TA's name (circle one) Paul Beale, Satyan Bhongale, Justin Freeman, David Hall,

1. You throw a ball straight up with initial velocity 8 m/s. How high will it go?
1. 3.2 m
2. 0.8 m
3. 1.3 m
4. 8.0 m
5. 10. m

vf^2=vi^2+2ad, so 0=8^2-2*10*d, d=(64m^2/s^2)/(20m/s^2)=3.2m

2. An astronaut on the moon (where there is no air) throws a projectile as shown.

(Note: Point A is already after the object has been released.

Point B is the very top of the trajectory.

Point C is higher than Point A, but the object is on its way down there)

At which of the labeled points on the trajectory is the speed the largest?

1. A
2. B
3. C
4. same at all three points.

x component is same everywhere, y component decreases with height, speed is a scalar (apways a positive #)

3. At which of the labeled points on the trajectory is the acceleration of the object the largest?
1. A
2. B
3. C
4. same at all three points

Acceleration is g everywhere, or in this case g_moon.

4. You push a 2.0 kg box horizontally along the ground. It moves with a constant speed of 3.5 m/s. There is friction resisting the motion, the coefficient of friction is. What force must you be exerting on the box?
1. 0 N
2. 7 N
3. 8 N
4. 10 N
5. 20 N

F_fr = mu_k*N = mu_k*m*g = 0.4*2*10=8 N. Since v is constant, the NET force on the box must be zero (by Newton II), so you must be exterting a force which exactly cancels F_fr.

5. In the previous problem, you now double the force you exert. The box
1. continues to move with the constant speed=3.5 m/s
2. moves with a new constant speed=7.0 m/s,
3. accelerates, with a=10. m/s^2
4. accelerates, with a=4. m/s^2
5. (motion is impossible to determine

F_net = 2*8-F_fr = 16N-8N=8 N = ma = 2kg*a, so a=4 m/s^2

The next 2 problems refer to the following scenario:

You pull a 0.33 kg cart along an air track with a constant force F=3N, at a fixed angle, =60 degrees, as shown.

The cart starts at rest.

(Neglect friction, and treat the cart as a point object)

6. What is the speed of the cart after it has moved horizontally through a distance =4.0 m?
1. 4.5 m/s
2. 9.1 m/s
3. 8.5 m/s
4. 6 m/s
5. Not enough information given

Work-energy says 1/2 m v^2 = work = F*d*cos(theta) = 3N*4m*cos(60)=6J. So,

v=Sqrt[2*6J/0.33 kg] = Sqrt[36 m^2/s^2] = 6 m/s

7. If I had continued applying the force F until the cart had moved horizontally a distance 16 m, (4 times farther than the previous problem), the final speed of the cart would have been
1. 1/4 the answer to the previous problem
2. the same as the previous problem
3. 2 times more than the answer to the previous problem.
4. 4 times more than the answer to the previous problem.
5. 16 times more than the answer to the previous problem.

As above, since v^2 goes like d, if d increases by 4, v increases by 2.

8. A particle moves with constant acceleration. Its velocity vectors at two different times (v₁ is earlier than v₂) are shown. The direction of the acceleration is
1. Right (along the +x direction)
2. Left (along the -x direction)
3. Down (along the -y direction)
4. Up (along the +y direction)
5. None of these

a = Delta v / Delta t, where Delta v = v2 - v1. This is an arrow pointing from the tip of v1 to the tip of v2, directly to the right

.

9. An elevator is being lifted up by a cable. The elevator rises at a constant speed, v=3 m/s. If the elevator has a total mass of 800kg, what is the tension in the cable?

Hint: Draw a force diagram. What is the acceleration of the elevator? (Neglect friction!)

1. zero
2. 11600 N
3. 2400 N
4. 3600 N
5. 8000 N

Constant speed means no acceleration. So, F_net=0. Thus, T=mg=800kg*10 m/s^2 = 8000 N

10. A person standing in an elevator has a mass of 60 kg. What is her apparent weight in each of the following two situations:

(the upward speed is a constant 3m/s) ; (the elevator accelerates upwards at 0.2g). (570 N) ; (600 N)

1. (570 N) ; (600 N)
2. (630 N) ; (600 N)
3. (600 N) ; (600 N)
4. (600 N) ; (480 N)
5. (600 N) ; (720 N)

Constant speed => NO CHANGE in apparent weight. (So, N=mg=60*10=600N) Upward acceleration means N-mg = ma = mg*0.2, orN=mg(1.2)=60*10*1.2=720 N

The next 2 problems refer to this scenario:

Suppose the space shuttle is in a circular orbit around the earth, a distance h=200 km above the surface of the earth.

