Systems and Center of Mass - AP Physics C Mechanics

A system of three particles moves such that particle 1 (mass $2m$ ) has velocity $\vec{v_1} = 3\hat{i}\text{ m/s}$ , particle 2 (mass $m$ ) has velocity $\vec{v_2} = -2\hat{i}\text{ m/s}$ , and particle 3 (mass $3m$ ) has velocity $\vec{v_3} = 1\hat{i}\text{ m/s}$ . What is the velocity of the center of mass?
Answer: $\frac{7}{6}\hat{i}\text{ m/s}$ The velocity of the center of mass is: $\vec{v}_{cm} = \frac{\sum m_i \vec{v}_i}{\sum m_i} = \frac{2m(3\hat{i}) + m(-2\hat{i}) + 3m(1\hat{i})}{2m + m + 3m} = \frac{6m\hat{i} - 2m\hat{i} + 3m\hat{i}}{6m} = \frac{7m\hat{i}}{6m} = \frac{7}{6}\hat{i}\text{ m/s}$ . Choice A incorrectly weights the mass contributions. Choice B uses an arithmetic error in the numerator calculation. Choice D represents an incorrect total mass calculation.
Based on the scenario, determine the motion of the system given its center of mass. Two pucks on frictionless ice are connected by a light string and can slide along the $x$ -axis; the string tension is internal to the system. The masses are $m_1=1.0\,\text{kg}$ at $x_1=-0.50\,\text{m}$ and $m_2=3.0\,\text{kg}$ at $x_2=+0.50\,\text{m}$ . A horizontal external force of $+8.0\,\text{N}$ is applied to puck 1, and no other external horizontal forces act. If the system starts from rest, find the CM acceleration immediately after the force is applied.
Answer: The CM accelerates at $a_{\text{CM}}=2.0\,\text{m/s}^2$ to the right. This question tests AP Physics C Mechanics skills: systems and center of mass, specifically understanding how to determine and analyze the center of mass in physical systems. The center of mass acceleration depends only on external forces, as internal forces (like string tension) cancel out when considering the system as a whole. In this scenario, the system consists of two pucks with masses 1.0 kg and 3.0 kg connected by a string, with only an 8.0 N external force applied to the first puck. Choice B is correct because a_CM = F_ext / M_total = 8.0 N / (1.0 + 3.0) kg = 8.0 N / 4.0 kg = 2.0 m/s² to the right. Choice C is incorrect because it assumes the internal tension force somehow cancels the external force, which violates Newton's laws for system analysis. To help students: Stress that internal forces always come in action-reaction pairs that cancel when analyzing the whole system. Practice identifying and separating internal from external forces before applying Newton's second law to the center of mass.
A composite system consists of a solid cylinder of mass $M$ and radius $R$ with its center at the origin, and a point mass $m$ located at distance $3R$ from the origin along the positive $x$ -axis. What condition must be satisfied for the center of mass to be located at $x = R$ ?
Answer: $m = \frac{M}{2}$ Setting the center of mass at $x = R$ : $R = \frac{M \cdot 0 + m \cdot 3R}{M + m} = \frac{3mR}{M + m}$ . Solving: $R(M + m) = 3mR$ , so $M + m = 3m$ , which gives $M = 2m$ or $m = \frac{M}{2}$ . Choice B would place the center of mass too far from the origin. Choice C overestimates the required point mass. Choice D makes the masses equal, shifting the center of mass too far right.
Two identical uniform rods, each of mass $M$ and length $L$ , are arranged to form a T-shape. One rod is horizontal and centered at the origin, while the other is vertical with its bottom end at the origin. What is the $y$ -coordinate of the center of mass of the T-shaped system?
Answer: $\frac{L}{4}$ The horizontal rod (centered at origin) has its center of mass at $(0, 0)$ . The vertical rod (bottom at origin) has its center of mass at $(0, L/2)$ . The combined center of mass is: $y_{cm} = \frac{M \times 0 + M \times L/2}{M + M} = \frac{ML/2}{2M} = \frac{L}{4}$ . Choice A represents the center of mass of just the vertical rod. Choice B uses incorrect weighting of the rod positions. Choice D underestimates the vertical displacement contribution.
A system of particles has a total mass of $12\text{ kg}$ and its center of mass is located at position $(3, 4)\text{ m}$ . If a particle of mass $4\text{ kg}$ is added to the system at position $(6, 1)\text{ m}$ , what is the new center of mass position?
