= y(x, t) =A cos (!t + kx)

Transcription

1 A harmonic wave propagates horizontally along a taut string of length L = 8.0 m and mass M = 0.23 kg. The vertical displacement of the string along its length is given by y(x, t) = 0. m cos(.5 t x), where x is measured in meters and t in seconds.. (7 points) What is the tension, T, in the string? a. T = 0.0 N b. T = 0.2 N c. T = 0.29 N d. T = N e. T = N v prop =! k = s T µ y! 2 L =) T = µ!2 k 2 = M L k 2 = =0.0 N x 2. (7 points) What is the maximum acceleration in the y-direction of any point along the string? a. a <=> =.50 m/s 2 b. a <=> = m/s 2 c. a <=> = 0.50 m/s 2 d. a <=> = 0.05 m/s 2 e. a <=> = m/s 2 y(x, t) =A cos(!t + kx) d 2 y dt 2 = a y(x, t) =! 2 A cos(!t + kx) =) a max,y =! 2 A =(.5) 2 0. =0.225 m/s 2 3. (5 points) In which direction is the wave propagating? a. +x direction b. x direction c. +y direction y(x, t) =A cos (!t + kx) When time (t) increases, then the position (x) must decrease in order for the value of the cosine to remain constant.

2 A block (which you should treat like a point particle) has mass m =.55 kg and slides with initial speed v B in the +x direction. It collides with a rod of length L =.3 m and mass M = 30.3 kg, which is initially at rest and oriented perpendicular to the path of the block. The block hits the rod a distance D = 0.39 m from the center of the rod. After the collision, the block is at rest and the rod spins with angular velocity of ω E = 28 rad/s. Everything is on top of a horizontal frictionless table, and the rod has a fixed frictionless pivot through its center that allows it to rotate freely but keeps its center from moving. 4. (7 points) Which of the following statements best describes the collision? a. Neither the angular momentum about the pivot nor the linear (translational) momentum are conserved. b. The angular momentum about the pivot is conserved, but the linear (translational) momentum is not conserved. c. The angular momentum about the pivot and the linear (translational) momentum are both conserved. d. The linear (translational) momentum is conserved, but the angular momentum about the pivot is not conserved. e. There is not enough information provided to answer this question. The pivot exerts an external force on the block/rod system to keep the center of the rod from sliding to the right upon impact of the block. 5. (7 points) What was the initial speed of the block? a. v B = m/s b. v B = m/s c. v B = m/s d. v B = m/s e. v B = m/s L i = L f mv i D = I rod! f = =) v i = 2 ML2! f md = m/s 2 ML2! f = 2 (30.3)(.3)2 28 (.55)(0.39) 6. (5 points) Suppose the experiment is repeated using new block that has the same mass and initial speed as in the original situation, but is made of a different material so that it bounces back and ends up moving in the x direction after the collision. How would the final angular velocity of the rod in this new case, ω HIJ, compare to the final angular velocity of the rod in the original case ω E? a. ω HIJ = ω E b. ω HIJ > ω E c. ω HIJ < ω E In the second case, the block will have a negative value of angular momentum (if it had positive before the collision), so to conserve angular momentum, the rod must spin faster in the second case.