11. True (A) or False (B): The magnitude of the acceleration of the shuttle depends on its mass.

F_shuttle=m_shuttle * a=GM_earth m_shuttle / R^2. The m_shuttle cancels out, a doesn't depend on it

12. Suppose the orbit of the shuttle is increased to h'=400 km above the surface of the earth. (Note: the radius of the earth is about 6000 km) Once it is in this new orbit, the acceleration of the shuttle is
1. the same as before
2. larger by a factor of 2
3. smaller by a factor of 2
4. smaller by a factor of 4
5. none of the above.

Using the equation above, a=G M_earth/R^2. R is the distance to the *center* of the earth, NOT the distance to the surface. So, R has gone from 6200 km to 6400 km, which does make a change in the acceleratio,n but not a whole lot (the change is approximately 6%)

13. A wheel is angularly accelerated (by a constant torque) from rest up to a final angular speed of = 0.20 rad/sec. During this period of acceleration, the wheel makes exactly three revolutions. How much time did it take to get up to the final angular speed?
1. 5 sec
2. 15 s
3. 31 s
4. 94 s
5. 188 s

Several ways to get this. Easiest is AVERAGE omega = (omega_f + omega_i)/2 = Delta theta/delta t.

So, (0+0.2 rad/s)/2 = 3*(2*pi rad) / Delta t. Solve for delta t = 6*pi/(/0.1 rad/s)=188 s

14. A heavy truck and a light car are moving at the same speed (opposite directions) when they collide into one another head on. After the collision, they stick together. Neglect any outside forces (there is no friction with the road.) Which ONE of the following statements is TRUE? (All statements are in the frame of reference of a viewer standing on the roadside.)
1. During the collision, the magnitude of the force of the car on the truck exactly equals the magnitude of the force of the truck on the car.
2. Mechanical energy is conserved during this collision.
3. The total momentum of the system before the collision is zero.
4. Immediately after the collision, the wreck is at rest.
5. The collision is elastic.

Note: Moment is initially NOT zero - the truck has more mass, same v, thus more mv. So adding the two momenta (car and truck) does not give zero. Crashes where things stick are *never* elastic, some energy must be lost to heat, noise, etc.

15. A small (5 kg) mass is suspended by two ropes as shown. Cord #1 is attached to the roof, and tilts down with a 30 degree angle. Cord #2 is perfectly horizontal.

(The mass is in equilibrium)

What is the tension in cord #1? (the tilted cord)

1. 25 N
2. 100 N
3. 29 N
4. 50 N
5. 87 N

T_1 has two components. The upward component, T_1*sin(30) must balance mg, because there are no other vertical forces on the mass, and the NET force on it must be zero. To T_1*0.5 = 5*10, or T_1=100 N.

16. A rigid bar of length L has a hinge at the left end, about which it can rotate. Four forces, A, B, C, or D, can be applied to the bar. All 4 forces have the same magnitude F, but are applied in different locations at different angles, as shown in the figure. (There is NO gravity in this problem)

Which force (A, B, C, or D) produces a torque around the hinge with the greatest magnitude?

(answer E if two or more forces tie for greatest magnitude of torque)

Torque = r_perp*F. F is the same everywhere, so you just need to figure out r_perp. It's L/4 for A,

L/2 for B, and zero for D. For C, it's L*sin(45) = 0.7 L. That's the biggest! It's C

The next 4 problems refer to the following scenario:

A mass on a spring oscillates back and forth in simple harmonic motion. The system originally has spring constant k, amplitude of motion , period T, maximum speed , and total energy E.

17. If the amplitude of motion is doubled (i.e. amplitude=2), the period becomes
1. T
2. 2 T
3. 4 T
4. (1/2) T
5. (1/4) T

T=2Pi*Sqrt[m/k]. Amplitude is irrelevant.