Answer: $(3.75, 3.25)\text{ m}$ The original system can be treated as a single particle of mass $12\text{ kg}$ at $(3, 4)$ . The new center of mass is: $x_{cm} = \frac{12 \cdot 3 + 4 \cdot 6}{12 + 4} = \frac{36 + 24}{16} = 3.75\text{ m}$ and $y_{cm} = \frac{12 \cdot 4 + 4 \cdot 1}{16} = \frac{48 + 4}{16} = 3.25\text{ m}$ . Choice A uses incorrect mass weighting. Choice C reverses the coordinate calculations. Choice D uses equal weighting instead of mass weighting.
Three identical masses $m$ are connected by rigid massless rods to form an equilateral triangle with side length $a$ . If the system rotates about an axis perpendicular to the plane and passing through one of the masses, what is the distance from this axis to the center of mass?
Answer: $\frac{a\sqrt{3}}{3}$ For an equilateral triangle, the center of mass (centroid) is located at the intersection of the medians. The distance from any vertex to the centroid is $\frac{2}{3}$ of the median length. The median (altitude) of an equilateral triangle with side length $a$ is $\frac{a\sqrt{3}}{2}$ . Therefore, the distance from vertex to centroid is $\frac{2}{3} \times \frac{a\sqrt{3}}{2} = \frac{a\sqrt{3}}{3}$ . Choice A ignores the geometric factor. Choice B represents $\frac{1}{3}$ of the altitude. Choice D represents half the side length.
A system consists of two point masses: $m_1 = 2.0\text{ kg}$ at position $\vec{r_1} = (1, 2, 3)\text{ m}$ and $m_2 = 4.0\text{ kg}$ at position $\vec{r_2} = (4, -1, 2)\text{ m}$ . What is the position vector of the center of mass?
Answer: $(3.0, 0, 2.33)\text{ m}$ The center of mass position is: $\vec{r}_{cm} = \frac{m_1\vec{r_1} + m_2\vec{r_2}}{m_1 + m_2}$ . Computing each component: $x_{cm} = \frac{2.0 \times 1 + 4.0 \times 4}{6.0} = \frac{18}{6.0} = 3.0$ , $y_{cm} = \frac{2.0 \times 2 + 4.0 \times (-1)}{6.0} = \frac{0}{6.0} = 0$ , $z_{cm} = \frac{2.0 \times 3 + 4.0 \times 2}{6.0} = \frac{14}{6.0} = 2.33$ . Therefore $\vec{r}_{cm} = (3.0, 0, 2.33)\text{ m}$ . The other choices use incorrect mass weightings or arithmetic errors.
A system consists of two identical spheres, each of mass $m$ , connected by a massless rod. Initially, the spheres are at positions $(0, 0)$ and $(4, 0)$ . If one sphere moves to position $(2, 3)$ while the other remains fixed, what is the new position of the center of mass?
Answer: $(3.0, 1.5)$ With equal masses, the center of mass is the average of the positions: $x_{cm} = \frac{x_1 + x_2}{2} = \frac{2 + 4}{2} = 3.0$ and $y_{cm} = \frac{y_1 + y_2}{2} = \frac{3 + 0}{2} = 1.5$ . So the center of mass is at $(3.0, 1.5)$ . Choice A uses incorrect weighting of the coordinates. Choice B places the center of mass at the moving sphere's $x$ -coordinate. Choice D incorrectly averages only one coordinate properly.
A system consists of three point masses: $m_1 = 2.0\text{ kg}$ at position $x_1 = 0\text{ m}$ , $m_2 = 3.0\text{ kg}$ at position $x_2 = 4.0\text{ m}$ , and $m_3 = 1.0\text{ kg}$ at position $x_3 = 6.0\text{ m}$ . What is the $x$ -coordinate of the center of mass of this system?
Answer: $2.7\text{ m}$ The center of mass is calculated using $x_{cm} = \frac{\sum m_i x_i}{\sum m_i} = \frac{(2.0)(0) + (3.0)(4.0) + (1.0)(6.0)}{2.0 + 3.0 + 1.0} = \frac{0 + 12.0 + 6.0}{6.0} = \frac{18.0}{6.0} = 2.7\text{ m}$ . Choice A uses an incorrect calculation. Choice C represents the simple average of positions without mass weighting. Choice D is the position of the middle mass, not the center of mass.
Two objects with masses $m_1 = 4.0\text{ kg}$ and $m_2 = 8.0\text{ kg}$ are connected by a light rigid rod of negligible mass. If the center of mass of the system is located $2.0\text{ m}$ from $m_1$ , what is the distance between the two masses?
