Seat: PHYS 1500 (Spring 2007) Exam #3, V1 Name: 5 pts 1. A pendulum is made with a length of string of negligible mass with a 0.25 kg mass at the end. A 2nd pendulum is identical except the mass is 0.50 kg at the end. The period of the first pendulum (a) is the same as that for the second pendulum. (b) is shorter than that for the second pendulum. (c) is longer than that for the second pendulum. (d) can not be obtained relative to the second pendulum without knowing the length of the string. 5 pts 2. Stars originate as large bodies of slowly rotating gas. Because of gravity, these clumps of gas slowly decrease in size. What happens to the angular speed of a star as it shrinks? Explain. Since there are no external torques the total angular momentum is conserved. This means that mrv for all of the atoms must not change when added up. Since the size of the star is shrinking, the velocity must increase. This means the angular speed of the star must increase because ω = v/r. 5 pts 3. You place 200 g of ice at 0 C into a cup that is at 40 C. The cup has a mass of 2 kg. After a time the ice melts and the ice and cup arrive at the same temperature, 10 C. What is the specific heat of the cup? (For water, c=4.0 10 3 J/(kg C); L f = 3.5 10 5 J/kg; L v = 2.2 10 6 J/kg, ρ = 1000 kg/m 3, e = 0.1.) For this situation, Q 1 + Q 2 = 0. Take Q 1 to be the heat into the ice/water. Then, Q 1 = m i L f + m i c w T w. The expression for Q 2 = m c c c T c = Q 1. Q 1 = 0.2 kg 3.5 10 5 J J +0.2 kg 4.0 103 kg kg C 10 C = 7 10 4 J+8 10 3 J = 7.8 10 4 J Q 2 = 2 kg c c ( 30 C) = 6 10 1 kg C c c c c = 7.8 104 J 6 10 1 kg C = 1.3 J 103 kg C (V 1) Q 1 = 0.2 kg 3.5 10 5 J J +0.2 kg 4.0 103 kg kg C 20 C = 7 10 4 J+1.6 10 4 J = 8.6 10 4 J Q 2 = 2 kg c c ( 43 C) = 8.6 10 1 kg C c c c c = 8.6 104 J 8.6 10 1 kg C = 1.0 103 J kg C (V 2)
10 pts 4. You have a cube of Aluminum with an edge of 10 cm. It floats in a fluid so that 8 cm is below the fluid. What extra force needs to be exerted on the cube so that it is completely submerged and stationary? There are several clever ways of solving this problem. The method I describe is not one of them. You have a metal cube that floats on a mystery fluid with 80% submerged. This lets you find the density of the fluid from the fact that the force from gravity on the cube plus the buoyant force must add up to 0. ρ met V met g + ρ liq [0.8 V met ]g = 0. This means ρ liq = (5/4)ρ met. When you are holding the metal cube under the liquid, the forces are ρ met V met g + ρ liq V met g + F = 0. Using the expression for the density of the liquid gives the force that needs to be exerted: F = (ρ met ρ liq )V m etg = (1/4)ρ met V met g. The force is negative meaning that the force has to be directed downward. F = 1 kg 3000 4 m (0.1 3 m)3 10 m s = 30 2 4 N = 7.5 N (V 1) F = 1 kg 8000 4 m (0.1 3 m)3 10 m s = 80 2 4 N = 20 N (V 2)
