Question 1
Identical blocks \( 1 \) and \( 2 \) are placed on a horizontal surface at points \( A \) and \( E \), respectively, as shown. The surface is frictionless except for the region between points \( C \) and \( D \), where the surface is rough. Beginning at time \( t_A \), block \( 1 \) is pushed with a constant horizontal force from point \( A \) to point \( B \) by a mechanical plunger. Upon reaching point \( B \), block \( 1 \) loses contact with the plunger and continues moving to the right along the horizontal surface toward block \( 2 \). Block \( 1 \) collides with and sticks to block \( 2 \) at point \( E \), after which the two-block system continues moving across the surface, eventually passing point \( F \).
(b) The plunger is returned to its original position, and both blocks are removed. A uniform solid sphere is placed at point \( A \), as shown. The sphere is pushed by the plunger from point \( A \) to point \( B \) with a constant horizontal force that is directed toward the sphere’s center of mass. The sphere loses contact with the plunger at point \( B \) and continues moving across the horizontal surface toward point \( E \). In which interval(s), if any, does the sphere’s angular momentum about its center of mass change? Check all that apply.
Most-appropriate topic codes (AP Physics 1):
• Topic \( 4.2 \) — Change in Momentum and Impulse (Part \( \mathrm{(a)} \))
• Topic \( 4.3 \) — Conservation of Linear Momentum (Part \( \mathrm{(a)} \))
• Topic \( 6.3 \) — Angular Momentum and Angular Impulse (Part \( \mathrm{(b)} \))
• Topic \( 2.9 \) — Circular Motion (Part \( \mathrm{(b)} \), rotational change about the center of mass)
• Topic \( 1.3 \) — Representing Motion (Part \( \mathrm{(a)} \))
▶️ Answer/Explanation
(a)
The center-of-mass speed graph should have these pieces:
\( \bullet \) From \( t_A \) to \( t_B \): a straight line starting at \( 0 \) and increasing linearly
\( \bullet \) From \( t_B \) to \( t_C \): a horizontal, nonzero segment
\( \bullet \) From \( t_C \) to \( t_D \): a straight line decreasing linearly
\( \bullet \) From \( t_D \) to \( t_F \): a horizontal, nonzero segment at a lower value than the segment from \( t_B \) to \( t_C \)
A correct sketch is shown below.
Explanation:
From \( A \) to \( B \), the plunger applies a constant external force to the two-block system, so the center of mass has a constant acceleration and its speed increases linearly.
From \( B \) to \( C \), there is no external horizontal force, so the center-of-mass speed remains constant.
From \( C \) to \( D \), friction acts externally on the system, so the center-of-mass speed decreases linearly.
From \( D \) to \( F \), the surface is frictionless again, so the center-of-mass speed stays constant.
There is no jump in the center-of-mass speed at \( t_E \), because the collision between the blocks is internal to the two-block system. Internal forces can change the motion of individual blocks, but not the motion of the center of mass of the whole system.
(b)
\( \boxed{C \text{ to } D} \)
The sphere’s angular momentum about its center of mass changes only during the interval from \( C \) to \( D \).
From \( A \) to \( B \), the plunger force is directed through the sphere’s center of mass, so it exerts no torque about the center of mass. Therefore, it does not change the sphere’s angular momentum about its center of mass.
From \( B \) to \( C \) and from \( D \) to \( E \), the surface is frictionless, so there is no external torque about the center of mass that would change the sphere’s rotation.
In the rough interval from \( C \) to \( D \), friction acts at the contact point with the surface. That friction force produces an external torque about the center of mass, so the sphere’s angular momentum changes there.
