial wave (real) on the string is represented with a dashed line and its reflected wave (imaginary) is represented with a dotted line. Initial time, t = 0 s v v Amplitude (centimeter) Distance (meter) After 5 s the positions of the two waves are have both moved 5 meters in opposite directions. The resultant sum of the two waves is the light gray line. Superposition, at t = 5 s Amplitude (centimeter) Distance (meter) Amplitude (centimeter) Resultant, at t = 5 s Distance (meter) Question Part of 5. 0 points. Assume: When the disk lands on the surface it does not bounce. The disk has mass 8 kg and outer radius 50 cm with a radial mass distribution (which may not be uniform) so that its moment of inertia is 5 9 m R. The disk is rotating at angular speed 7 rad/s around its axis when it touches the surface, as shown in the figure below. The disk is carefully lowered onto a horizontal surface and released at time t 0 with zero initial linear velocity along the surface. The coefficient of friction between the disk and the surface is 0.0. The kinetic friction force between the surface and the disk slows down the rotation of the disk and at the same time gives it a horizontal acceleration. Eventually, the disk s linear motion catches up with its rotation, and the disk begins to roll (at time t rolling ) without slipping on the surface. The acceleration of gravity is 9.8 m/s. 8 kg 50 cm, radius 7 rad/s I= 5 9 m R µ = 0.0 How long t = t rolling t 0 does it take for the ball to roll without slipping?

9 final 0 TAYLOR, JEFFREY W Due: May 0 006, :00 pm s correct.. s s s s s Let : r = 50 cm = 0.5 m, ω 0 = 7 rad/s, m = 8 kg, and µ = 0.0. From the perspective of the surface, let the speed of the center of the disk be v surface. Using the frictional force f, we can determine the acceleration Since f = µ m g, and Fsurface = m a, or m a = µ m g, so a = µ g, and α = µ g R. () ω surface = α t, we have = µ g R t. () After pure rolling begins at t rolling there is no longer any frictional force and consequently no acceleration. From the perspective of the center of the disk, let the tangential velocity of the rim of the disk be v disk and the angular velocity be ω ; the angular acceleration is τ = I α, so α = τ I = µ m g R 5 9 m R = 9 µ g 5 R = 9 (0.0) (9.8 m/s ) 5 (0.5 m) = 0.58 rad/s. () The time dependence of ω is ω = ω 0 α t = ω 0 9 µ g 5 R t. (4) When the disk reaches pure rolling, the velocity from the perspective of the surface will be the same as the velocity from the perspective of the center of the disk; that is, there will be no slipping. Setting the velocity ω disk from Eq. 4 equal to ω surface from Eq. gives ω disk = ω surface ω 0 9 µ g 5 R t = µ g R t, or 4 5 µ g t = R ω 0, so t = 5 4 = 5 4 R ω 0 µ g. (5) (0.5 m) (7 rad/s) (0.0) (9.8 m/s ) =.755 s. Question Part of 5. 0 points. Once the disk rolls without slipping, what is its angular speed?.. rad/s rad/s..4 rad/s 4..5 rad/s correct rad/s rad/s Using Eqs. and 5, we have ω rot = α t = µ g R t = µ g T ( 5 4 ) ω 0 R µ g = 5 4 ω 0, (6) = 5 (7 rad/s) 4 =.5 rad/s.