3 A block of mass M slides on a frictionless surface with a velocity V. It strikes an identical block of mass M that is at rest and is attached to an ideal spring of spring constant k. Before the collision, the spring is not compressed and the collision takes place very quickly before the spring has time to compress appreciably. After the collision, the blocks stick together. 7. (7 points) Immediately after the collision, the blocks slide to the right and the spring is compressed. By how much is the spring shortened, ΔL, before the blocks are pushed to the left? a. ΔL = 2MV P /k b. ΔL = MV P /(2k) c. ΔL = MV P /k d. ΔL = 4MV P /k P i = P f =) MV =(2M)V f =) V f = V/2 =) e. ΔL = MV P /(4k) 2 k( L)2 = 4 MV 2 r M =) L = 2k V 8. (7 points) After the collision, the blocks undergo simple harmonic motion. How much time does it take after the collision for the blocks to return to their starting position (spring at equilibrium length) for the first time? (Hint: the blocks will oscillate through this position repeatedly. The problem refers to the first occurrence.) M =) E tot = 2 (2M) V 2 V 2 = 4 MV 2 M k a. Δt = 4π 2M k b. Δt = 2π k 2M c. Δt = 2π 2M k d. Δt = 2π k 2M! =2 f = 2 r P = r Mtot k M tot =) P =2 k r 2M t = P/2= k e. Δt = π 2M k 9. (5 points) After the collision, the blocks undergo simple harmonic motion. If the initial velocity V is increased, the frequency of the oscillation a. decreases. b. stays the same. c. increases. The initial speed of the block will affect the oscillation amplitude. However, for simple harmonic motion, oscillation frequency does NOT depend on oscillation amplitude.

4 A T-shaped object is made by combining two identical, uniform rods of equal mass M and length L. The thickness of each rod is negligible. The object can be rotated around the four different axes shown by the dashed lines. 20. (5 points) Rank in increasing order the moments of inertia, I T to I U for rotation about the dashed lines. a. I P < I T < I U < I V b. I T < I P < I U < I V c. I T < I P < I V < I U 2. (7 points) Now consider the situation depicted above in diagram #2 and imagine that the object is oscillating as a pendulum around the axis shown. Gravity acts downward as indicated. What is the correct expression for the oscillation frequency, ω? a. ω = 3g 2L b. ω = 2g 3L I = 2 ML2 I 2 = 3 ML2 ~g I 3 = 3 ML2 + ML 2 = 4 3 ML2 I 4 = 3 ML2 + M(L/2) 2 = 7 2 ML2 Rod at top of T does not contribute to oscillation frequency because it has no moment of inertia about the given axis (width of each rod is neglected). c. ω = 2g L d. ω = 2 g L e. ω = 2 g L 22. (7 points) Now consider the situation depicted above in diagram #3. If the mass of each rod is M = 2. kg and each has a length of L =.7 m, what is the moment of inertia of the object around the axis shown? a. I V = 5. kg m 2 b. I V = 6. kg m 2 c. I V = 8. kg m 2 d. I V = 7.4 kg m 2 e. I V = 4.0 kg m 2 I 3 = I top,3 + I vert,3 = 0+MD ML2 D = L =) I 3 = ML ML2 = 4 3 ML2 = 4 3 (2. kg)(.7 m)2 =8. kgm 2

5 A uniform rod of length d and mass M is inclined at 45 with respect to the horizontal. On one end of the rod is an ideal pivot. A massless string is attached to the other end of the rod. The string runs vertically upward, then over a frictionless pulley, and a block of mass m is hung from the other end of the string. In this configuration, the system is in equilibrium. 23. (7 points) What is the mass of the block, m? a. m = 3M b. m = 2M c. m = 2M/3 d. m = M/2 e. m = M/4 24. (5 points) What is the horizontal force on the pivot? a. 0 b. Mg/ 2 c. Mg d Mg cos 45 = T (d cos 45 ) 2 d T = mg =) Mg cos 45 = mg (d cos 45 ) 2 =) m = M/2 The tension in the rope acts vertically, so the rope exerts no horizontal force on the rod. Thus, the pivot can exert no horizontal force or the rod s CM would accelerate horizontally. d 45 M g m 25. (7 points) Consider the case where the mass of the rod is M = 30 kg, and its length is d = 3 m. At a particular moment, the string is cut and the rod is free to rotate about the pivot. What is the angular acceleration, α, of the beam immediately after the string is cut? a. α = 3.9 rad/s 2 b. α = 3.47 rad/s 2 c. α = 4.9 rad/s 2 d. α = 6.94 rad/s 2 e. α = 2.45 rad/s 2 = = Mg d 2 cos 45 I rod,end 3 Md2 = 3g 2 p 2 d = p 3g = d 3 p 8 =3.47 rad/s 2

