PY105 Final Exam Makeup

December 20, 1999

Name:_______________________ Student Number:_________________

Signature:___________________________

Section: [ ] A - Duffy [ ] B -Goldberg [ ] C-Rothschild [ ] D- Narain

The test is 120 minutes long. There are eight questions, each worth 15 points, for a total of 120 points. Write your answers in the space provided. If you need more space, write on the back of the pages. Showing your work clearly will maximize your chances of receiving partial credit. If you do not show your work, you may not be given full credit, even if your answer is correct. Remember to state all answers with the appropriate number of significant digits, and with appropriate units.

This is a closed-book test, but an equation sheet is provided. Good luck!

For staff use only: scores on each problem

Question 1 (/15): _____________

Question 2 (/15): _____________

Question 3 (/15): _____________

Question 4 (/15): _____________

Question 5 (/15): _____________

Question 6 (/15): _____________

Question 7 (/15): _____________

Question 8 (/15): _____________

Total (/120): ______________

[15 points] Problem 1. Pulleys and Newton's Laws

N

T

T

1. Two masses are attached together by a massless rope that runs over a massless pulley as shown to the right. M₁ = 10kgs and M₂ = 15kg. The coefficient of kinetic friction between M₂ and the incline plane is m_k= 0.2.

T

a. [5 pts ] Draw the free body diagram for the two masses.

M₂g

b. [5 pts] Write down Newton's second law equations for each of the masses.

For M1:

M₁g – T = M₁ a (1)

For M2:

M₂g cos(30^o) = N (2)

T-M₂g sin(30^o) - m_k N = M₂ a (3)

Or

T – M₂g(sin (30^o) + m_k cos(30^o) = M₂a (4)

c. [3 pts] Calculate the acceleration of the mass M₁.

(1) + (4) è M₁g – M₂g(sin (30^o) + m_k cos(30^o) = (M₁+M₂) a

a =( M₁g – M₂g(sin (30^o) + m_k cos(30^o) ) / (M₁+M₂) = 3.2 m/s²

11

11

[15 points] Problem 2. Springs, slides and collisions

A massless spring attached to a wall with spring constant K equal to 2.0 N/m is compressed a distance of 1 meter from its equilibrium position by a mass m₁ = 2.0 kg. After the spring is released, m₁ moves across a frictionless, horizontal surface, slides down a frictionless slope, which drops a height h =0.5 m and then collides with a second mass, m₂ = 4.0 kg, which is at rest.

1. [5 points] What is the speed of m₁ just before it begins to go down the slope.

Potential energy of the spring transfer to Kinetic energy of the mass:

½ K A² = ½ m₁ v²

è v = sqrt(K A²/m₁) = 1 m/s

2. [5 points] What is the speed of m₁ when it reaches the bottom of the slope?

Gravity potential energy is added to the kinetic energy of the mass:

½ m₁ v²_f = ½ m₁ v²_i + m₁gh

è vf = sqrt(v²_i + 2gh) = sqrt(1 + 2 *0.5 *9.8) = 3.3 m/s

3. [5 points] Assume that m₁ sticks to m₂ when they collide, what is their speed after the collision?

In the collision, the moment is conserved.

m₁ v_i = (m₁ + m₂) v_f è v_f = m₁v_i / (m₁+m₂) = 1*3.3/(2 + 4) = 1.1 m/s

[15 points] Problem. 3. Rotational dynamics

A)[ 6 points]

i)The earth revolves around the sun in an elliptical orbit. As the earth moves closer to the sun, the orbiting speed of the earth

[ ] does not change [ X ] increases [ ] decreases

[ ] it changes, but impossible to tell which way

* Potential energy for the earth-sun system P = - Mm/R, when R is smaller, P is smaller, to conserve the energy,

The kinetic energy have to increase , so that the speed increase.

ii)As a rocket moves further away from the surface of the earth, the weight of the rocket

[ ] does not change [ ] increases [ X ] decreases

[ ] it changes, but impossible to tell which way

* outside of the shell of the earch, mg = GmM/R, R increases, mg decreases

iii) The moment of inertia of a solid cylinder about its axis is given by 0.5 MR².