10 final 0 TAYLOR, JEFFREY W Due: May 0 006, :00 pm 0 or using Eqs. 4 and 5, we have ω rot = ω 0 9 µ g 5 R t ( 9 µ g = ω 0 5 R ) ( 5 4 ) R ω 0 µ g = 5 4 ω 0. (6) Question 4 Part of 5. 0 points. How far s does the ball slide until it begins to roll without slipping? m m m m m correct m Starting at rest, the ball slides at a constant acceleration. Using Eq. 5, we have s = a t = µ g 5 4 ] R ω 0 µ g = 5 R ω0 9 µ g = 5 (0.5 m) (7 rad/s) 9 (0.0) (9.8 m/s ) = m. Question 5 Part 4 of 5. 0 points. (7) Through what angle θ does the disk rotate while sliding before it begins to roll without slipping? rev rev rev rev rev correct rev The ball spins at a constant deceleration. Using Eq., we have θ = ω 0 t α t ( ) 5 R ω 0 = ω 0 4 µ g ( ) ( 9 µ g 5 5 R 4 5 = 4 45 ] R ω 0 9 µ g = 95 R ω0 9 µ g = 95 (0.5 m) (7 rad/s) 9 (0.0) (9.8 m/s ) = rad = The number of revolutions is θ rev = θ 60 ( rad) = 60 = rev. ) R ω 0 µ g (8) Note: When comparing Part to Part 4, the ball spins more than it slides due to the fact it is slipping on the surface R θ > s R ω0 > 5 µ g 9 R ω 0 µ g. Question 6 Part 5 of 5. 0 points. What is the ratio of the final kinetic energy (after pure rolling occures) to the initial kinetic energy?. K f K 0 = 5. K f K 0 = 7. K f K 0 = 5 4 correct

11 final 0 TAYLOR, JEFFREY W Due: May 0 006, :00 pm 4. K f K 0 = K f K 0 = 4 6. K f K 0 = 9 9 h 8.8 cm light liquid 6 kg/m Using Eq. 6, ω disk = ω rolling = 5 4 ω 0, heavy liquid 000 kg/m K 0 = I ω 0, and K f = I ω rolling + m v rolling = I ω rolling + m R ωrolling = I ω rolling + m ] R I = ] 5 I 4 ω ] 5 = ] ] 5 4 I ω = 5 4 K 0, so K f K 0 = 5 4. Question 7 Part of. 0 points. A simple U-tube that is open at both ends is partially filled with heavy liquid. The density of heavy liquid is 000 kg/m. A liquid of density 6 kg/m is then poured into one arm of the tube, forming a column 8.8 cm in height, as shown in the following diagram. What is the difference h in the heights of the two liquid surfaces?..955 cm..008 cm..059 cm cm 5..5 cm 6..4 cm correct Let : l = 8.8 cm, ρ l = 6 kg/m, ρ h = 000 kg/m. and Basic Concepts: gauge pressure, variation of pressure with depth Because the liquid in the U-tube is static, the pressure exerted by the heavy liquid column of height l h in the left branch of the tube must balance the pressure exerted by the liquid of height h poured into the right branch. Therefore, P 0 + (l h) ρ h g = P 0 + l ρ l g. Solving for h, h = l ρ ] l ρ h = (8.8 cm) (6 ] kg/m ) (000 kg/m ) =.4 cm.

12 final 0 TAYLOR, JEFFREY W Due: May 0 006, :00 pm Question 8 Part of. 0 points. The cylindrical disk has mass 7 kg and outer radius 8 cm with a radial mass distribution (which may not be uniform) so that its moment of inertia is 5 m r. The disk rolls (perpendicularly to the axis) without slipping in a cylindrical trough, see figure below. The acceleration of gravity is 9.8 m/s. 40 cm 8 cm Determine (for small displacements from equilibrium) the period of harmonic oscillation which the disk undergoes s..464 s s s correct s s Basic Concepts: Let r be the radius of the disk and R be the radius of the cylindrical trough. The disk is rolling without slipping. Choose the point of contact as our axis. Around this point, the rotational inertia of the disk is, using parallel axis theorem, I = 5 m r + m r = 8 m r. () Solution: Let the angle of rotation around this instantaneous contact point be φ and the angle the center of the disk makes from the center of the trough to the vertical be θ. S is the common arc length S = r (φ + θ) S = R θ θ R r Note: The dotted curve is a hypocycloid denoting the path of the contact point at equilibrium as the disk rolls back and forth in the cylindrical trough. The arc length along the disk (which rolls back and forth) must equal the arc length along the cylindrical trough (both arc lengths are labeled S in the figure). Therefore, φ and θ are related by r (φ + θ) = R θ, or r φ = (R r) θ, so r d φ d θ = (R r) dt dt, from the rolling without slipping condition. Now the torque equation around the point of contact is d φ τ : m g r sin θ = I dt I d φ dt + m g r θ = 0, since sin θ θ. Substituting θ in place of φ, we have I R r r S φ S R θ r d θ dt + m g r θ = 0 d θ dt + m g r I (R r) θ = 0. Substituting I from Eq., we have d θ dt d θ dt m g r (R r) m r θ = 0 g (R r) θ = 0. ()