Question 2
Most-appropriate topic codes (AP Physics 1):
• Topic \( 2.5 \) — Newton’s Second Law (Part \( \mathrm{(a)} \), Part \( \mathrm{(c)} \), Part \( \mathrm{(d)} \), Part \( \mathrm{(e)} \))
• Topic \( 2.6 \) — Gravitational Force (Part \( \mathrm{(a)} \), Part \( \mathrm{(b)} \), Part \( \mathrm{(c)} \), Part \( \mathrm{(d)} \), Part \( \mathrm{(e)} \))
• Topic \( 5.6 \) — Newton’s Second Law in Rotational Form (Part \( \mathrm{(e)} \), through the effect of pulley inertia)
• Topic \( 3.A \) / Scientific Questioning and Argumentation (Part \( \mathrm{(d)} \))
▶️ Answer/Explanation
(a)(i)
The acceleration is approximately \( \boxed{0} \), or extremely small.
If block \( A \) is much more massive than block \( B \), then block \( A \) has a very large inertia and is hard to accelerate. The hanging block is too light to produce much motion, so the whole system accelerates only a tiny amount.
In words: the heavy block on the table is so difficult to speed up that the system barely moves.
(a)(ii)
The acceleration is approximately \( \boxed{g} \) \( \big(\text{or about } 9.8\ \mathrm{m/s^2}\big) \).
If block \( A \) is much less massive than block \( B \), then block \( A \) offers very little resistance to the motion. The hanging block behaves almost like a freely falling mass, so the acceleration approaches \( g \).
It is not exactly equal to \( g \) unless the other mass becomes negligible, but it gets very close.
(b)
The correct free-body diagrams are:
For block \( A \):
\( \bullet \) Normal force \( F_N \) upward
\( \bullet \) Gravitational force \( F_g \) downward
\( \bullet \) Tension force \( F_T \) horizontally to the right
For block \( B \):
\( \bullet \) Tension force \( F_T \) upward
\( \bullet \) Gravitational force \( F_g \) downward
There is no friction force on block \( A \), because the tabletop is stated to have negligible friction.
(c)
Apply Newton’s second law separately to each block.
For block \( A \) \( (\text{horizontal direction}) \):
\( F_T = m_A a \)
For block \( B \) \( (\text{taking downward as positive}) \):
\( m_B g – F_T = m_B a \)
Substitute \( F_T = m_A a \) into the equation for block \( B \):
\( m_B g – m_A a = m_B a \)
\( m_B g = m_A a + m_B a \)
\( m_B g = (m_A + m_B)a \)
Solve for \( a \):
\( \boxed{a = \dfrac{m_B}{m_A + m_B}\,g} \)
This expression makes good physical sense: increasing \( m_B \) makes the pull stronger, while increasing either mass increases the inertia of the moving system.
(d)
\( \boxed{\text{Yes}} \)
The equation from part \( \mathrm{(c)} \) is \( a = \dfrac{m_B}{m_A + m_B}g \). When \( m_A \ll m_B \), the term \( m_A \) is negligible compared with \( m_B \), so
\( a \approx \dfrac{m_B}{m_B}g = g \)
Therefore the equation predicts that the acceleration approaches \( g \), which matches the reasoning in part \( \mathrm{(a)(ii)} \).
(e)
\( \boxed{T_2 > T_1} \)
When the pulley has nonnegligible mass, some of the net force effect goes into rotating the pulley as well as accelerating the blocks. That makes the acceleration of the blocks smaller than before.
For block \( B \), the forces are weight downward and tension upward. If the downward acceleration becomes smaller while \( m_B g \) stays the same, then the upward tension must be larger. So the tension in the vertical part of the string is greater with the massive pulley.
In other words, a rotationally inert pulley reduces the system’s acceleration, which means the hanging block’s net downward force is smaller, so the tension has to be closer to the block’s weight.
Question 3
| Quantity to be Measured | Symbol for Quantity | Equipment for Measurement |
|---|---|---|
Most-appropriate topic codes (AP Physics 1):
• Topic \( 3.4 \) — Conservation of Energy (Part \( \mathrm{(a)} \), Part \( \mathrm{(b)} \), Part \( \mathrm{(c)} \), Part \( \mathrm{(d)} \))
• Topic \( 3.1 \) — Translational Kinetic Energy (Part \( \mathrm{(a)} \), Part \( \mathrm{(b)} \), Part \( \mathrm{(c)} \), Part \( \mathrm{(d)} \))
• Topic \( 1.3 \) — Representing Motion (Part \( \mathrm{(d)} \))
• Topic \( 3.A \) — Scientific Questioning and Argumentation (Part \( \mathrm{(b)} \), Part \( \mathrm{(c)} \))
▶️ Answer/Explanation
(a)(i)
A valid principle is conservation of energy.
