Unit 2: Dynamics (Part 1)
Systems, FBDs, and Newton’s Fundamental Laws
Section 2.1
Systems and Center of Mass
A “system” is a single object or a collection of objects. The Center of Mass (CM) is the unique point representing the average position of the system’s mass.
Q1: Two masses, \( 2 \text{ kg} \) at \( x = 0 \) and \( 6 \text{ kg} \) at \( x = 4 \text{ m} \), form a system. Where is the center of mass located?
Use the discrete formula: \( x_{cm} = \displaystyle\frac{m_1x_1 + m_2x_2}{m_1 + m_2} \).
\( x_{cm} = \displaystyle\frac{(2)(0) + (6)(4)}{2 + 6} = \displaystyle\frac{24}{8} = 3 \text{ m} \).
Correct Answer: C
\( x_{cm} = \displaystyle\frac{(2)(0) + (6)(4)}{2 + 6} = \displaystyle\frac{24}{8} = 3 \text{ m} \).
Correct Answer: C
Q2: A thin rod of length \( L \) has a linear mass density \( \lambda(x) = kx \). What is the x-coordinate of the center of mass?
Total mass \( M = \int_{0}^{L} kx \, dx = \displaystyle\frac{1}{2}kL^2 \).
Moment \( \int x \, dm = \int_{0}^{L} kx^2 \, dx = \displaystyle\frac{1}{3}kL^3 \).
\( x_{cm} = \displaystyle\frac{1}{M} \int x \, dm = \displaystyle\frac{kL^3/3}{kL^2/2} = \displaystyle\frac{2L}{3} \).
Correct Answer: B
Moment \( \int x \, dm = \int_{0}^{L} kx^2 \, dx = \displaystyle\frac{1}{3}kL^3 \).
\( x_{cm} = \displaystyle\frac{1}{M} \int x \, dm = \displaystyle\frac{kL^3/3}{kL^2/2} = \displaystyle\frac{2L}{3} \).
Correct Answer: B
Q3: A 60 kg person stands at one end of a 10 m, 40 kg floating log. If the person walks to the other end, how far does the log move relative to the water?
The system’s CM remains stationary as there are no external forces.
\( m_p \Delta x_p + m_L \Delta x_L = 0 \).
Relative to log: \( \Delta x_p = \Delta x_L + 10 \).
\( 60(\Delta x_L + 10) + 40 \Delta x_L = 0 \Rightarrow 100 \Delta x_L = -600 \Rightarrow \Delta x_L = -6 \text{ m} \).
Correct Answer: B
\( m_p \Delta x_p + m_L \Delta x_L = 0 \).
Relative to log: \( \Delta x_p = \Delta x_L + 10 \).
\( 60(\Delta x_L + 10) + 40 \Delta x_L = 0 \Rightarrow 100 \Delta x_L = -600 \Rightarrow \Delta x_L = -6 \text{ m} \).
Correct Answer: B
Section 2.2
Forces and Free-Body Diagrams (FBD)
Q1: Which of the following should NOT appear on a block’s formal FBD?
A formal FBD must show only actual external forces acting on the object, not their mathematical components.
Correct Answer: C
Correct Answer: C
Q2: A 2 kg block is on a \( 30^{\circ} \) incline. What is the normal force? (Use \( g=10 \))
Normal force on an incline is \( F_N = mg \cos\theta \).
\( F_N = (2)(10) \cos 30^\circ = 20 \times 0.866 = 17.3 \text{ N} \).
Correct Answer: C
\( F_N = (2)(10) \cos 30^\circ = 20 \times 0.866 = 17.3 \text{ N} \).
Correct Answer: C
Section 2.3
Newton’s Third Law
Q1: A truck collides with a car. The magnitude of the force of truck on car is \( F_T \), car on truck is \( F_C \).
According to Newton’s Third Law, action and reaction forces are always equal in magnitude and opposite in direction, regardless of mass or speed.
Correct Answer: C
Correct Answer: C
Q2: A person of mass \( m \) stands in an elevator accelerating upward at \( a \). What is the normal force?
Applying Newton’s Second Law: \( F_N – mg = ma \).
Solving for normal force: \( F_N = m(g+a) \).
Correct Answer: B
Solving for normal force: \( F_N = m(g+a) \).