13 final 0 TAYLOR, JEFFREY W Due: May 0 006, :00 pm Equation is the differential equation for simple harmonic motion. The coefficient of θ is ω. Therefore 5 g ω = 8 (R r), and 8 (R r) T = π 5 g = π 8 5 =.466 s. The kinetic en- Alternative Solution: ergy, (0.4 m) (0.08 m) (9.8 m/s ) K = ( ) d x m + dt I = m (R r) ( d θ dt = m (R r) ( d θ dt ( d φ dt ) + 0 m r ) + m (R r) 0 ), = 4 5 m (R r) ( d θ dt ) ( ) d φ dt ( ) d θ dt since I disk = 5 m r, v = r ω, and ω = d θ dt. The potential energy is d E dt = 8 5 = constant, so m (R r) d θ dt d θ dt +m g (R r) d θ dt θ = 0. Therefore, d θ dt g (R r) θ = 0. () Equations and are the same equation for simple harmonic motion. Question 9 Part of. 0 points. The figure below shows a complex wave pattern on a string moving towards a rigid hook at the wall on the right. After some time, the wave is reflected from the wall. v Select the wave pattern for the reflected wave.... v v correct U = m g h = m g (R r) ( cos θ) m g (R r) θ, 4. v since cos θ θ. Energy is conserved, therefore E = K +U = constant, and we have 5. v E = 4 ( ) d θ 5 m (R r) dt + m g (R r) θ Consider the wave pattern image reflected about the rigid hook on the wall.

14 final 0 TAYLOR, JEFFREY W Due: May 0 006, :00 pm 4 v After the time it takes for the wave to be reflected from the wall, this image is the wave pattern traveling to the left along the string. Note: Reflection about a point (hook) is the same as reflection about the y-axis (wall) followed by reflection about the x-axis (string). The leading part of the wave must remain in front and the wave is flipped over. This is the first wave pattern of four possible wave patterns presented by this question. v. h W + h W = F b. b W + b W = F h. b 4 W + b W = F h correct 4. (W + W ) h = F b 5. (W + W ) b = F h 6. h 4 W + h W = F b Question 0 Part of. 0 points. Consider a uniform ladder leaning against a smooth wall and resting on a smooth floor at point P. There is a rope stretched horizontally, with one end tied to the bottom of the ladder essentially at P and the other end to the wall. The top of the ladder is at a height is h up the wall and the base of the ladder is at a distance b from the wall. The weight of the ladder is W. Jill, with a weight W, is one-fourth the way ( d = l 4 ) up the ladder. The force which the wall exerts on the ladder is F. T P ivot W W N f F θ Fx : T F = 0, () P l d W T b W F h θ Note: Figure is not to scale. The torque equation about P is given by Fy : N f W W = 0, and () l τp : W d cos θ + W cos θ () F l sin θ = 0, where d is the distance of the person from the bottom of the ladder. Therefore F l sin θ = W d cos θ + W l cos θ. Since sin θ = h l and cos θ = b, the torque l equation about P is given by b 4 W + b W = F h. (4)