The elastic potential energy stored in the compressed spring is converted into the kinetic energy of the sphere at launch, neglecting friction and other losses.
(a)(ii)
If the spring is compressed a distance \( \Delta x \), then using conservation of energy:
\( \dfrac{1}{2}k(\Delta x)^2 = \dfrac{1}{2}mv^2 \)
Solving for \( k \),
\( \boxed{k = \dfrac{mv^2}{(\Delta x)^2}} \)
Here, \( m \) is the sphere’s mass, \( v \) is its launch speed, and \( \Delta x \) is the compression distance. These can all be measured with standard lab equipment.
(b)
One valid experimental design is to launch the sphere horizontally and measure its launch speed with a motion sensor.
| Quantity to Be Measured | Symbol for Quantity | Equipment for Measurement |
|---|---|---|
| Mass of sphere | \( m \) | Triple-beam balance |
| Spring compression distance | \( \Delta x \) | Ruler or meterstick |
| Launch speed of sphere | \( v \) | Motion sensor |
Procedure:
Measure the mass \( m \) of the sphere with a triple-beam balance. Set the launcher on a level surface aimed horizontally toward a motion sensor. Compress the spring to pin position \( A \), and measure the corresponding compression distance \( \Delta x_A \) with a ruler, relative to the spring’s unstretched position.
Release the pin and record the launch speed \( v_A \) of the sphere from the motion sensor. Repeat the launch at least three times for pin position \( A \), and average the speeds to reduce random error.
Repeat the same process for pin positions \( B \) and \( C \), measuring \( \Delta x_B \), \( \Delta x_C \), and the corresponding average launch speeds \( v_B \) and \( v_C \).
Keep the launcher angle the same for all trials and use the same sphere each time. Repeating trials and averaging improves reliability, and measuring each compression directly avoids assuming equally spaced pin positions.
Another acceptable method would be to launch the sphere horizontally and determine \( v \) from range and fall time, or to launch vertically and determine \( v \) from maximum height.
(c)
For each pin position, calculate the spring constant using
\( k = \dfrac{mv^2}{(\Delta x)^2} \)
so that you obtain values \( k_A \), \( k_B \), and \( k_C \).
Then compare the three calculated values. If the values of \( k \) agree within experimental uncertainty, the hypothesis is supported. If the values differ by more than the uncertainty, the hypothesis is not supported.
A nice check is to compare percent differences between the calculated \( k \)-values and see whether those differences are comparable to the experimental uncertainty from the speed and compression measurements.
(d)
Since each sphere is launched from the same pin position \( A \), the spring compression is the same each time, so the stored spring energy is the same:
\( \dfrac{1}{2}k(\Delta x)^2 = \dfrac{1}{2}mv^2 \)
Thus,
\( v = \sqrt{\dfrac{k(\Delta x)^2}{m}} \)
so launch speed decreases as sphere mass increases. Because \( v \propto \dfrac{1}{\sqrt{m}} \), the graph is a decreasing curve that is concave up.
A correct sketch is shown below.
Question 4
Most-appropriate topic codes (AP Physics 1):
• Topic \( 3.2 \) — Work (Part \( \mathrm{(a)} \), Part \( \mathrm{(b)} \))
• Topic \( 5.8 \) — Electric Circuits with Energy Transfer Ideas (Part \( \mathrm{(b)} \))
• Topic \( 3.B \) — Apply a law, definition, theoretical relationship, or model to make a claim (Part \( \mathrm{(b)} \))
• Topic \( 3.C \) — Justify or support a claim using physical principles or laws (Part \( \mathrm{(b)} \))
▶️ Answer/Explanation
(a)
The block is lifted at constant speed, so the motor’s mechanical work goes into increasing the block’s gravitational potential energy.