If this cylinder rolls without slipping, the ratio of it rotational kinetic energy to its translational kinetic energy is

[ ] 1:1 [ X ]1:2 [ ] 2:1 [ ]1:3

Et = ½ m v²

Er = 1/2 I w² = ½ (0.5)mR² w² = ¼ m v² = ½ Et

Er/Et = 1/2

(v = wR)

B) A cylindrical shaped communications satellite of mass 1350 kg is set spinning at 2.21 rev/s about the center axis and then launched from the shuttle cargo bay. The dimensions of the satellite are : diameter = 1.56 m and length = 1.75 m. Calculate the satellite’s

a) [ 3 points] Rotational Inertia about the rotation axis:

I = ½ M R² = 0.5 * 1350 * (1.56/2)² = 410 kg m²

b) [ 6 points ] Rotational kinetic energy

w = 2 p / T = 2 p f = 2 * 3.14 * 2.21 = 13.9 /s

E = ½ I w² = 0.5 * 410 * (13.9)² = 3.96 * 10⁴ J

[15 points] Problem. 4. Static equilibrium - Walking the plank

Problem 4 [15 points] – Walking the plank

(a) [8 points] If you have a mass of 50 kg, what maximum distance x beyond the roof can you stand on the beam without tipping it over?

The edge is the rotation center, so all torques are related to that point.

At the maximum distance x, the normal force is only at the edge, which gives no torque. So only gravity of the beam, bucket and man should be considered. Obviously the center of the beam is 0.5 meter to the right of the edge. So we have the following equation:

40 * 2 = 80 * 0.5 + 50 * x

è x = 0.8 m

(b) [4 points] With you standing on the beam, what is the magnitude of the normal force applied by the roof on the beam?

Consider the beam, bucket and man together, all weight adding together is equal to the normal force.

N = (40 + 50 + 80) g = 1.7 * 10³ N

(c) [3 points] With you standing on the beam at x, the maximum distance point you determined in part (a), where is the normal force applied by the roof on the beam?

[ X ] at the very edge of the roof [ ] at the beam’s center-of-gravity

[ ] evenly distributed over the 2.0 m of the beam in contact with the roof

[ ] 1.0 m to the left of the edge of the roof

[15 points] Problem. 5. Hydrodynamics

A fluid flows through a pipe as shown in the diagram above. Assume the fluid is nonviscous and incompressible.

1. [ 5 points] If the velocity of the fluid is 10 m/s at point 2, which has a cross sectional area of 2.0 cm² , what is the velocity of the fluid in the pipe at point 1 where it is wider with a cross sectional area 10 cm².

Mass is conserved in the pipe.

r A₁V₁ = r A₂V₂

è A₁V₁ = A₂V₂

è V1 = A₂V₂/A₁ = 2 m/s

2. [5 points] If the difference in pressure between point 1 and 2 (P₁ - P₂).is 50,000 N/m², what is the density of the fluid?

Bernoulli’s equation:

½ r V₁² + P1 = ½ r V₂² + P2

è r = 2*(P1-P2) /(V₂² – V₁²) = 1.04 * 10³ Kg/m³

3. [5 points] A small tube which connects points 1 and 2, as shown in the diagram, is partially filled with a fluid. The fluid on the right side of this small tube rises to a level h higher than on the left side. Explain how you would find the density of this fluid, r, using the information given in this problem (write an expression for r as part of your answer).

The pressure difference P1-P2 have to be canceled by rgh so that this fluid can be stable

è P1-P2 = r g h è r = (P1-P2)/(gh)

[15 points] Problem 6. Simple Harmonic motion

A) [2 point] A mass at the end of a spring oscillates both on the moon and the earth. It’s

period on the moon compared to that on earth is

[ ] larger [ ] smaller [ X ] same

T only depends on mass and k.

B) [ 4 points] A mass is attached to a vertical spring which executes simple harmonic motion between two points A and B. Where is the mass located

i) When it’s kinetic energy is a maximum

[ ] At either A or B [ X ] midway between A and B

[ ] quarter of the way between A and B, measured from either A or B

[ ] None of the above

ii) When the spring potential energy is a maximum

[ X ] At either A or B [ ] midway between A and B

[ ] quarter of the way between A and B, measured from either A or B

[ ] None of the above

C) [2 point] A simple pendulum’s period is measured on the earth and the moon. It’s

period on the moon compared to that on the earth is

[ X ] larger [ ] smaller [ ] same

T = 2 p sqrt(l/g), g decreases, T increases

D) [2 points] The pendulum of a grandfather’s clock is 1.5m long. It’s period on earth is