15 final 0 TAYLOR, JEFFREY W Due: May 0 006, :00 pm 5 Question Part of. 0 points. A flexible chain weighing 44. N hangs between two hooks located at the same height. At each hook, the tangent to the chain makes an angle θ = 5.8 with the horizontal. θ Find the magnitude of the force each hook exerts on the chain N N N N correct N N Basic Concept: In equilibrium, F = 0 Solution: In equilibrium F = 0. By symmetry each hook supports half the weight of the chain, so Therefore T y = T e sin θ = W T e = W sin θ = 44. N sin 5.8 = N Question Part of. 0 points. Hint: For this part, make a free-body diagram for half the chain. Find the tension in the chain at its midpoint N N correct..597 N N N N At the midpoint of the chain, there is only a horizontal component of the tension. Since the chain is in equilibrium, the tension at the midpoint must equal the horizontal component of the force of a hook. Therefore T e cos θ = T m T m = T e cos θ = ( N) cos(5.8 ) = N Question Part of. 0 points. A string of linear mass density µ =.94 g/m is stretched by the weight of an adjustable mass m as shown on the picture below. vibrator.8 m µ =.94 g/m m 8 Note: The wave pattern show above is for illustrative purposes only. Near the end of the string a vibrator is attached at a constant but unknown frequency; the length of the string which vibrates is L =.8 m. For some values of the mass m n, the string resonates with the vibrator s frequency and develops visible standing waves with n nodes (not counting the nodes as each end) and n anti-nodes between the vibrator and the pulley. A lab student who performed this experiment recorded two consecutive resonances for m n =.5 kg and for m n =.56 kg. Unfortunately, he forgot to record the actual node

16 final 0 TAYLOR, JEFFREY W Due: May 0 006, :00 pm 6 numbers for the resonances but only wrote that there were no resonances for any masses between.5 kg and.56 kg. How many antinodes did the resonant wave have for m =.5 kg? correct A resonant standing wave with (n ) nodes between two fixed ends of the string which are also nodes occupies length of n half-wavelengths, hence the resonance occurs whenever L = n λ n = λ n = L n. The wavelength λ depends on the oscillator s frequency f and the speed F m g v = µ = µ of the transverse waves on the string, f λ = v. Consequently, the resonance happens when f L n = f λ mn g n = v n = µ, i. e., for the string-stretching mass m n = n 4 L f µ g Now consider the ratio of masses for two consecutive resonances m n m n = n (n ), which does not depend on anything but n. Consequently, n.56 kg = (n ).5 kg =.778, n n =.778 =.06667, n = = 5.. Therefore, for the mass m 6 =.5 kg, the resonant wave has n = 6 antinodes and n = 5 nodes (not counting the two nodes at each end), as seen in the figure below. vibrator.8 m µ =.94 g/m Question 4 Part of. 0 points..5 kg What is the frequency of the oscillator? Hz correct Hz Hz Hz Hz Hz In the first part of the problem, we derived f L n = f λ n = v n = mn g µ. Now that we know n, we use this formula to evaluate the frequency f = n mn g L µ = Hz. Question 5 Part of. 0 points. Needing help, the secretary of the United States Department of Agriculture asked your teacher, If a chicken-and-ahalf can lay an egg-and-a-half in a dayand-a-half, how many days will it take

17 final 0 TAYLOR, JEFFREY W Due: May 0 006, :00 pm 7 two chickens to lay twenty-four eggs? Please help your teacher select the correct answer to the secretary s question.. Two chickens will lay twenty-four eggs in twenty-two days.. Two chickens will lay twenty-four eggs in ten days.. Two chickens will lay twenty-four eggs in eighteen days. correct 4. Two chickens will lay twenty-four eggs in twenty-one days. 5. Two chickens will lay twenty-four eggs in twelve days. 6. Two chickens will lay twenty-four eggs in twenty days. Basic Concept: The information given in the question is the rate of egg production in one instance and you must make this rate compatible with another instance. The rate of egg production is constant. The number of eggs per chicken per day is a constant. Solution: Since it takes a chicken-and-a-half a day-and-a-half to lay an egg-and-a-half, it will take one chicken one-and-a-half days to lay one egg. Therefore, to lay twenty-four eggs it will take two chickens eighteen days. Alternative Method: Unit analysis is basic to every physics problem and is central to this problem. The rate of egg production is the number of eggs produced per chicken per day. In the given instance the rate is {/ eggs} rate = {/ chickens} {/ days} = eggs chickens days. () In the requested instance, the number of chickens is ( chickens) and the number of eggs is (4 eggs). The number of days N is to be determined. Therefore in the requested instance, the rate is rate = {4 eggs} { chickens} {N }. () The rate is constant, so equating the rates Eqs. () and (), we have eggs chicken days = 4 eggs chickens N Solving for the number of days N, we have 4 eggs N = chickens = 8 days. chickens days eggs The correct answer: Two chickens will lay twenty-four eggs in eighteen days. Note: The early chicken catches the worm. Question 6 Part of. 0 points. Assume: h = m, L = 6 m, and θ = 60, and that the cross-sectional area at A is very large compared with that at B. Assume: y = 0 at B. Measure the height from the top of the angled tube. The figure below shows a water tank with a valve at the bottom. The acceleration of gravity is 9.8 m/s. m Valve A 6 m B h max 60 Figure: Not drawn to scale. If this valve is opened, what is the maximum height h max attained by the water stream coming out of the spigot on the right side of the figure?