18. If the amplitude of motion is doubled (i.e. amplitude=2), the maximum speed becomes
1. 
2. 2
3. 4
4. (1/2)
5. (1/4)

Energy conservation: 1/2 m v_max^2 = 1/2 k A^2, so v_max is proportional to A.

19. If the amplitude of motion is doubled (i.e. amplitude=2), the total energy becomes
1. E
2. 2E
3. 4E
4. (1/2)E
5. (1/4)E

Energy = 1/2 k A^2, so doubling A quadruples Energy.

20. If the spring constant is quadrupled (i.e. it becomes 4k) but everything else is left as it originally was, the period becomes
1. T
2. 2T
3. 4T
4. (1/2)T
5. (1/4)T

T=2Pi*Sqrt[m/k], so if k goes up by 4, T goes down by Sqrt[4] = 2.

21. A grandfather clock is 0.4 m long. The pendulum's mass is 5.0 kg. What is the clock's period on the moon, where the acceleration of gravity is 1/6 that of earth?
1. 3.1 s
2. 4.4 s
3. 7.5 s
4. 1.3 s
5. 0.21 s

T=2Pi*Sqrt[L/g] = 2Pi*Sqrt[0.4m/(10/6 m/s^2)] = 3.1 s.

(Note: I made a little mistake here; REAL grandfather clocks tick only once a second (i.e. period is 2 sec) on earth, and the #'s above don't quite work out. I announced this during the exam - since you can compute the answer, it doesn't matter...)

22. Your grandfather clock is losing time (it's running too slow). What should you do?
1. Decrease the weight of the pendulum
2. Shorten the length of the pendulum.
3. Increase the amplitude of the swing.
4. Increase the weight of the pendulum.
5. None of the above will help.

T=2Pi*Sqrt[L/g], so smaller L means smaller period means it ticks faster.

23. A wave moves along a string with wavelength and frequency f. A second wave on an identical string (with identical tension) has twice the wavelength. What is the frequency of the second wave?
1. f
2. 2f
3. f/2
4. f/4
5. (not enough information given)

lambda*f=v.( Identical string and identical tension here tells you speeds are also identical), so if lambda goes up, f must go down proportionally.

24. The diagram shows a snapshot of a wave at some given time. The frequency of this wave is 120Hz. What is the speed of the wave?
1. 540 m/s
2. 4.5 m/s
3. 3.0 m/s
4. 360 m/s
5. (not enough information given)

v = lambda*freq. Here, 3/2 lambda = 4.5, or lambda = 3.0 m, so v = 3.0 m * 120 /s = 360 m/s.

25. Suppose the "A" string on a piano is one meter long, and has a mass of 0.008 kg. The frequency of the fundamental of this string is 440 Hz. What tension in the wire is needed?

HINT: Piano strings are fixed at both ends.What is the wavelength of the fundamental vibration?

(It's NOT 1m!)

1. 3,100 N
2. 6,200 N
3. 1.2E4 N
4. 5.5E4 N
5. 790 N

lambda is 2 m. (Fundamental has lambda = 2L) So v = Sqrt[T/(m/L)] = lambda*freq = 2m*440/s.

Thus, T = (m/L)*(880 m/s)^2 = (0.008 kg)*880^2 m^2/s^2 / (1 m) = 6200 N.

26. A guitar string has a fundamental frequency of 500Hz. Which of the following frequencies would most likely set the string into resonant vibration?
1. 1750 Hz
2. 100 Hz
3. 750 Hz
4. 250 Hz
5. 1000 Hz The fundamental is 500 Hz, and resonances (standing waves) occur at integer multiples of the fundamental, i.e. n*f_1. (So, 1000 Hz, 1500 Hz, 2000 Hz, etc) e is the only answe that fits this description.

You got a summer job as "crown tester" on a remote Greek island. Use the following chart to answer the following three questions. (Remember to use g=10 m/s^2 when needed)

Material	Density (in kg/m^3)	Heat Capacity (in J/(kg deg C)
Aluminum	2.7 E3	900.
Copper	8.9 E3	390.
Steel	7.8 E3	455.
Lead	11.3 E3	130.
Gold	19.3 E3	230.
Water	1.0 E3	4186.