5 pts 5. Roughly, one atmosphere of pressure is (a) 10 5 N/m (b) 10 5 N/m 2 (c) 10 5 N/m 3 (d) 10 9 N m (e) 10 9 N m 2 (f) 10 9 N m 3 5 pts 6. The masses and volumes of 6 objects are shown. Rank them, from greatest to least, on the basis of buoyant the force on the blocks by the water. Explain your ranking. Remember to indicate any ties. A B C D E F M= 60 g V = 40 cm 3 M= 80 g V = 40 cm 3 M= 60 g V = 10 cm 3 M= 60 g V = 10 cm 3 M= 80 g V = 10 cm 3 M= 60 g V = 40 cm 3 For this case, the buoyant force only depends on the volume of the object. So B, D, E are the same and larger than, A, C, F. 5 pts 7. When the temperature is 20 C, a copper rod is 2.00000 m long. To what temperature should the copper be at for the rod to be 2.00030 m long? The relation that governs this is L = L 0 α T. You are given L 0, L, α and T 0. Using the relationship above, you can find T and add it to T 0. T = T = 4 10 4 m 2 m2 10 5 C 1 = 10 C T f = 30 C (V 1) 3 10 4 m 2 m3 10 5 C 1 = 5 C T f = 45 C (V 2)
10 pts 8. The weight m 1 = 5.0 kg is resting on a frictionless surface. The weight m 2 = 10 kg. The moment of inertia of the pulley is 3.0 kg m 2 and has a radius of 0.5 m. If the system starts at rest, how far has m 1 traveled when it reaches the speed 2 m/s? This problem can be solved several ways but the easiest is to use conservation of energy ideas. When the system starts, the KE is 0 and we can define the PE to be 0. After the weight has moved a distance d, the KE = 1/2m 1 v 2 + 1/2m 2 v 2 + 1/2Iω 2 where ω = v/r for the pulley. The potential energy decreases and is given by m 2 gd. Since KE + P E = 0, this means m 2 gd = KE. This can be rearranged to solve for d = KE/(m 2 g). ω = 2 m/s 0.5 m = 4rad s KE = 1 2 15 kg(2 m/s)2 + 1 2 3 kg m2 (4 rad/s) 2 = 54 J d = 54 J = 0.54 m (V 1) 10 kg 10 m/s2 ω = 2 m/s 0.5 m = 4rad s KE = 1 2 15 kg(2 m/s)2 + 1 2 2.5 kg m2 (4 rad/s) 2 = 50 J d = 50 J = 1.0 m (V 2) 5 kg 10 m/s2
5 pts 9. You add 335 J of heat to 8 kg of ice at 0 C and 1 g of the ice melts. If you add 335 J of heat to 4 kg of ice at 0 C, (a) 1/2 g of ice melts. (b) 1 g of ice melts. (c) 2 g of ice melts. (d) 4 g of ice melts. 5 pts 10. At the Joint Institute for Laboratory Astrophysics, they perform extremely sensitive experiments that depend on (for example) the floor remaining perfectly horizontal. During an experiment, they noticed that in the morning the building was slightly tilted away from vertical toward the west and in the evening the building was slightly tilted away from vertical toward the east. The effect only occured on sunny days. Explain. The side of the building that the sun is on heats up compared to the other side. This causes a slight expansion in the material on that side. This means the building tilts away from that side. 5 pts 11. Water flows with a mass flow rate of 400 kg/s in a pipe. The pipe is 10 km long. Find the water speed where the pipe has a square shape that is 20 cm on edge. (Water properties: density 1.00 10 3 kg/m 3, latent heat of fusion 33.0 10 4 J/kg, latent heat of vaporization 22.0 10 5 J/kg). The mass flow rate is ρva. For a square A = L 2. v = 400 kg/s 1000 kg/m 3 (0.2 m) = 0.4 2 0.04 m s = 10 m/s (V 2) v = 800 kg/s 1000 kg/m 3 (0.2 m) 2 = 0.8 0.04 m s = 20 m/s (V 2)