The work done on the block is
\( W = MgH \)
Mechanical power is work done per unit time, so
\( P = \dfrac{W}{\Delta t} \)
Therefore,
\( \boxed{P = \dfrac{MgH}{\Delta t}} \)
Since \( M \) and \( H \) are fixed, decreasing \( \Delta t \) means the motor must deliver a larger power.
(b)
\( \boxed{\text{In parallel with } R_1} \)
To decrease \( \Delta t \), the motor must lift the same block through the same height in less time. From part \( \mathrm{(a)} \), that means the required mechanical power must increase:
\( P = \dfrac{MgH}{\Delta t} \)
The problem states that the mechanical power of the motor increases when the potential difference across the motor increases. Therefore, to reduce \( \Delta t \), we must increase the voltage across the motor.
The motor and \( R_1 \) are originally in series, so the battery voltage is shared between them. If a resistor \( R_2 \) is added in parallel with \( R_1 \), the equivalent resistance of that branch becomes smaller than \( R_1 \) alone:
\( R_{\text{eq}} = \dfrac{R_1 R_2}{R_1 + R_2} \)
so \( R_{\text{eq}} < R_1 \).
This reduces the resistance of the non-motor part of the series circuit, which increases the total current. Because the motor is still in series with the rest of the circuit, the current through the motor also increases. Since the motor has constant resistance, its voltage drop is
\( \Delta V_{\text{motor}} = I R_{\text{motor}} \)
so a larger current means a larger potential difference across the motor.
A larger voltage across the motor means a larger mechanical power output, so the motor lifts the block faster. Therefore, the time interval \( \Delta t \) decreases.
The other options do not help in the same way: putting \( R_2 \) in series would increase total resistance and reduce current, while putting it in parallel with the motor would reduce the motor branch voltage/current behavior needed to raise the motor’s power.
4. Question: (7 points, suggested time 13 minutes)
A motor is a device that when connected to a battery converts electrical energy into mechanical energy. The motor shown above is used to lift a block of mass M at constant speed from the ground to a height H above the ground in a time interval Δt. The motor has constant resistance and is connected in series with a resistor of resistance R1 and a battery.
Mechanical power, the rate at which mechanical work is done on the block, increases if the potential difference (voltage drop) between the two terminals of the motor increases.
(a) Determine an expression for the mechanical power in terms of M, H, Δt, and physical constants, as appropriate.
(b) Without M or H being changed, the time interval Δt can be decreased by adding one resistor of
resistance R 2 , where R 2 > R 1, to the circuit shown above. How should the resistor of resistance R 2 be added to the circuit to decrease Δt ?
___ In parallel with ___ In parallel ___ In parallel with ___ In series with the battery,
the battery with R1 the motor R1, and the motor
In a clear, coherent, paragraph-length response that may also contain figures and/or equations, justify why your selection would decrease Δt.
Answer/Explanation
Ans:
(a) Mechanical power = \(\frac{M_{g}H}{\Delta t}\) Work = Fd
(b) _√__ In parallel
with R1
If R2 is added in parallel with R, the total resistance of the circuit, excluding the motor, would be less than R1 because the total resistance of a parallel circuit is less than the smallest resistor. According to kirchoff’s loop rule, the voltage drop across an entire circuit must equal zero. When R2 is added in parallel with R1, the voltage drop across that section decreases, which means the voltage drop across the motor most increase assuming the battery stays the same. Mechanical power increases because of this, so time most decrease assuming M and H are held constant.
Question
A rod with a sphere attached to the end is connected to a horizontal mounted axle and carefully balanced so that it rests in a position vertically upward from the axle. The center of mass of the rod-sphere system is indicated with a \( \otimes \), as shown in Figure \( 1 \). The rod-sphere system rotates clockwise with negligible friction about the axle due to the gravitational force.
Note: these frames are not equally spaced apart in time.
