Dynamics MCQ Study Guide: AP Physics 1

Spring Dynamics and Equilibrium on Inclined Planes

In a scenario involving an object of mass $m$ attached to a spring on a frictionless inclined plane that makes an angle $\theta$ with the horizontal, the system undergoes specific mechanical behavior upon release from rest. When the spring is in its initial unstretched position, the displacement $x$ is zero. As the object is released, it moves along the plane and subsequently oscillates about an equilibrium position. This equilibrium occurs at a specific displacement denoted as $x_0$.

To determine the spring constant $k$ of the spring, one must analyze the forces acting on the object at the equilibrium position $x_0$. At equilibrium, the net force $\sum F$ acting on the mass is zero. The forces acting parallel to the surface of the inclined plane include the component of the gravitational force pulling the block down the incline and the restoring force of the spring pulling the block up the incline. The gravitational component is calculated as $F_{gx} = m \times g \times \sin(\theta)$. The spring force is defined by Hooke's Law as $F_s = k \times x_0$. Setting these forces equal to satisfy the equilibrium condition yields:

$k \times x_0 = m \times g \times \sin(\theta)$

Solving for the spring constant results in the expression:

$k = \frac{m \times g \times \sin(\theta)}{x_0}$

Identification of Action-Reaction Force Pairs

Newton's Third Law of Motion states that for every action, there is an equal and opposite reaction. These forces always act on different objects. In a configuration where a stick is used to hit a ball at an angle above the horizontal, several forces are at play. When the stick exerts a force on the ball (the action), the ball simultaneously exerts a force of equal magnitude and opposite direction back on the stick (the reaction).

In free body diagrams (FBDs), these forces must be identified by their interaction. For instance, the force labeled as the stick pushing the ball and the force labeled as the ball pushing the stick constitute a valid action-reaction pair. Other forces present, such as gravity acting on the ball or the normal force from the ground, do not form an action-reaction pair with the stick's force because they involve different interactions (e.g., the Earth’s gravity interacts with the ball's mass, not the stick).

Statics and Hooke’s Law in Multi-Spring Systems

When a system involves multiple springs supporting a single load, such as a board hung from two identical springs, the load is distributed across both supports. In the provided case, each spring has a spring constant $k = 10,000\,N/m$. When a person of mass $m = 50\,kg$ sits on the board and the system reaches equilibrium, the total downward force is the weight of the person (assuming the board's mass is negligible or already accounted for in the initial state). The total weight $W$ is calculated as:

$W = m \times g = 50\,kg \times 10\,m/s^2 = 500\,N$

Because the board is in equilibrium and supported by two springs, the total upward force provided by the springs must equal the downward weight. If each spring stretches by a distance $x$, the total upward force is $F_{total} = 2 \times k \times x$. The equilibrium equation is:

$2 \times k \times x = m \times g$

$2 \times (10,000\,N/m) \times x = 500\,N$

$20,000\,x = 500$

$x = \frac{500}{20,000} = 0.025\,m$

Thus, each spring will stretch by $0.025\,m$.

Internal Forces and the Impossibility of Self-Propulsion

A student may claim that they can use their own hands to apply a constant force on their own body to cause it to move or fall backward. This claim is evaluated through the lens of Newton’s Laws. According to Newton’s Second Law, an acceleration of the center of mass requires a net external force. When a person uses their hands to push their own body, the hands and the body are parts of the same system. The force exerted by the hands on the body and the force exerted by the body on the hands are internal forces. According to Newton’s Third Law, these forces are equal in magnitude and opposite in direction ($F_{hands \rightarrow body} = -F_{body \rightarrow hands}$), meaning they cancel each other out within the system. Consequently, they cannot exert a net force on the system’s center of mass to cause a change in motion. The claim is incorrect because the student’s body will exert a force of equal magnitude back on the student’s hands, resulting in no net external force.

Experimental Analysis of Gravitational Acceleration (Atwood Machines)

In an Atwood machine experiment designed to determine the value of $g$ (acceleration due to gravity), two blocks of masses $M_1$ and $M_2$ are connected by a string of negligible mass over a pulley of negligible mass. The acceleration $a$ of the blocks is measured as they are released from rest.

Applying Newton’s Second Law to the system, the net force is the difference in the weights of the two blocks: $F_{net} = (M_1 - M_2) \times g$. The total mass being accelerated is $M_{total} = M_1 + M_2$. The relationship is expressed as:

$a = g \times \frac{M_1 - M_2}{M_1 + M_2}$

If the measured acceleration is significantly less than the expected value of $g$, the data provides a reasonable determination of $g$ if the analysis accounts for the net force on the blocks-string system. However, if the calculated value of $g$ is consistently too small, it suggests the presence of an additional unaccounted force acting on the system, such as friction in the pulley or air resistance.