27. You are given a solid cube of metal (raw material for a crown.) The cube is 10.0 cm on a side, and it weighs 89.0 N. Which material above is the cube most likely to be?
1. Aluminum
2. Copper
3. Steel
4. Lead
5. Gold

Density = mass/volume = (89N/10 m/s^2)/(0.1m)^3 = 8900kg/m^3, which is Copper.

28. You are given a crown, which weighs 40.0 N. You submerge it in water, and its apparent weight becomes 37.93 N. Which material above is the crown most likely to be?
1. Aluminum
2. Copper
3. Steel
4. Lead
5. Gold

Call T the normal weight, and T'' the apparent weight in water.

You have T=mg, and T''=mg-F_b= mg-(rho_water*V*g). Combining these gives

T/(T-T'') = mg/[mg-(mg-rho_water*V*g)]=mg/(rho_wather*V*g).

But, m = rho_stuff*V, so the V and the g cancels leaving

rho_stuff/rho_water = T/(T-T'') = 40/(40-37.93) = 19.3.

This is Gold.

29. You are given another crown, with a mass of 5.1 kg. It is at room temperature, 20.0 C. You submerge it in 0.01 m^3 of cold water (at 1.0 C) The water is well insulated, so heat can only flow between the crown and water. The final temperature of the crown+water is 2.0 C .

Which material above is the crown most likely to be?

1. Aluminum
2. Copper
3. Steel
4. Lead
5. Gold

m_water*C_water*|Delta T_water| = m_stuff*c_stuff*|Delta T_stuff|

So (0.01 m^3)*(1000 kg/m^3)*4186 J/(kg *degree) * (2-1 degrees) = 5.1 kg * c_stuff*(20-2) degrees.

Solving for c_stuff gives 0.01*1000*4186*1/(5.1*18) = 455 K/kg/degree, which is Steel.

30. A steel lift column in a service station is 4.0 m tall, and 0.2m in diameter. The Young's modulus (ratio of stress/strain) for steel is 2.0E11 N/m^2. By how much does the column shrink when a 5000. kg truck is put on it?
1. 3.2E-5 m
2. 0.05 m
3. 7.8E-6 m
4. 3.2E-6 m
5. 0 m

Delta L /L_0 = (1/E) * F/A = (1/2E11 N/m^2) * mg) / (Pi * R^2) where R=0.1m. (radius, not diam!)

So Delta L = 4.0 m * 500kg 0*10 m/s^2/(Pi(*0.1m)^2) * (1/2E11N/m^2) = 3.2E-5 m

31. I have a mass of 50 kg, and the bottom of each of my feet has an area of 250 cm^2. If I put on a pair of snowshoes, each with an area of 0.25 m^2, which of the following is most correct?
1. The pressure exerted on the snow becomes 10 times greater
2. The pressure exerted on the snow is the same.
3. The pressure exerted on the snow is 10 times less
4. The force exerted on the snow is 10 times greater
5. The force exerted on the snow is 10 times less.

250 cm^2 ( 1m/100cm)^2 = 0.025 m^2. So, The snowshoes have increased your effective surface area by a factor of 10. Your weight is unchanged, (F is unchanged), but Pressure = F/A drops by 10.

32. Two rectangular fish tanks are filled to the same depth with water. Tank A has a bottom surface area of 2 m^2, tank B has a bottom surface area of 4 m^2. Which pair of statements about forces and pressures at the bottom of the tanks is most correct? (Let Fa be the total force of water on the bottom of tank A, let Pa be the pressure of water at the bottom of tank A, etc)
1. Fa < Fb, and Pa = Pb
2. Fa = Fb, and Pa = Pb
3. Fa > Fb, and Pa = Pb
4. Fa = Fb, and Pa > Pb
5. Fa < Fb, and Pa < Pb

Pressure is always the same under equal heights, so Pa=Pb. Pressure = F/A, so if Area_a is smaller, then F_a must also be smaller. So, F_a < F_b.

33. A water barometer in Boulder has a column height of 8.5m. What is the air pressure?

(The density of water is 1000 kg/m^3)

1. 1 atm
2. 1 N/m^2
3. 85 N/m^2
4. 85,000 N/m^2
5. 85,000 atm

Pressure = rho*g*h = 1000 kg/m^3 * 10 m/s^2 * 8.5 m = 85000 kg m/s^2 / m^2 = 85000 N/m^2.

For the next 2 problems, consider the following scenario.