Most-appropriate topic codes (AP Physics 1):
• Topic \( 6.1 \) — Rotational Kinetic Energy (Part \( \mathrm{(a)(ii)} \), Part \( \mathrm{(b)(i)} \), Part \( \mathrm{(b)(ii)} \))
• Topic \( 6.2 \) — Torque and Work (Part \( \mathrm{(a)(i)} \), Part \( \mathrm{(b)} \))
• Topic \( 3.4 \) — Conservation of Energy (Part \( \mathrm{(b)(i)} \), Part \( \mathrm{(b)(ii)} \))
▶️ Answer/Explanation
(a)(i)
The angular acceleration is greatest in Frame \( \mathrm{C} \).
Angular acceleration depends on the net torque:
\( \alpha \propto \tau \)
The torque due to gravity is greatest when the lever arm is greatest, or equivalently when the line from the axle to the center of mass is perpendicular to the gravitational force. That happens when the rod-sphere system is horizontal, which is shown in Frame \( \mathrm{C} \).
So Frame \( \mathrm{C} \) has the greatest angular acceleration.
(a)(ii)
The rotational kinetic energy is greatest in Frame \( \mathrm{E} \).
As the system rotates downward, gravity does positive work the entire time, so the rod-sphere system speeds up continuously. By the time it reaches the bottom position in Frame \( \mathrm{E} \), the system has its greatest angular speed.
Since rotational kinetic energy is
\( K_{\mathrm{rot}} = \dfrac{1}{2}I\omega^2 \),
the frame with the greatest angular speed also has the greatest rotational kinetic energy. Therefore, the rotational kinetic energy is greatest in Frame \( \mathrm{E} \).
(b)(i)
Start with conservation of energy for the rod-sphere-Earth system:
\( E_i = E_f \)
or
\( U_{g,i} + K_i = U_{g,f} + K_f \)
The system starts from rest in Frame \( \mathrm{A} \), so
\( K_i = 0 \)
Therefore,
\( \Delta K = K_f – K_i = U_{g,i} – U_{g,f} \)
The center of mass is initially \( \dfrac{3}{4}L \) above the axle and finally \( \dfrac{3}{4}L \) below the axle, so the total vertical drop is
\( \Delta y = \dfrac{3}{4}L + \dfrac{3}{4}L = \dfrac{3}{2}L \)
Thus the decrease in gravitational potential energy, and therefore the increase in kinetic energy, is
\( \Delta K = Mg\Delta y \)
\( \Delta K = Mg\left(\dfrac{3}{2}L\right) \)
\( \boxed{\Delta K = \dfrac{3}{2}MgL} \)
This is also the final kinetic energy in Frame \( \mathrm{E} \), since the initial kinetic energy is zero.
(b)(ii)
Even if Earth is not included in the system, the rod and sphere still gain kinetic energy because gravity is an external force on the rod-sphere system.
That external gravitational force does positive work on the rod-sphere system as it rotates downward. External work can increase the kinetic energy of a system, so the rod and sphere gain kinetic energy even when Earth is not part of the chosen system.
5. Question: (7 points, suggested time 13 minutes)
A tuning fork vibrating at 512 Hz is held near one end of a tube of length L that is open at both ends, as shown above. The column of air in the tube resonates at its fundamental frequency. The speed of sound in air is 340 m/ s.
(a) Calculate the length L of the tube.
(b) The column of air in the tube is still resonating at its fundamental frequency. On the axes below, sketch a graph of the maximum speed of air molecules as they oscillate in the tube, as a function of position x, from x = 0 (left end of tube) to x = L (right end of tube). (Ignore random thermal motion of the air molecules.)
(c) The right end of the tube is now capped shut, and the tube is placed in a chamber that is filled with another gas in which the speed of sound is 1005 m s. Calculate the new fundamental frequency of the tube.
Answer/Explanation
Ans:
(a)
\(v = \frac{1}{2}\lambda \)
λ = 2L v = λf v = 2Lf
\(\frac{v}{2f} = L = \frac{3f0}{512.2} = \)
0.33 m
(b)
(c)
\(\frac{1}{4}\lambda = L\) λ = AL
vf= λf v = ALf
\(\frac{\sqrt{}}{4L} = f = \frac{1005}{4.(0.33)} = 761.36 H_{2}\)