Correct Answer: B
Dynamics Formulas Recap
Discrete CM\( x_{cm} = \displaystyle\frac{\sum m_ix_i}{\sum m_i} \)
Calculus CM\( x_{cm} = \displaystyle\frac{1}{M} \int x \, dm \)
System Law\( \sum \vec{F}_{ext} = M\vec{a}_{cm} \)
Newton II\( \vec{F}_{net} = m\vec{a} \)
Newton III\( \vec{F}_{AB} = -\vec{F}_{BA} \)
Equilibrium\( \sum \vec{F} = 0 \)
Unit 2: Dynamics (Part 2)
Newton’s Second Law, Universal Gravitation, and Friction
Section 2.5
Newton’s Second Law
The core of mechanics: \( \sum \vec{F} = m\vec{a} \). In AP Physics C, we extend this to cases where forces vary with time or velocity (resistive forces).
Q1: A mass \( m \) is subject to a resistive force \( F_r = -kv \). If released from rest under gravity, what is its terminal velocity?
At terminal velocity, \( a = 0 \). Thus the net force is zero:
\( mg – kv_t = 0 \Rightarrow kv_t = mg \Rightarrow v_t = \displaystyle\frac{mg}{k} \).
Correct Answer: B
\( mg – kv_t = 0 \Rightarrow kv_t = mg \Rightarrow v_t = \displaystyle\frac{mg}{k} \).
Correct Answer: B
Q2: A 2 kg object has net force \( F(t) = 3t^2 \). If starting from rest, what is the speed at \( t = 2 \text{ s} \)?
\( a(t) = \displaystyle\frac{F(t)}{m} = \displaystyle\frac{3t^2}{2} = 1.5t^2 \).
\( v(2) = \int_0^2 1.5t^2 dt = [0.5t^3]_0^2 = 0.5(8) – 0 = 4 \text{ m/s} \).
Correct Answer: A
\( v(2) = \int_0^2 1.5t^2 dt = [0.5t^3]_0^2 = 0.5(8) – 0 = 4 \text{ m/s} \).
Correct Answer: A
Q3: Mass \( m \) pulling mass \( M \) on a frictionless table via a pulley. Acceleration of the system?
The pulling force is the weight of the hanging mass: \( F_{net} = mg \).
The total mass accelerated is both blocks: \( m_{total} = m + M \).
\( a = \displaystyle\frac{F_{net}}{m_{total}} = \displaystyle\frac{mg}{m+M} \).
Correct Answer: C
The total mass accelerated is both blocks: \( m_{total} = m + M \).
\( a = \displaystyle\frac{F_{net}}{m_{total}} = \displaystyle\frac{mg}{m+M} \).
Correct Answer: C
Q4: Person \( m \) in elevator accelerating downward at \( a < g \). What is the Normal force?
Sum of forces (down positive): \( mg – F_N = ma \).
Rearrange for normal force: \( F_N = mg – ma = m(g – a) \).
Correct Answer: B
Rearrange for normal force: \( F_N = mg – ma = m(g – a) \).
Correct Answer: B
Q5: Force \( F \) increases speed from \( v \) to \( 2v \) over distance \( d \). If force is doubled, what is the required distance?
From Work-Energy Theorem: \( \Delta K = W = Fd \).
If the change in kinetic energy \( \Delta K \) is kept the same, then \( F \times d = \text{constant} \).
If \( F \) is doubled, \( d \) must be halved to \( d/2 \).
Correct Answer: C
If the change in kinetic energy \( \Delta K \) is kept the same, then \( F \times d = \text{constant} \).
If \( F \) is doubled, \( d \) must be halved to \( d/2 \).
Correct Answer: C
Section 2.6
Gravitational Force
Q1: If distance between two masses is tripled, the force becomes:
Inverse Square Law: \( F \propto \displaystyle\frac{1}{r^2} \).
If \( r \rightarrow 3r \), then \( F \rightarrow \displaystyle\frac{1}{(3)^2} = \displaystyle\frac{1}{9} \).
Correct Answer: D
If \( r \rightarrow 3r \), then \( F \rightarrow \displaystyle\frac{1}{(3)^2} = \displaystyle\frac{1}{9} \).
Correct Answer: D
Q2: A planet has \( 2M \) and \( 2R \) of Earth. Surface \( g’ \) is?