[ ]1.0s [ ]1.5 s [ ]2.0s [ X ]2.5s

E) In an electric shaver the blade moves back and forth over a distance of 2.4mm. If this motion is described as a simple harmonic motion with a frequency of 110 Hz, then find :

a) [2 points] Amplitude:

A = 1.2mm (moves between –A and A

b) [3 points] The maximum acceleration of the blade

Maximum acceleration happens at the maximum position:

F = kA = ma è a = A k/m = w²A = (2p f)² A = 570 m/s²

[15 points] Problem 7. Calorimetry and thermal expansion

A) A 1.5kg mass of ice cubes at -25 ^oC is added to a 1.0kg mass of water at +25 ^oC. The specific heat of ice is 2100 J/kg ^oC. The specific heat of water is 4186 J/kg^oC. The latent heat of fusion is 3.33 x10⁵J/kg

a. [2 pts] How much heat is required to warm the ice to 0 ^oC?

Q1 = m c DT = 1.5 * 2100 * 25 = 78750

b. [2 pts] How much heat is lost in cooling the water to 0 ^oC?

Q2 = m c DT = 1 * 4186 * 25 = 104650 J

c. [3 pts] When the ice and water are allowed to reach thermal equilibrium, what fraction is ice?

The extra heat that can melt the ice is

Q2 – Q1 = (104650-78750) = Dm L = 25900

è Dm = 0.78 kg

So now we only have 1.5 – 0.78 = 0.72 kg ice, which is

0.72 / (1.5+1) = 0.29

B) On the P vs. V diagram below,

a. [2 pts] Draw two isotherms, and label one hot and one cold.

( PV = nRT )

b. [2 pts] Choose a point A on the hotter isotherm and draw an isothermal expansion to a point B. What is the relation between the work done by the gas and the heat input during this part of the cycle?

Q = DU + W

Since DU = 0, Q= W

c. [2 pts] Draw an adiabatic expansion to the colder isotherm, and describe the relationship between the internal energy of the gas and the work done by the gas during this part of the cycle.```

Q = DU + W

Since for adiabatic expansion , Q = 0, so W = -DU. That means the work done by the gas is equal to the decrease of internal energy (DU is negative).

d. [2 pts] Complete the cycle with an isothermal compression followed by an adiabatic compression back to the point A. Write down the expression for the efficiency of this cycle in terms of the temperature of the hot and cold isotherms.

e = 1 – T_c / T_h

[15 points] Problem 8. Thermodynamics- A heat pump

A heat pump is a device that can extract heat from cold air and use

it to heat a house very efficiently. The P-V graph for one cycle of a heat pump is shown.

Two of the processes take place at constant volume, while the other two take place at constant pressure.

(a) [3 points] For a complete cycle of the heat pump, the work done by the system is …

[ ] positive [ X ] negative [ ] zero

Because it expands at low pressure, but is compressed at high pressure. In terms formula, it

Is P1 DV – P4 DV = -(P4-P1) DV < 0, since P4-P1 > 0.

Q = 360 J of heat is added to move the system from state 1 to state 2 at constant pressure.

(b) [4 points] How much work is done by the system in this process?

We can not calculate work until we know V2. V2 can be got by the following procedure:

First we have the state equation, so we have

V1/T1 = V2/T2 (1)

We don’t know T2 yet, but we have energy conservation equation

Q = DU + W (2)

And we know

DU = 3/2 n R DT = 3/2 n R (T2 – T1) (3)

and

W = P1 DV = P1 (V2 – V1) (4)

nR can be calculated from

PV = nRT è nR = P1V1/T1 (5)

= 1.2 * 350*10^-6 / 423 = 10^-6

Plug (3) and (4) and (5) into (2), we have

Q = 3/2 P1V1/T1*(T2-T1) + P1(V2-V1) = 1.5 P1V1(T2/T1-1)+P1(V2-V1) (6)

But from (1), we can have

T2/T1 = V2/V1 (7)

Plug (7) into (6), we have

Q = 1.5 P1V1(V2/V1-1)+P1(V2-V1)=2.5 P1(V2-V1) (8)

è V2-V1 =Q / (2.5 P1) = 360 / (2.5*1.2*10⁵) = 1200 cm³

è è V2 = V1 + 1200 = 1550 cm³

Now plug V2 into (7) to get T2:

T2 = T1 * V2/V1 = 1873K

(c) [4 points] What is the temperature, T, of the system in state 2?

(d) [4 points] What is the volume, V, of the system in state 2?