Vertical Dynamics and Tension

Consider a block of mass $m = 2\,kg$ suspended from a rope. If the tension $T$ in the rope is $10\,N$, the acceleration of the block can be determined by analyzing the vertical forces. The weight of the block is $F_g = m \times g = 2\,kg \times 10\,m/s^2 = 20\,N$, acting downward. The tension acts upward. Let upward be the positive direction:

$\sum F = T - mg = m \times a$

$10\,N - 20\,N = 2\,kg \times a$

$-10\,N = 2\,kg \times a$

$a = -5\,m/s^2$

An acceleration of $-5\,m/s^2$ indicates the block is accelerating downward at $5\,m/s^2$.

In a system where two masses $m$ and $M$ are suspended vertically and lowered with a constant downward acceleration $a$, the tension in the string above both masses (at point P) must support both masses. The total mass is $(m + M)$. The equation of motion for the system is:

$(m + M) \times g - T_P = (m + M) \times a$

Rearranging to solve for the tension $T_P$ at point P:

$T_P = (m + M)(g - a)$

Force Distribution in Connected Block Systems

When two blocks A and B (masses $M_a$ and $M_b$) are connected by a light string and pulled across a frictionless surface by a constant horizontal force $F$, they share the same acceleration $a$. The force $F$ acts on the entire system of mass $(M_a + M_b)$, such that $F = (M_a + M_b) \times a$.

The tension $T$ in the string pulling block B forward is the force responsible for accelerating block B alone. Therefore, $T = M_b \times a$. Because $F$ must accelerate the combined mass of both blocks while $T$ only accelerates the trailing block, it is a physical requirement that $F$ is greater than $T$. Specifically, if $F$ is the leading force applied to block A, then $F$ must be greater than the tension $T$ in the coupling string.

Dynamics of Running and Locomotion

When a person runs on a track, the force that propels the runner forward is the force of friction exerted by the ground on the person. As the runner's foot pushes backward against the ground, the ground exerts an equal and opposite frictional force forward on the foot. Without this friction—such as on a perfectly frictionless surface like smooth ice—the runner would be unable to generate the horizontal force necessary for forward acceleration. The normal force, while present, acts vertically and does not provide forward propulsion.

Net Force Calculations on Inclined Planes

A box of mass $m$ on a rough inclined plane at angle $\theta$ is subjected to an external force $F$ applied at an angle $\Phi$ relative to the plane. As the block moves up the plane, several forces must be considered along the axis parallel to the incline:

1. The component of the applied force parallel to the plane: $F \times \cos(\Phi)$.
2. The component of gravity acting down the plane: $m \times g \times \sin(\theta)$.
3. The frictional force opposing the motion: $f$.

The magnitude of the net force acting on the box as it moves up the plane is expressed as:

$F_{net} = F \times \cos(\Phi) - m \times g \times \sin(\theta) - f$

Gravity and Weight Consistency

During an elevator ride from the first floor to the fifth floor, the person’s apparent weight (the normal force) changes as the elevator accelerates and decelerates. However, the true weight of the person—defined as the gravitational force exerted by Earth on the person—remains constant throughout the ride. According to Newton’s Third Law, the gravitational force exerted by the person on Earth is the reaction pair to the Earth's gravitational pull on the person; therefore, the gravitational force exerted by the person on Earth is always equal to the person's weight at all times during the journey.

Forces on Multi-Object Systems in Contact

In a system where three blocks (A, B, and C) are pushed by a constant force $F$, the acceleration is $a = \frac{F}{M_A + M_B + M_C}$. The normal force between A and B ($N_{AB}$) is responsible for accelerating blocks B and C. The normal force between B and C ($N_{BC}$) is responsible for accelerating only block C.

If a small object is moved from the top of block B to the top of block C:

1. The total mass of the system remains the same, so the overall acceleration $a$ stays the same.
2. The mass being pushed by $N_{AB}$ (blocks B and C plus the small object) remains the same because the small object is still part of the load pushed by block A. Thus, $N_{AB}$ stays the same.
3. The mass being pushed by $N_{BC}$ (block C plus the small object) increases. Since the acceleration is constant and the mass of the "C + object" unit has increased, the force $N_{BC}$ must increase to maintain that acceleration.

