2012 Ap Physics C Mechanics Frq

Diving into the 2012 AP Physics C Mechanics Free Response Questions (FRQ) offers invaluable insight into the core concepts tested in this challenging exam. These problems are designed to assess a student's understanding of Newtonian mechanics, including kinematics, work, energy, momentum, rotational motion, and oscillations, all within the framework of calculus.

Question 1: Projectile Motion and Circular Motion

This question masterfully combines projectile motion with circular motion, requiring students to apply principles of kinematics, energy conservation, and centripetal force.

(a) Determining the Horizontal Distance

A block of mass m is launched horizontally from a height H above the ground with an initial speed v₀. To determine the horizontal distance R it travels before hitting the ground, we first need to find the time it takes for the block to fall. Since the initial vertical velocity is zero, the time t can be found using the kinematic equation:

H = (1/2)gt²

Solving for t:

t = √(2H/g)

The horizontal distance R is then simply the product of the horizontal velocity v₀ and the time t:

R = v₀√(2H/g)

(b) Deriving the Speed at Impact

To derive the speed v of the block just before it hits the ground, we can use the principle of energy conservation. The initial total energy of the block is its potential energy mgh plus its kinetic energy (1/2)mv₀². Just before impact, all of this energy is converted into kinetic energy (1/2)mv². Therefore:

mgh + (1/2)mv₀² = (1/2)mv²

Solving for v:

v = √(v₀² + 2gH)

(c) Analyzing the Circular Path

Now, consider the situation where the block is attached to a string of length L and swung in a vertical circle.

(i) Minimum Speed at the Top

At the top of the circular path, the minimum speed v_top required to keep the string taut is when the tension in the string is zero. In this case, the centripetal force is provided solely by gravity:

mg = mv_top²/ L

Solving for v_top:

v_top = √(gL)

(ii) Tension at the Bottom

To determine the tension T in the string at the bottom of the circle, we can again use energy conservation. The block's kinetic energy at the bottom is equal to its initial potential energy at the top plus the work done by gravity as it moves from the top to the bottom:

(1/2)mv_bottom² = mg(2L) + (1/2)mv_top²

Substituting v_top = √(gL):

(1/2)mv_bottom² = mg(2L) + (1/2)m(gL) = (5/2)mgL

v_bottom² = 5gL

At the bottom, the tension T minus the weight mg provides the centripetal force:

T - mg = mv_bottom²/ L

T = mg + m(5gL)/ L = 6mg

Therefore, the tension in the string at the bottom of the circle is 6mg.

(d) Effect of Increasing Mass

If the mass of the block is doubled, the minimum speed at the top of the circle, v_top = √(gL), remains unchanged because the mass cancels out in the equation. However, the tension at the bottom of the circle, T = 6mg, will double because the tension is directly proportional to the mass.

Question 2: Experimental Design - Determining the Coefficient of Friction

This question delves into experimental design, focusing on determining the coefficient of kinetic friction between a block and a horizontal surface.

(a) Experimental Procedure

To experimentally determine the coefficient of kinetic friction µ_k between the block and the surface, one could perform the following procedure:

1. Setup: Place the block on the horizontal surface and attach a string to it. Pass the string over a pulley and attach a hanging mass m_h to the other end.
2. Measurement: Measure the mass of the block m_b using a balance.
3. Execution: Release the hanging mass and allow the block to accelerate across the surface.
4. Data Collection: Measure the acceleration a of the block using a motion sensor or by measuring the time it takes to travel a known distance.
5. Repetition: Repeat the experiment several times with different hanging masses to obtain multiple data points.

(b) Measurements and Equipment

The following measurements and equipment are required:

• Mass of the block (m_b): Balance
• Hanging mass (m_h): Balance
• Acceleration of the block (a): Motion sensor, meter stick, and timer

(c) Derivation of the Equation for µ_k

Applying Newton's second law to the hanging mass:

m_h g - T = m_h a

Where T is the tension in the string.

Applying Newton's second law to the block:

T - f_k = m_b a

Where f_k is the kinetic friction force. Also, f_k = µ_k N, where N is the normal force. Since the surface is horizontal, N = m_b g. So, f_k = µ_k m_b g.

Substituting T from the first equation into the second equation:

(m_h g - m_h a) - µ_k m_b g = m_b a

Rearranging to solve for µ_k:

µ_k = (m_h g - (m_h + m_b) a) / (m_b g)

(d) Identifying a Source of Error

A significant source of error in this experiment could be the assumption that the pulley is massless and frictionless. In reality, the pulley has mass and friction, which would reduce the acceleration of the block. To mitigate this, one could use a pulley with very low friction bearings or account for the rotational inertia of the pulley in the calculations.