6 Mass m T initially has a speed of V \ in the +x direction along a frictionless horizontal floor. It collides with mass m P, which is initially at rest. Immediately after the collision m T is moving in the x direction with speed V TE = V \ /4 and m P is moving in the +x direction with speed V PE = V \ / (5 points) Is this collision elastic? a. There is not enough information provided to determine whether the collision is elastic. b. No, the collision is not elastic. c. Yes, the collision is elastic. 27. (7 points) Which of the following correctly expresses the relationship between the masses? a. m T = 2m P b. m T = m P /2 c. m T = (4/3)m P V approach = V 0 V separate = V 0 /4+V 0 /2=3V 0 /4 =)V approach >V separate m V 0 = m 2 V 0 /2 m V 0 /4=) 5 4 m = 2 m 2 d. m T = (3/4)m P e. m T = (2/5)m P =) m = 2 5 m (7 points) Mass m P continues to the right and slides up a frictionless ramp. It momentarily comes to rest at a maximum height h above its starting height on the floor before sliding back down. Which of the following correctly expresses the maximum height h reached by m P? a. h = V \ P /(2g) b. h = 4V \ P /(3g) c. h = V \ P /(8g) d. h = 3V \ P /(4g) e. h = V \ P /(4g) E m2,after collision = E m2,top of ramp 2 2 m V0 2 = m 2 gh 2 =) h = V 2 0 8g

7 Block A has mass M^ = 4 kg and slides on a frictionless inclined plane. It is connected to a hanging block B of mass M _ by a massless string that runs over a frictionless pulley. The incline makes an angle of θ = 30 with horizontal. The acceleration of block A up the incline is a = 2. m/s (7 points) As block A moves a distance.3 m along the incline, what is the work done on it by the tension in the string, W^,b? a. W^,b = J b. W^,b = 4.59 J c. W^,b = 55. J d. W^,b = 0.92 J e. W^,b = 0 J T M A g sin = M A a =) T = M A (a + g sin ) = (4 kg)( sin 30 ) m/s 2 = kg m/s 2 W A,T = T d = = J 30. (5 points) As block A moves a distance.3 m up the incline, how does the magnitude of the work done on it by the tension in the string W^,b, compare to magnitude of the work done on it by gravity, W^,d? a. W^,b = W^,d b. W^,b > W^,d c. W^,b < W^,d W net = W A,T W A,g = K>0 =) W A,T >W A,g 3. (7 points) Now assume that there is friction between block A and the plane and that both blocks have the same mass M^ = M _ = M. For the geometry shown, what is the minimum coefficient of static friction, μ f, needed to hold block A stationary if it is initially released from rest? a. μ g = 0.74 b. μ g = 0.67 c. μ g = 0.82 d. μ g = 0.58 e. μ g = 0.49 Block B: T = Mg Block A: T = Mgsin + f s =) f s = Mg( sin ) µ s N = µ s Mgcos = Mg( sin ) =) µ s = sin cos = =0.58

8 A child is playing with a new toy that consists of a ball of mass m = 0.87 kg tied to the end of a string of length ll = 0.3 m. The child swings the toy above her head so that it moves with uniform circular motion in a horizontal plane. The string makes an angle of 63 with respect to the vertical arm of the child. The ball should be treated as a point particle. 32. (7 points) What is the tension in the string? a. T = 8.8 N b. T = N c. T = 8.53 N d. T = 4.35 N e. T = 9.58 N F y =0=) T cos = mg T = mg = cos cos 63 = 8.8 N 33. (5 points) Say the speed of the ball in the original problem is V \. Suppose we increase mass of the ball but keep the length of the string the same. If we want the angle that the string makes with the vertical to be the same as in the original problem, what would the new speed of the ball have to be? a. V HIJ > V \ b. V HIJ < V \ c. V HIJ = V \ F x = ma cent =) T sin = m V 2 R =) mg tan = m V 2 ` sin V 2 = `g sin tan (independent of m) 34. (7 points) Now consider once again the geometry shown in the figure above. Assuming the ball travels with tangential speed V \ as it travels along the circular path shown, what is the magnitude of the angular momentum of the ball as it orbits the central vertical axis? a. L = m ll V \ sin θ b. L = m ll V \ cos θ c. L = < ll V \ P sin θ d. L = < ll V \ P tan θ e. L = m ll V \ tan θ ~L = ~p ~ R =) L = mv 0 R = mv 0` sin