18 final 0 TAYLOR, JEFFREY W Due: May 0 006, :00 pm m m m m m correct m Let us first compute the speed with which the water leaves the tank at B by applying Bernoulli s equation between A and B. Since the cross-sectional area at A is much larger than the one at B, we can neglect the speed of the water at A compared to that at B. The pressure at both points is going to be equal to the atmospheric pressure P atm. The equation is then P atm + ρ w g (h L sin θ) = P atm + ρ w v B, and so v B = g (h L sin θ). Now the problem reduces to that of projectile motion, for which the maximum height is given by h max = v B sin θ g = h L sin θ] sin θ = ( m) (6 m) sin(60 )] sin (60 ) = m. Question 7 Part of. 0 points. The system shown below is released from rest and moves 94.7 cm in.5074 s. The acceleration of gravity is 9.8 m/s. kg µ = 0 M What is the value of the mass M? Assume all surfaces are frictionless kg correct kg kg kg kg kg Given : m = kg, y 0 = 0 m, y = 94.7 cm = m, t 0 = 0 s, t =.5074 s, and µ = 0. N a m m g T From kinematics, we have M T M g y y 0 = v 0 + a t = a t a = y t The only external force acting on the system is the weight M g suspended from the rope, so from Newton s second law we have M g = (M + m) a M = m a g a m y = t t g y t t a () = m y g t y ( kg) (0.947 m) = (.5074 s) (9.8 m/s ) (0.947 m) = kg.

19 final 0 TAYLOR, JEFFREY W Due: May 0 006, :00 pm 9 Question 8 Part of. 0 points. The system is in equilibrium and the pulleys are frictionless and massless. The acceleration of gravity is 9.8 m/s. At the mass m, T acts up, and m g and T act down, so T = m g + T = m g + m g. At pulley, T acts up on either side of the pulley and T 4 acts down, so 9 kg T Find the force T N. 5.6 N. 5.4 N N N correct N 7 kg 5 kg Let : m = 5 kg, m = 7 kg, m = 9 kg. and At the mass m, T 4 = T = m g + 4 m g. T 4 = T + m g T = T 4 m g = ( m + 4 m m ) g = (7 kg) + 4 (5 kg) (9 kg)] (9.8 m/s ) = 45 N. Question 9 Part of. 0 points. An object of mass m is moving with speed v 0 to the right on a horizontal frictionless surface, as shown, when it explodes into two pieces. Subsequently, one piece of mass 0 m moves with a speed v 0/ = v 0 to the left. 5 T 4 m T T T T T m T T m The mass m defines the tension T : T = m g. At pulley, T acts down on either side of the pulley and T acts up, so T = T = m g. v m m v 0 before after m v / What is the speed v / of the other piece of the object?. v / = 5 v 0. correct. v / = 9 0 v 0.. v / = v 0.