\( g’ = \displaystyle\frac{G(2M)}{(2R)^2} = \displaystyle\frac{2GM}{4R^2} = \displaystyle\frac{1}{2} \left( \displaystyle\frac{GM}{R^2} \right) = g/2 \).
Correct Answer: C
Correct Answer: C
Q3: At what height \( h \) is acceleration equal to \( g/4 \)?
\( g(r) \propto \displaystyle\frac{1}{r^2} \). For \( g/4 \), the distance from center must be \( 2R \).
Since \( r = R + h \), we have \( 2R = R + h \Rightarrow h = R \).
Correct Answer: A
Since \( r = R + h \), we have \( 2R = R + h \Rightarrow h = R \).
Correct Answer: A
Q4: If satellite mass is doubled, the gravitational force on it:
\( F_g = G \displaystyle\frac{Mm}{r^2} \). The force is directly proportional to the mass of the satellite (\( m \)).
Correct Answer: B
Correct Answer: B
Q5: Which graph describes \( g \) vs \( r \) for a uniform spherical planet?
Inside the planet (\( r < R \)), gravity increases linearly with distance: \( g \propto r \). Outside (\( r > R \)), it follows the inverse square law: \( g \propto 1/r^2 \).
Correct Answer: B
Correct Answer: B
Section 2.7
Kinetic and Static Friction
Q1: Block \( m \) at rest on incline \( \theta \). Magnitude of static friction?
At rest, the net force is zero. Friction must balance the component of gravity acting down the ramp: \( f_s = mg \sin\theta \).
Correct Answer: D
Correct Answer: D
Q2: 5 kg block against vertical wall (\( \mu_s = 0.4 \)). Minimum force to prevent sliding? (Use \( g=10 \))
Vertical: \( f_s = mg = 5(10) = 50 \text{ N} \).
Friction limit: \( f_s \le \mu_s F_N \). Here \( F_N = F_{push} \).
\( 50 \le 0.4F \Rightarrow F \ge 50/0.4 = 125 \text{ N} \).
Correct Answer: C
Friction limit: \( f_s \le \mu_s F_N \). Here \( F_N = F_{push} \).
\( 50 \le 0.4F \Rightarrow F \ge 50/0.4 = 125 \text{ N} \).
Correct Answer: C
Q3: Car rounds unbanked curve \( R \) at speed \( v \). Required \( \mu_s \) to not slip?
Centripetal force is provided by friction: \( \mu_s mg = \displaystyle\frac{mv^2}{R} \).
Solving for coefficient: \( \mu_s = \displaystyle\frac{v^2}{gR} \).
Correct Answer: A
Solving for coefficient: \( \mu_s = \displaystyle\frac{v^2}{gR} \).
Correct Answer: A
Q4: Pushing down at angle \( \theta \). Kinetic friction force?
Normal force is \( F_N = mg + F \sin\theta \). Since the normal force increases, and \( f_k = \mu_k F_N \), the friction force increases.
Correct Answer: A
Correct Answer: A
Q5: Which relationship is usually true for contact surfaces?
It generally requires more force to start an object moving than to keep it in motion. Thus, the static coefficient is typically greater than the kinetic coefficient.
Correct Answer: B
Correct Answer: B
Dynamics Formulas Recap
Newton II\( \sum \vec{F} = m\vec{a} \)
Resistive\( \vec{F}_r = -k\vec{v} \)
Gravitation\( F_g = G \displaystyle\frac{m_1m_2}{r^2} \)
Field\( g = \displaystyle\frac{GM}{r^2} \)
Static\( F_s \le \mu_s F_N \)
Kinetic\( F_k = \mu_k F_N \)
Unit 2: Dynamics (Part 3)
Springs, Resistive Forces, and Advanced Circular Motion
Section 2.8
Spring Forces
Springs exert a restoring force proportional to their displacement: \( F_s = -kx \). Work done on or by a spring is found by integrating the force over the distance stretched or compressed.
Q1: A mass \( m \) is attached to a vertical spring with constant \( k \). At equilibrium, what is the vertical stretch \( \Delta y \)?
At equilibrium, the magnitude of the upward spring force must balance the downward gravitational force:
\( k\Delta y = mg \Rightarrow \Delta y = \displaystyle\frac{mg}{k} \).
Correct Answer: B
\( k\Delta y = mg \Rightarrow \Delta y = \displaystyle\frac{mg}{k} \).