Conservation of Momentum and Center of Mass Systems

A person of mass $40\,kg$ standing on a raft of mass $120\,kg$ in a lake (with negligible drag) constitutes a system where momentum is conserved. If the person begins to walk to the right at a speed $v$ relative to the water, the raft must move to the left to keep the system's total momentum at zero (assuming it started at rest).

Using conservation of momentum ($p_{initial} = p_{final} = 0$):

$m_{person} \times v_{person} + m_{raft} \times v_{raft} = 0$

$40 \times v + 120 \times v_{raft} = 0$

$120 \times v_{raft} = -40 \times v$

$v_{raft} = -\frac{40}{120} \times v = -\frac{1}{3} \times v$

This indicates the raft moves to the left at a speed less than $v$.

Pulley System Dynamics on Horizontal Surfaces

For a system consisting of a $4.0\,kg$ block on a frictionless table connected by a string over a pulley to a hanging $1.0\,kg$ block, the acceleration is determined by the external force of gravity on the hanging mass.

The net force on the system is the weight of the hanging block: $F_{net} = 1.0\,kg \times 10\,m/s^2 = 10\,N$.

The total mass of the system is the sum of the sliding and hanging masses: $M_{total} = 4.0\,kg + 1.0\,kg = 5.0\,kg$.

The acceleration of the system is:

$a = \frac{F_{net}}{M_{total}} = \frac{10\,N}{5.0\,kg} = 2.0\,m/s^2$

Rate of Change of Center of Mass Speed

When a block of mass $M_2$ enters a frictional surface with coefficient $\mu$, after colliding with mass $M_1$, the only external horizontal force acting on the two-block system is the kinetic friction on block $M_2$. This force is $f_k = \mu \times M_2 \times g$.

The rate of change of the speed of the center of mass ($a_{cm}$) for the system is found by dividing the net external force by the total mass:

$a_{cm} = \frac{\sum F_{ext}}{M_{total}} = \frac{\mu \times M_2 \times g}{M_1 + M_2}$

Vector Components in Static Tension Systems

Suppose a box of mass $m$ hangs from two massless strings forming a $90^\circ$ angle between them. The strings make angles $\theta_1$ and $\theta_2$ with the ceiling. Given that the angle between the strings is $90^\circ$, it follows that $\theta_1 + \theta_2 = 90^\circ$, which implies $\sin(\theta_1) = \cos(\theta_2)$ and $\cos(\theta_1) = \sin(\theta_2)$.

For the system to be in equilibrium, the horizontal components of tension must be equal:

$T_{1x} = T_{2x}$

$T_1 \times \cos(\theta_1) = T_2 \times \cos(\theta_2)$

Based on identity substitution, $T_{2x} = T_1 \times \cos(\theta_1)$.

The vertical component of the tension in string 2, $T_{2y}$, is given by:

$T_{2y} = T_2 \times \sin(\theta_2)$

Substituting $T_2 = T_1 \frac{\cos(\theta_1)}{\cos(\theta_2)}$ into the expression for $T_{2y}$:

$T_{2y} = T_1 \times \frac{\cos(\theta_1)}{\cos(\theta_2)} \times \sin(\theta_2) = T_1 \times \cos(\theta_1) \times \tan(\theta_2)$

Questions & Discussion

Question: A student calculates a cart's mass using $F/a$ to be $0.8\,kg$, but the balance shows $1.0\,kg$. What could explain this? Response: The calculated mass is $m = F_{net}/a$. If the student assumes the applied force is the net force ($F_{applied} = F_{net}$) and ignores friction, they will derive an incorrect mass. If friction $f$ is present, then $F_{applied} - f = m_{true} \times a$. By ignoring $f$, the student uses a larger numerator than the true net force, resulting in an inaccurate mass calculation. Rough bearings in the wheels causing significant friction is a likely cause for this discrepancy.

Question: A ball is moved to a planet with double the gravitational acceleration of Earth. How does the gravitational force compare? Response: The gravitational force, or weight, is defined as $F_g = m \times g$. Since mass $m$ is an intrinsic property and remains constant, doubling the gravitational acceleration $g$ will result in the gravitational force on the ball being exactly double the force on the ball when it is on Earth.

Question: A student rolls balls of different masses ($2\,kg$ and $5\,kg$) into a wall with different forces. Which ball receives a greater force from the wall? Response: According to Newton's Third Law, the force the wall exerts on a ball is equal in magnitude to the force the ball exerts on the wall. Since the $2\,kg$ ball hit the wall with $10,000\,N$ and the $5\,kg$ ball hit with $5,000\,N$, the wall exerts a greater force ($10,000\,N$) on the $2\,kg$ ball.