20 final 0 TAYLOR, JEFFREY W Due: May 0 006, :00 pm 0 4. v / = 0 v v / = v v / = v 0. The horizontal component of the momentum is conserved, so 0 + m v 0 = 0 m v 0/ m v 0 = 0 ( m v 0 5 m v 0 = 0 ) m v / + m v / 65 m v 0 + m v / ( 65 v / = ) v 0 65 v / = v 0 v / = v / = 5 v 0. v 0 Question 0 Part of. 0 points. A police car is traveling at a speed, v c, to the right. A truck is traveling at a speed, v t, to the left. The frequency of the siren on the police car is f c. The speed of sound in air is v a. Let v t be the speed of the observer in the truck, and v c be the speed of the source, the police car. v c Police v t Truck What is the frequency, f t, heard by an observer in the moving truck?. f t = v a v t v a v c f c. f t = v a v t v a + v c f c. f t = v a + v t v a v c f c correct 4. f t = v a + v t v a + v c f c Basic Concepts: The Doppler shifted frequency, f, heard in the truck is f = v a ± v 0 v a v s f, () where v a is the speed of sound in air, v o is the speel of the observer, and v s is the speed of the source, The upper sign is used when the relative velocities are toward one-another, and vice versa. Solution: The relative velocity of the observer is towards the source so the upper sign is used in the numerator (± +), and the relative velocity of the source is towards the observer so the upper sign is used in the denominator ( ). Therefore Eq. becomes f t = v a + v t v a v c f c. This is version three of four versions. Question Part of. 0 points. A police car is traveling at a speed, v c, to the right. A truck is traveling at a speed, v t, to the right. A wind is blowing in the opposite direction as that of the truck with a speed, v w, to the left. The frequency of the siren on the police car is f c. The speed of sound in air is v a. v c Police wind v w v t Truck What is the frequency, f t, heard by an observer in the moving truck?. f t = v a v t v w v a + v c + v w f c

21 final 0 TAYLOR, JEFFREY W Due: May 0 006, :00 pm. f t = v a v t + v w v a v c v w f c. f t = v a + v t + v w v a + v c + v w f c 4. f t = v a v t v w v a v c v w f c correct 5. f t = v a + v t v w v a + v c + v w f c 6. f t = v a + v t + v w v a v c v w f c The problem must be worked in the frame of reference relative to the air. v t + v w is the relative velocity of the truck (observer), v o. v c + v w is the relative velocity of the car (source), v s, therefore f t = v a ± (v t + v w ) v a (v c + v w ) f c. () The relative velocity of the observer is away from the source so the lower sign is used in the numerator (± ), and the relative velocity of the source is towards the observer so the upper sign is used in the denominator ( ). Therefore Eq. becomes f t = v a (v t + v w ) v a (v c + v w ) f c, = v a v t v w v a v c v w f c. so T T kg kg.8 m 6 kg T. cm What is the tension T in the string between the block with mass kg and the block with mass kg (on the left-hand side of the pulley)? N. 4. N N N N N correct Let : R =. cm, m = kg, m = kg, m = 6 kg, h =.8 m, v = ω R, I = M R, ω and K disk = I ω = 4 M v. Consider the free body diagrams This is version four of eight versions. T T T Question Part of. 0 points. a kg kg 6 kg a The pulley is massless and frictionless. A massless inextensible string is attached to these masses: kg, kg, and 6 kg. The acceleration of gravity is 9.8 m/s. m g m g T m g

22 final 0 TAYLOR, JEFFREY W Due: May 0 006, :00 pm Basic Concept : For each mass in the system F net = m a. Solution : Since the string changes direction around the pulley, the forces due to the tensions T and T are in the same direction (up). The acceleration of the system will be down to the right (m > m + m ), and each mass in the system accelerates at the same rate (the string does not stretch). Let this acceleration rate be a and the tension over the pulley be T T = T. In free-body diagram for the lower left-hand mass m the acceleration is up and T m g = m a. () In free-body diagram for the upper left-hand mass m the acceleration is up and T T m g = m a. () In free-body diagram for the right-hand mass m the acceleration is down and T + m g = m a. () Adding Eqs. (), (), and (), we have (m m m ) g = (m + m + m ) a. (4) Therefore a = m m m g (5) m + m + m 6 kg kg kg = kg + kg + 6 kg g = kg kg (9.8 m/s ) = (9.8 m/s ) = m/s. The tension in the string between block m and block m (on the left-hand side of the pulley) can be determined from Eq. (). ( ) T = m + g (6) ( ) = ( kg) (9.8 m/s ) ( ) 6 = kg (9.8 m/s ) =.077 N. Question Part of. 0 points. What is the magnitude of the acceleration of the block kg? m/s correct m/s..4 m/s 4..6 m/s m/s m/s The acceleration is the same for every mass, since the string is inextensible. See Part, Eq. (5). Question 4 Part of. 0 points. All angles are measured in a counterclockwise direction from the positive x-axis. A hiker makes four straight-line walks (A, B, C, and D) in random directions and lengths starting at position (4 km, 4 km), listed below and shown below in the plot. A A 7 km at 46 B km at 5 C km at 06 D 9 km at 9 B Figure: Drawn to scale. How far from the starting point is the hiker after these four legs of the hike?.. km..045 km..46 km D C