Correct Answer: B
Q2: Two identical springs (each with constant \( k \)) are connected in parallel. What is the effective spring constant of the system?
In a parallel arrangement, both springs undergo the same displacement \( x \). The total restoring force is the sum of individual forces:
\( F_{total} = kx + kx = 2kx \). Therefore, \( k_{eff} = 2k \).
Correct Answer: C
\( F_{total} = kx + kx = 2kx \). Therefore, \( k_{eff} = 2k \).
Correct Answer: C
Q3: A mass \( m \) on a frictionless surface is compressed by a distance \( A \) against a spring \( k \) and released. What is its acceleration at the moment of release?
At the point of maximum compression, the restoring force is at its maximum: \( F_{max} = kA \).
Using Newton’s Second Law: \( a = F/m = \displaystyle\frac{kA}{m} \).
Correct Answer: B
Using Newton’s Second Law: \( a = F/m = \displaystyle\frac{kA}{m} \).
Correct Answer: B
Section 2.9
Resistive Forces
Resistive forces (drag) oppose motion and are typically dependent on the object’s velocity. Terminal velocity is reached when the drag force magnitude equals the driving force (e.g., gravity).
Q1: A mass \( m \) falls from rest through a fluid with a drag force \( F_D = -bv \). What is the differential equation describing its motion?
Gravity acts downward while the resistive force acts upward (opposing velocity).
Sum of forces: \( \sum F = mg – bv \).
By \( F = ma \): \( mg – bv = m \displaystyle\frac{dv}{dt} \).
Correct Answer: A
Sum of forces: \( \sum F = mg – bv \).
By \( F = ma \): \( mg – bv = m \displaystyle\frac{dv}{dt} \).
Correct Answer: A
Q2: What is the terminal velocity \( v_t \) for an object experiencing quadratic drag \( F_D = -cv^2 \) while falling under gravity?
Terminal velocity occurs when acceleration is zero (\( a=0 \)).
\( mg – cv_t^2 = 0 \Rightarrow cv_t^2 = mg \Rightarrow v_t = \sqrt{\displaystyle\frac{mg}{c}} \).
Correct Answer: B
\( mg – cv_t^2 = 0 \Rightarrow cv_t^2 = mg \Rightarrow v_t = \sqrt{\displaystyle\frac{mg}{c}} \).
Correct Answer: B
Section 2.10
Circular Motion
Circular motion requires a centripetal force directed toward the center of rotation. In non-uniform circular motion, objects also experience tangential acceleration.
Q1: What is the ideal speed \( v \) for a car on a **banked** curve (angle \( \theta \), radius \( R \)) where no friction is required?
The horizontal component of the normal force provides the centripetal force: \( F_N \sin\theta = \displaystyle\frac{mv^2}{R} \).
The vertical component balances weight: \( F_N \cos\theta = mg \).
Dividing the equations gives \( \tan\theta = \displaystyle\frac{v^2}{gR} \Rightarrow v = \sqrt{gR \tan\theta} \).
Correct Answer: C
The vertical component balances weight: \( F_N \cos\theta = mg \).
Dividing the equations gives \( \tan\theta = \displaystyle\frac{v^2}{gR} \Rightarrow v = \sqrt{gR \tan\theta} \).
Correct Answer: C
Q2: A ball is swung in a **vertical** circle. What is the minimum speed at the very top of the path to keep the string from going slack?
At the highest point, both tension \( T \) and gravity \( mg \) point toward the center.
\( T + mg = \displaystyle\frac{mv^2}{R} \).
The string goes slack when \( T = 0 \), so \( mg = \displaystyle\frac{mv^2}{R} \Rightarrow v = \sqrt{gR} \).
Correct Answer: A
\( T + mg = \displaystyle\frac{mv^2}{R} \).
The string goes slack when \( T = 0 \), so \( mg = \displaystyle\frac{mv^2}{R} \Rightarrow v = \sqrt{gR} \).
Correct Answer: A
Dynamics Formulas Recap
Hooke’s Law\( F_s = -kx \)
Spring Work\( W = \displaystyle\frac{1}{2}kx^2 \)
Linear Drag\( F_D = -bv \)
Terminal Velocity\( v_t = \displaystyle\frac{mg}{b} \)
Radial Accel\( a_c = \displaystyle\frac{v^2}{r} \)
Banked Curve\( \tan\theta = \displaystyle\frac{v^2}{gR} \)