Answer: $3.0\text{ m}$ Using the center of mass definition with $m_1$ at the origin: $x_{cm} = \frac{m_2 d}{m_1 + m_2}$ where $d$ is the separation. Since $x_{cm} = 2.0\text{ m}$ : $2.0 = \frac{8.0 \cdot d}{4.0 + 8.0} = \frac{8.0d}{12.0}$ . Solving: $d = 3.0\text{ m}$ . Choice A equals the distance from $m_1$ to the center of mass. Choice C would be correct if the masses were equal. Choice D assumes an incorrect mass ratio calculation.
A uniform semicircular wire of mass $M$ and radius $R$ lies in the $xy$ -plane with its diameter along the $x$ -axis and center at the origin. What is the $y$ -coordinate of the center of mass?
Answer: $\frac{2R}{\pi}$ For a uniform semicircular wire, using polar coordinates where $y = R\sin\theta$ and $\theta$ ranges from $0$ to $\pi$ . The linear mass density is $\lambda = M/(\pi R)$ . The $y$ -coordinate of the center of mass is: $y_{cm} = \frac{\int_0^\pi R\sin\theta \cdot \lambda \cdot R d\theta}{\int_0^\pi \lambda \cdot R d\theta} = \frac{\lambda R^2 \int_0^\pi \sin\theta d\theta}{\lambda R \cdot \pi} = \frac{R[-\cos\theta]_0^\pi}{\pi} = \frac{R(1-(-1))}{\pi} = \frac{2R}{\pi}$ . Choice A uses half the correct result. Choice C represents the geometric centroid of a semicircle incorrectly. Choice D includes an unnecessary $\sqrt{2}$ factor.
Two blocks of masses $m_1 = 4.0 \text{ kg}$ and $m_2 = 6.0 \text{ kg}$ are connected by a light rope and rest on a frictionless horizontal surface. A horizontal force $F = 30 \text{ N}$ is applied to the $6.0 \text{ kg}$ block. What is the magnitude of the tension in the rope connecting the blocks?
Answer: $12 \text{ N}$ Since the blocks are connected, they have the same acceleration. For the system: $a = \frac{F}{m_1 + m_2} = \frac{30}{4.0 + 6.0} = 3.0 \text{ m/s}^2$ . For the 4.0 kg block alone: $T = m_1 a = 4.0 \times 3.0 = 12 \text{ N}$ . Choice B (15 N) might result from incorrectly using half the applied force. Choice C (18 N) could come from using the wrong mass ratio. Choice D (20 N) might result from incorrect force analysis on the 6.0 kg block.
A projectile of mass $m$ is fired at an angle $\theta$ above the horizontal with initial speed $v_0$ . At the highest point of its trajectory, the projectile explodes into two fragments of equal mass. One fragment continues horizontally with speed $2v_0 \cos\theta$ , and the other fragment lands at the same horizontal distance from the launch point as the original projectile would have landed. What is the horizontal component of velocity of the second fragment immediately after the explosion?
Answer: $0$ This problem combines projectile motion with conservation of momentum during an explosion. When you see fragments created at the highest point of a trajectory, focus on applying momentum conservation in both horizontal and vertical directions. At the highest point, the original projectile has only horizontal velocity $v_0 \cos\theta$ (vertical velocity is zero). After the explosion, momentum must be conserved. Let's call the horizontal velocity of the second fragment $v_x$ . For horizontal momentum conservation: $m(v_0 \cos\theta) = \frac{m}{2}(2v_0 \cos\theta) + \frac{m}{2}(v_x)$ Simplifying: $mv_0 \cos\theta = mv_0 \cos\theta + \frac{m}{2}v_x$ This gives us $v_x = 0$ . The key insight is that the second fragment lands at the same horizontal distance as the original projectile would have. Since it starts from the peak with zero horizontal velocity, it falls straight down and then must travel the remaining horizontal distance through some other means - but the problem states it lands at the original impact point, which is only possible if its horizontal velocity is zero throughout its fall. Choice A ( $v_0 \cos\theta$ ) would violate momentum conservation. Choice C ( $-v_0 \cos\theta$ ) would also violate momentum conservation, giving the wrong total momentum. Choice D ( $2v_0 \cos\theta$ ) would mean both fragments have the same velocity, which wouldn't conserve momentum given the initial conditions. Remember: In explosion problems, always write out momentum conservation equations immediately. The math often reveals the answer more quickly than trying to visualize the motion.