23 final 0 TAYLOR, JEFFREY W Due: May 0 006, :00 pm km correct km km A E Scale: 0 km = Note: E = km and θ e = a x = (7 km) cos 46 =.809 km, a y = (7 km) sin 46 =.88 km, b x = ( km) cos 5 =.597 km, b y = ( km) sin 5 = 5.65 km, c x = ( km) cos 06 = 7.64 km, c y = ( km) sin 06 = 0.57 km, d x = (9 km) cos 9 = 8.5 km, d y = (9 km) sin 9 = 4.74 km, x 0 = 4 km, starting point y 0 = 4 km, starting point a x = (.809 km) + (4 km) = km, a y = (.88 km) + (4 km) = 5.88 km, b x = (.597 km) + (5.809 km) = km, b y = ( 5.65 km) + (5.88 km) = km, c x = (7.64 km) + ( km) = km, c y = ( 0.57 km) + ( km) = km, d x = ( 8.5 km) + (9.044 km) = 74.5 km, d y = ( 4.74 km) + (7.549 km) =.75 km, Therefore E = { (74.5 km) (4 km)] B θ e D C + (.75 km) (4 km)] } / = km, { } (.75 km) (4 km)] θ E = arctan (74.5 km) (4 km)] = Question 5 Part of. 0 points. A particle of mass m moves along the x axis. Its position varies with time according to x = (6 m/s ) t + ( 5 m/s ) t. What is the work done by the force from t = 0 to t = t?. W = 8 m t (6 m/s 6 ) t ] +( 8 m/s 5 ) t + ( m/s 4 ). W = m t (44 m/s 6 ) t ] +( 0 m/s 5 ) t + (5 m/s 4 ). W = m t (8 m/s 6 ) t ] +( 90 m/s 5 ) t + (5 m/s 4 ) correct 4. W = 9 m t (5 m/s 6 ) t ] +( 40 m/s 5 ) t + (6 m/s 4 ) 5. W = 9 m t (8 m/s 6 ) t ] +( 6 m/s 5 ) t + (4 m/s 4 ) 6. W = m t (6 m/s 6 ) t ] +( 60 m/s 5 ) t + (5 m/s 4 ) Let : x = a t + b t, a = 6 m/s, and b = 5 m/s, Since the force is time dependent W xf x i F d x

24 final 0 TAYLOR, JEFFREY W Due: May 0 006, :00 pm 4 = = m = m = m xf x i xf m a dx x i xf x i vf v i d v dt dx d v dx v dv d x dt dx = m v f m v i. The velocity of the particle is v = d x dt = d a t + b t ] dt ] = a t + b t ] = (6 m/s ) t + ( 5 m/s ) t = (8 m/s ) t + ( 0 m/s ) t. Therefore work done on the particle is the change in kinetic energy. For this case, W = K = K f K i = m (v v 0 ) = m v = m a t + b t ] = m 9 a t a b t + 4 b t ] = m 9 (6 m/s ) t 4 +6 (6 m/s ) ( 5 m/s ) t +4 ( 5 m/s ) t = m t (8 m/s 6 ) t ] +( 90 m/s 5 ) t + (5 m/s 4 ). ]