(e) Sketching a Graph

If the experiment is repeated with different hanging masses, a graph of a (acceleration) versus m_h (hanging mass) can be plotted. The data points should form a roughly linear relationship. The slope of the best-fit line through the data points can be related to µ_k.

Question 3: Simple Harmonic Motion and Energy Conservation

This question examines simple harmonic motion (SHM) in the context of a spring-mass system and requires the application of energy conservation principles.

(a) Deriving the Expression for the Block's Speed

A block of mass m is attached to a spring with spring constant k. The block is displaced a distance x_m from its equilibrium position and released. To derive an expression for the block's speed v as a function of position x, we can use energy conservation.

The total energy E of the system is constant and is equal to the potential energy stored in the spring at maximum displacement:

E = (1/2)kx_m²*

At any position x, the total energy is the sum of the kinetic energy of the block and the potential energy stored in the spring:

E = (1/2)mv² + (1/2)kx²

Setting these two expressions for E equal to each other:

(1/2)kx_m²* = (1/2)mv² + (1/2)kx²

Solving for v:

v = √((k/m)(x_m² - x²))

(b) Determining the Period of Oscillation

The period T of oscillation for a spring-mass system is given by:

T = 2π√(m/k)

(c) Analyzing the Collision and Subsequent Motion

Now, consider the situation where a clay blob of mass m is dropped from a height H onto the block when it is at its lowest point.

(i) Speed of the Clay Blob Just Before Impact

The speed v_clay of the clay blob just before impact can be found using kinematics:

v_clay = √(2gH)

(ii) Speed of the Block/Clay System Immediately After Impact

The collision between the clay blob and the block is inelastic. Therefore, we can use conservation of momentum to find the speed v_new of the combined system immediately after impact. The momentum of the clay blob just before impact is m v_clay. The momentum of the block just before impact is m v_bottom, where v_bottom can be determined from energy conservation.

Using energy conservation before the collision to find the block's speed at the bottom: (1/2)kx_m²* = (1/2)mv_bottom²* v_bottom = √((k/m) x_m²) = x_m *√((k/m))

Applying conservation of momentum:

m v_clay + m v_bottom = (2m) v_new

v_new = (v_clay + v_bottom) / 2 = (√(2gH) + x_m *√((k/m))) / 2

(iii) Amplitude of the Resulting Motion

To find the amplitude A of the resulting motion, we can use energy conservation again. The total energy of the system after the collision is the sum of the kinetic energy of the combined mass and the potential energy of the spring. At the maximum displacement (amplitude A), the kinetic energy is zero. So:

(1/2)(2m*)v_new² + (1/2)kx²* = (1/2)kA²*

Now, to simplify, let's consider the system at the equilibrium position after the clay is added. In this case, let's also consider the potential energy change due to gravity of the additional mass. The new equilibrium position will shift down by a distance Δx, such that kΔx = mg, then Δx = mg/k.

Using the conservation of energy in terms of the new equilibrium position, we know:

(1/2)(2m)v_new^2 = (1/2)k(A^2)

where v_new is (√(2gH) + x_m *√((k/m))) / 2, and A is the new amplitude

A = sqrt( (2m v_new^2)/k ) = sqrt ( (2m * ((√(2gH) + x_m *√((k/m))) / 2)^2) / k )

A = sqrt ( (m/2k)*(√(2gH) + x_m *√((k/m)))^2 )

(d) Effect on the Period

The period T of oscillation after the clay blob is dropped is given by:

T_new = 2π√((2m)/k*)

Since the mass has doubled, the new period is √2 times the original period.

Strategies for Success

1. Master Fundamental Concepts: Ensure a strong grasp of kinematics, Newton's laws, energy conservation, momentum, rotational motion, and simple harmonic motion.
2. Practice Problem-Solving: Regularly solve a variety of problems, including those from previous AP Physics C exams.
3. Understand Calculus: Be proficient in applying calculus to solve physics problems.
4. Manage Time Effectively: During the exam, allocate your time wisely and avoid spending too much time on any one question.
5. Show Your Work: Clearly show all steps in your solutions, as partial credit is often awarded.
6. Use Units: Always include appropriate units in your answers.
7. Review and Correct: If time permits, review your answers and correct any mistakes.

By thoroughly understanding the concepts tested in the 2012 AP Physics C Mechanics FRQ and practicing effective problem-solving strategies, students can improve their performance on the AP exam and deepen their understanding of classical mechanics. These questions serve as a valuable tool for reinforcing fundamental principles and developing critical thinking skills in physics.