10 pts 12. Two blocks, each 10 kg, are resting on a horizontal frictionless surface. Between the two is squeezed a spring. The spring is compressed 20 cm from its unstrained length and is not permanently attached to either block. When the spring is released both blocks move apart with a speed of 4 m/s. What is the spring constant of the spring? The simplest way to solve this problem is to use conservation of energy. The energy before the spring is released is all PE. After the spring is released, the energy is all KE. Since there are two blocks KE = 2 (1/2)mv 2. The potential energy is only from the spring and is given by P E = (1/2)kx 2. The spring constant can be found from setting these two equal to each other k = KE/([1/2]x 2 ) KE = 10 kg (4 m/s) 2 = 160 J k = KE = 9 kg (2 m/s) 2 = 36 J k = 160 J = 8000 N/m (V 1) (1/2)(0.2 m) 2 36 J = 7200 N/m (V 2) (1/2)(0.1 m) 2
Equations Material density Linear Expansion Coef Therm Conductivity Aluminum 3000 kg/m 3 3 10 5 C 1 250 J/(s m C) Steel 8000 kg/m 3 1 10 5 C 1 80 J/(s m C) Copper 9000 kg/m 3 2 10 5 C 1 400 J/(s m C) Basic Mathematic Formulas sin θ = h o /h cos θ = h a /h tan θ = h o /h a h 2 = h 2 o + h 2 a A circ = πr 2 Circum of circ = 2πr V sph = 4π 3 r3 A sur of sph = 4πr 2 cos 37 = 4/5 sin 37 = 3/5 cos 53 = 3/5 sin 53 = 4/5 Chapter 2 Avg speed = distance/elapsed time g = 10 m/s 2 x = x f x i v = x/ t ā = (v f v i )/(t f t i ) = v/ t v = v 0 + at x = 1 2 (v 0 + v)t x = v 0 t + 1 2 at2 v 2 = v0 2 + 2a x Chapter 3 A x = A cos θ A y = A sin θ A = A 2 x + A 2 y tan θ = A y /A x r = r f r i v av = r/ t a av = ( v)/ t v x = v 0x + a x t x = 1 2 (v 0x + v x )t x = v 0x t + 1 2 a xt 2 vx 2 = v0x 2 + 2a x x v y = v 0y + a y t y = 1 2 (v 0y + v y )t y = v 0y t + 1 2 a yt 2 vy 2 = v0y 2 + 2a y y Chapter 4 F = m a Fg = G m 1m 2 r 2 11 N m2 G = 6.67259 10 kg 2 w = mg F 12 = F 21 f s µ s n f k = µ k n Chapter 5 W = (F cos θ) x KE = 1 2 mv2 W net = KE f KE 0 W g = mg(y i y f ) PE = mgy W nc = KE f KE i + PE f PE i P = W t P = F v Chapter 6 p = m v I = F t F t = p = m vf m v i m 1 v 1i + m 2 v 2i = m 1 v 1f + m 2 v 2f
ω av = θ f θ i t f t i = θ t Chapter 7 α av = ω f ω i t f t i = ω t ω = ω i + αt θ = 1 2 (ω i + ω)t θ = ω i t + 1 2 αt2 ω 2 = ω 2 i + 2α θ v t = rω a t = rα a c = v2 r = rω2 T 2 = 4π2 GM S r 3 Chapter 8 τ = rf sin θ I mr 2 τ = Iα KEr = 1 2 Iω2 W nc = KE t + KE r + P E Chapter 9 L Iω τ = L t ρ = M V P = F A B = ρ fl V fl g ρ 1 A 1 v 1 = ρ 2 A 2 v 2 P 1 + 1 2 ρv2 1+ρgy 1 = P 2 + 1 2 ρv2 2+ρgy 2 Chapter 10 T C = T 273 T F = 9 5 T C + 32 L = αl 0 T A = γa 0 T V = βv 0 T Chapter 11 Q = mc T Qk = 0 Q = ±ml P = ka T h T c L P net = σae(t 4 T 4 0 ) Chapter 12 W = P V U = U f U i = Q + W W eng = Q h Q c e = W eng Q h COP (cool) = Q c W COP (heat) = Q h W e c = 1 T c T h S = Q r T Chapter 13 and 14 F s = kx P E s = 1 m 2 kx2 T = 2π k f = 1 T ω = 2πf x = A cos(ωt) L v = Aω sin(ωt) a = Aω 2 cos(ωt) T = 2π g v = fλ v = F µ f n = nv 2L