Problem 1 (2.5 points)

How many times did the Yankees win the world series?

1. 22

2. 24

3. 26

4. 28

Unfortunately, this number will not change this year.

Problem 2 (2.5 points)

List the four basic forces in order of strength (start with the strongest force and end with the weakest force).

1. Strong > Electromagnetic > Gravitational > Weak.

2. Electromagnetic > Strong > Weak > Gravitational.

3. Strong > Electromagnetic > Weak > Gravitational.

4. Strong > Electromagnetic > Gravitational > Weak.

Problem 3 (2.5 points)

Which of the following statements is false?

1. The buoyant force in liquid is much larger than the buoyant force in air.

2. The buoyant force is a result of small differences in molecular density of the air/liquid across the surface of the object.

3. The buoyant force is a result of differences in the average molecular velocity of the air/liquid molecules across the surface of the object.

4. The magnitude of the buoyant force increases with increasing temperature, due to the corresponding increase in the average molecular velocities.

Problem 4 (2.5 points)

The only force acting on a body as it moves along the x axis varies as shown in the following Figure. The body has a positive velocity when it is located at x = 0 m.

At what x position between x = 0 m and x = 8 m will the kinetic energy of the body reach a maximum value?

1. x = 0 m

2. x = 1 m

3. x = 2 m

4. x = 3 m

5. x = 8 m.

Problem 5 (2.5 points)

Which of the following statements is correct?

1. Sound propagation requires a medium; light propagation requires a medium.

2. Sound propagation does not require a medium; light propagation requires a medium.

3. Sound propagation requires a medium; light propagation does not require a medium.

4. Sound propagation does not require a medium; light propagation does not require a medium.

Problem 6 (2.5 points)

A constant force is exerted on a cart that is initially at rest on an air track. Friction between the cart and the track is negligible. The force acts for a short time interval and gives the cart a certain speed. To reach the same final speed with a force that is only half as big, the force must be exerted on the cart for a time interval

1. four times as long as

2. twice as long as

3. equal to

4. half as long as

5. a quarter of

that for the stronger force.

Problem 7 (2.5 points)

The Figure on the right shows various possible trajectories of an object launched from the surface of the earth. What can you say about the total energy of the object for the various trajectories shown in the Figure?

1. E_{elliptical trajectory} > 0 J, E_{hyperbolic trajectory} = 0 J, and E_{parabolic trajectory} < 0 J.

2. E_{elliptical trajectory} > 0 J, E_{hyperbolic trajectory} < 0 J, and E_{parabolic trajectory} = 0 J.

3. E_{elliptical trajectory} = 0 J, E_{hyperbolic trajectory} > 0 J, and E_{parabolic trajectory} < 0 J.

4. E_{elliptical trajectory} = 0 J, E_{hyperbolic trajectory} < 0 J, and E_{parabolic trajectory} > 0 J.

5. E_{elliptical trajectory} < 0 J, E_{hyperbolic trajectory} = 0 J, and E_{parabolic trajectory} > 0 J.

6. E_{elliptical trajectory} < 0 J, E_{hyperbolic trajectory} > 0 J, and E_{parabolic trajectory} = 0 J.

Problem 8 (2.5 points)

In part (a) of the figure, an air track cart attached to a spring rests on the track at the position x_equilibrium and the spring is relaxed. In part (b), the cart is pulled to a position x_start and released. It then oscillates about x_equilibrium. Which of the six graphs correctly represents the potential energy of the spring as a function of the position of the cart?

Problem 9 (2.5 points)

A mass attached to a spring oscillates back and forth as indicated in the position vs. time plot below. What is the velocity and the acceleration of the mass at point P?

1. Positive velocity and positive acceleration.

2. Positive velocity and negative acceleration.

3. Positive velocity and zero acceleration.

4. Negative velocity and positive acceleration.

5. Negative velocity and negative acceleration.

6. Negative velocity and zero acceleration.

7. Zero velocity and non-zero acceleration (positive or negative).

8. Zero velocity and zero acceleration.

Problem 10 (2.5 points)

Two satellites A and B of the same mass are going around Earth in concentric orbits. The distance of satellite B from Earth’s center is twice that of satellite A. What is the ratio of the centripetal force acting on B to that acting on A?

1. 1/8

2. 1/4

3. 1/2

4. 1/√2

5. 1

Problem 11 (25 points)

Consider a cube of side d. A spherical mass m is located at each corner of the cube. The cube is positioned such that its center is located at the origin of the coordinate system and its sides are parallel to the coordinate axes.

a. What is the net force that acts on a spherical mass M (M > m) located at the origin of the coordinate system (specify magnitude and direction)?

Answer:

Mass M is located in the center of the cube and has the same distance from each of the corner masses m. The magnitude of the gravitational force exerted by each mass m on mass M is thus the same. The total force on mass M is the vector sum of the 8 individual forces. Due to the symmetry of the configuration, the net force on mass M is zero (e.g. the force due to the mass located at (d/2, d/2, d/2) cancels the force due to the mass located at (-d/2, -d/2, -d/2), the force due to the mass located at (d/2, d/2, -d/2) cancels the force due to the mass located at (-d/2, -d/2, d/2), etc.).