9 Two houses are located on opposite sides of a river as shown. The width of the river is W = 247 m, and the river flows in the +y direction with speed V m,n =.3 m/s relative to the houses. A boat sets off from the left house, moving with velocity V _,m relative to the river. The x-component of V _,m is 2. m/s and the y-component of V _,m is 3. m/s. 35. (7 points) What is the speed of the boat as measured in the reference frame of the houses? a. V _,n = 3.74 m/s b. V _,n = 2.77 m/s c. V _,n = 4.88 m/s d. V _,n = 4.4 m/s e. V _,n = 2.47 m/s ~V B,H = ~ V B,R + ~ V R,H x : (V B,H ) x =(V B,R ) x +(V R,H ) x =2. m/s + 0 y : (V B,H ) y =(V B,R ) y +(V R,H ) y =3. m/s +.3 m/s = 4.4 m/s V B,H = p =4.88 m/s 36. (7 points) When the boat reaches the other side of the river, how far is it from the nearest house? a m b m c. 2.7 m d m e m t cross = W = 247 (V B,H ) x 2. = 7.62 s D downstream =(V B,H ) y t cross = = m 37. (5 points) Say the time it takes the boat to get to the other side in the above scenario is T \. Now suppose the driver of the boat steers in such a way that the boat moves along the x-axis, directly from the house on the left to the house on the right. If the speed of the boat relative to the water is the same as in the original problem, compare the time it takes the boat to get across the river in the new case, T HIJ, to T \. a. T HIJ > T \ b. T HIJ < T \ c. T HIJ = T \ The time it takes for the boat to cross the river is dictated by the x-component of its velocity relative to the houses (or river, since the water flows in the y-direction). In the first scenario, the x-component of the boat s velocity relative to the river is 2. m/s and the y-component is 3. m/s. In the second scenario, the boat travels directly across the river, which means the y-component of the boat s velocity relative to the water is only.3 m/s. Since the y-component of the boat s velocity relative to the water is smaller than in the first case, but its total speed relative to the water is the same as before, then its x-component of velocity must be larger and it crosses the river more quickly.

10 A football is kicked across a level field. The ball spends Δt = 4.8 seconds in the air and lands a distance X = 30 meters from the point where it was kicked. The initial speed of the ball is V \. You should ignore air resistance. 38. (7 points) What is the speed of the ball at the top of its trajectory? a. v = m/s b. v = 3.06 m/s c. v = m/s V x = X t d. v = 6.25 m/s e. There is not enough information given to determine this. = 30/4.8 =6.25 m/s 39. (7 points) What is the maximum height reached by the ball? a. h = m b. h = 7.5 m c. h = 56.5 m d. h = 5 m e. h = 0 m t peak = V y (V 0 ) y = gt peak =) (V 0 ) y = gt peak h =(V 0 ) y t peak 2 gt2 peak = gt 2 peak t 2 =) h = g t 2 8 = m 2 gt2 peak = 2 gt2 peak 40. (5 points) After the kick, but before the ball hits the ground, which of the following statements best describes the acceleration vector of the ball? a. It always points toward the right. b. It points upward before the ball reaches its maximum height, then downward. c. It always points downward.