Solid chunks of pine, lead, and aluminum, each exactly 0.5 cm^3 in volume, are dropped over the side of a boat in the middle of a deep fresh water lake. The pine floats easily, the two metals sink.

34. When the lead sits at the bottom, compare the buoyant force on the wood to that on the lead:
1. The wood feels a greater buoyant force.
2. The lead feels a greater buoyant force.
3. They feel identical buoyant forces.
4. Can't answer: it depends on the exact numerical values of the densities.

F_B = Weight of water displaced. Pine is not all underwater so it displaces less volume => Less F_B.

35. When they both sit at the bottom, compare the buoyant force on the lead to that on the aluminum.
1. The lead feels a greater buoyant force.
2. The aluminum feels a greater buoyant force.
3. They feel identical buoyant forces.
4. Can't answer: it depends on the exact numerical values of the densities.

F_B = weight of water displaced. The bricks have equal volume=> displace equal amounts of water => feel equal buoyant forces.

36. My dog, Sasha, is barking, delivering 1 milli-Watt of power. How loud is her bark at a distance of 5m? (Hint: sound spreads out evenly in all directions, and the surface area of a sphere is )
1. 65 dB
2. 55 dB
3. -55 dB
4. 60 dB
5. 53 dB

Loudness = 10Log(I/I0) = 10 Log( Power/Area/10^-12 W/m^2)

= 10 Log(10^-3 W/(4 Pi( 5m)^2)/10^-12 W/m^2) = 65 dB

37. The intensity level of the sound from a jet plane at a distance of 40m is 120 dB. What is the level at 400m?
1. 1.2 dB
2. 100 dB
3. 110 dB
4. 140 dB
5. 12 dB

Distance increases by factor 10 => Intensity decreases by factor 10^2 = 100. A decrease in intensity of 100 means 10 Log(I/I0) goes down by 10*Log(100) = 20, so 120 dB goes down to 100 dB.

38. Two musical strings each have a fundamental standing wave with the exact same wavelength, =0.25m, but the speed of waves on one string is 300 m/s, and on the other is 302 m/s.

True(A) or False(B) If the two strings are played together, you will hear precisely 2 beats/sec.

Beat freq = |f1 - f2| = (v_1/lambda - v_2/lambda) The lambda is the same in both cases, 0.25, so

Beat frequency = 302/(.25) - 300/(.25) = 8 /s, which is not 2.

39. The expansion coefficient of aluminum is 25.E-6/, and for iron it is 12.E-6/. Of the following, which expands the MOST?
1. 1 meter of aluminum, when the temperature increases by 2 C.
2. 2 meters of iron, when the temperature increases by 2 C.
3. 4 meters of iron, when the temperature increases by 0.25 C

Delta L = L_0*alpha*Delta T. For a) 1 m*25E-6*2C = 50E-6m

For b) 2m*12E-6*2C = 48E-6 m (less)

For c) 4 m*12E-6*0.25C = 12E-6m (less still) So, a is the biggest.

40. Sweat evaporates, absorbing 2.2E6 J/kg in the process. How much sweat must evaporate to cool a 66 kg body by 1 C? The heat capacity of a body is roughly that of water, 4200 J/(kg C)
1. 3E-5 kg
2. 2.8E5 kg
3. 0.13 kg
4. 7.9 kg
5. 0.2 kg

66 kg * 4200 J/kg degree*(1 C) = heat required = 2.2E6 J/kg * X kg. Solve for X

X = 66*4220/2.2E6 = 0.13

41. A water heater puts 5000. W of power into a tank with 150. kg of water. (Assume all the energy goes into warming the water) How long does it take to raise the water temperature from 20. C to 60. C? (The heat capacity of water is 4200 J/(kg C).)
1. 84 min
2. 8.4 min
3. 120 min
4. 130 min
5. 42 min

150 kg * 4200 J/kg C * (60-20C) = heat required = 5000 W * time. Solve for time:

t = 150*4200/(40)/5000 = 5040 seconds * (1 min/60 sec) = 84 min.

(Hot tubs need BIG powerful water heaters, and they still take awhile to heat up!)

42. A steel ring is heated over a flame. As the ring heats, the diameter of the inner hole
1. increases
2. decreases
3. stays the same
4. (depends on how big the inner hole is to start with)