Now consider what happens when we replace the mass m located at position (d/2, d/2, d/2) with a sphere of mass M (see Figure below).

b. What is now the net force that acts on a spherical mass M (M > m) located at the origin of the coordinate system (specify magnitude and direction)?

Answer:

Mass M is located in the center of the cube and has the same distance from each of the corner masses. The magnitude of the gravitational force exerted by each corner mass m on the mass located at the origin is thus the same, but the gravitational force exerted by the corner mass M, located at (d/2, d/2, d/2), will be larger in magnitude since M > m. The total force on the mass located at the origin of the coordinate system is the vector sum of the 8 individual forces. Due to the symmetry of the configuration, the net force on center mass is the vector sum of the gravitational forces due to mass M located at (d/2, d/2, d/2) and mass m located at (-d/2, -d/2, -d/2); all other individual forces add up to 0 N. The net force on the center mass is thus

This problem can also be solved by using the principle of superposition. In the following figure we see that the configuration of the system studied here is the superposition of the configuration studied in part (a) and a simple system with just two masses:

The figure shows that the force on the mass located at the center of the coordinate system is equal to the force generated by the second configuration shown on the right-hand side of the Figure. Its magnitude is equal to

and it is directed along the unit vector that connects (0, 0, 0) and (d/2, d/2, d/2).

Express your answers in terms of the variables provided (m, M, and d) and the gravitational constant G.

Problem 12 (25 points)

Two springs, with spring constants k₁ and k₂ and with negligible mass, are connected as shown in the Figure below. The length of the double spring system is L. When a mass m is connected to the bottom spring, the equilibrium length of the two springs is increased by ∆L.

a. Determine the increase in length ∆L of the two springs.

Answer:

Consider that the length of springs 1 and 2 change by ∆x₁ and ∆x₂, respectively, when mass m is attached to the bottom spring. If the system is in equilibrium, the net force on the mass must be zero, and the net force on the point where the two springs connect to each other must be zero. This requires that

and

The first equation can be used to determine the change in the length of spring 2:

Substituting this in the second equation we can determine the change in the length of spring 1:

The total change in the length of the system is thus equal to

The last equation shows that we can replace the two springs by a single spring with a spring constant k_eff where

or

In terms of the effective spring constant, the change in length of the system is equal to

Using the relations obtained as part of this solution, we can relate the change in the lengths of spring 1 and 2 to the change in length of the total system:

b. What is the total energy of the system (the two springs and the mass) when it is in equilibrium? The gravitational potential energy is 0 J at the position indicated in the Figure.

Answer:

The total energy of the system is the sum of the potential energy of the springs and the gravitational potential energy of mass m. The kinetic energy of the system is zero since the system is in equilibrium (and all components are thus at rest). The total energy of the system is thus equal to

When the mass is in its equilibrium position, the mass is given a velocity v₀ (directed upwards) and the system will start to carry out harmonic motion.

c. What is the amplitude of the harmonic motion?

Answer:

In order to describe the harmonic motion of the system around its equilibrium position we consider the force acting on mass m when it is moved a distance w below its equilibrium position. At this position, the force acting on the mass m is equal to

The net force is thus proportional to the displacement from the equilibrium position, and harmonic motion can thus result when the mass is put into oscillation.

If we redefine the zero point of the potential energy of the spring system, we can set the potential energy to be zero when the system with mass m is in its equilibrium position. When the system is displaced by a distance A from this position, it potential energy will thus be equal to

Conservation of energy requires that the potential energy of the system when the mass is located a distance A from its equilibrium position is the same as the kinetic energy of the system when it is located at its equilibrium position:

or

d. What is the period of the harmonic motion?

Answer:

In the solution of part (c) we observed that the double spring system can be replaced by a single spring with a spring constant k_eff = k₁k₂/(k₁ + k₂). The frequency associated with the harmonic motion of such a system is

The period of the resulting harmonic motion is thus equal to

Express all your answers in terms of the variables provided (L, k₁, k₂, m, and v₀) and the gravitational acceleration g.

Problem 13 (25 points)

The Stanford Linear Accelerator Center (SLAC), located at Stanford University in Palo Alto, California, accelerates electrons through a vacuum tube of length L.

Electrons of mass m, which are initially at rest, are subjected to a continuous force F along the entire length of the tube and reach speeds very close to the speed of light.

a. Calculate the final energy of the electrons.

Answer:

The work done by the force F is FL, where L is the length of the accelerator. Using the work energy theorem, we can express the final energy of the electron in terms of its initial energy and the work done by F:

b. Calculate the final momentum of the electrons.

Answer:

The final momentum of the electron can be obtained using the following relation:

or

c. Calculate the final speed of the electrons.

Answer:

The final velocity of the electron can be found using the following relation:

or

d. Calculate the time required travel the distance L.

Answer:

The time required to complete the trip through the accelerator can be found by using the momentum principle:

Express all your answers in terms of the variables provided (m, F, and L) and the speed of light c.