A system consists of two particles connected by a spring. Particle A has mass $2m$ and particle B has mass $m$ . If the system is subject to no external forces and particle A has acceleration $\vec{a}_A = 3\hat{i} - 4\hat{j}$ at some instant, what is the acceleration of the center of mass of the system at that instant?
Answer: $\vec{a}_{cm} = 0$ When you encounter a system with no external forces, immediately think about Newton's first law and the center of mass. The key insight is that the center of mass of an isolated system (no external forces) cannot accelerate. Here's why: Newton's second law for the entire system states that $\vec{F}_{external} = M_{total}\vec{a}_{cm}$ , where $M_{total}$ is the total mass and $\vec{a}_{cm}$ is the center of mass acceleration. Since there are no external forces acting on this spring-particle system, $\vec{F}_{external} = 0$ . Therefore, $\vec{a}_{cm} = 0$ , regardless of what's happening internally between the particles. The spring force between particles A and B is an internal force. By Newton's third law, if the spring exerts a force on particle A, it exerts an equal and opposite force on particle B. These internal forces cause the individual particles to accelerate, but they cannot change the motion of the center of mass. Choice A ( $2\hat{i} - \frac{8}{3}\hat{j}$ ) incorrectly assumes the center of mass acceleration is proportional to particle A's acceleration. Choice B ( $\hat{i} - \frac{4}{3}\hat{j}$ ) attempts to weight the accelerations by mass but ignores that we don't know particle B's acceleration. Choice D ( $3\hat{i} - 4\hat{j}$ ) incorrectly sets the center of mass acceleration equal to particle A's acceleration. Study tip: Whenever you see "no external forces" in a mechanics problem, immediately check if the question asks about center of mass motion. The center of mass of an isolated system always moves at constant velocity (zero acceleration).
Three identical blocks, each of mass $m$ , are arranged in a line on a frictionless surface. Block 1 moves with velocity $v_0$ toward stationary blocks 2 and 3. After block 1 collides elastically with block 2, and block 2 subsequently collides elastically with block 3, what are the final velocities of the three blocks?
Answer: $v_1 = 0, v_2 = 0, v_3 = v_0$ When you encounter elastic collision problems with identical masses, think systematically about conservation laws and the special properties that make these collisions predictable. For elastic collisions between identical masses where one is initially at rest, the moving object stops completely and transfers all its momentum and kinetic energy to the stationary object. This is a key principle you should memorize. Let's trace through the collisions step by step. Initially: Block 1 has velocity $v_0$ , blocks 2 and 3 are at rest. First collision (Block 1 hits Block 2): Since the masses are equal and Block 2 is at rest, Block 1 stops completely ( $v_1 = 0$ ) and Block 2 acquires velocity $v_0$ . Block 3 remains at rest. Second collision (Block 2 hits Block 3): Again, equal masses with Block 3 initially at rest means Block 2 stops completely ( $v_2 = 0$ ) and Block 3 acquires velocity $v_0$ ( $v_3 = v_0$ ). Final result: $v_1 = 0, v_2 = 0, v_3 = v_0$ , which is answer D. Choice A incorrectly suggests partial momentum transfer and a negative velocity. Choice B assumes the momentum gets evenly distributed among all blocks, which would violate the sequential nature of the collisions. Choice C incorrectly has Block 2 retaining half its velocity after hitting Block 3. Remember this pattern: in elastic collisions between identical masses where one is at rest, momentum and energy transfer completely from the moving object to the stationary one. This creates a "domino effect" in multi-collision scenarios.
A uniform chain of mass $M$ and length $L$ hangs vertically from one end. A portion of length $x$ (measured from the bottom) is then lifted and placed horizontally on a frictionless table while the remaining portion of length $(L-x)$ continues to hang vertically. What is the position of the center of mass of the entire chain, measured vertically downward from the table surface?
Answer: $\frac{(L-x)^2}{2L}$ When you encounter center of mass problems involving objects with different orientations, you need to treat each portion separately and then combine them using the weighted average principle. The chain has two distinct portions: a horizontal section of length $x$ on the table, and a vertical hanging section of length $(L-x)$ . Since the chain is uniform, its linear mass density is $\lambda = M/L$ . For the horizontal portion (mass $\frac{Mx}{L}$ ), the center of mass is at table level, contributing zero to the vertical position. For the hanging portion (mass $\frac{M(L-x)}{L}$ ), the center of mass is at the midpoint of the hanging length, which is $\frac{L-x}{2}$ below the table. Using the center of mass formula: $y_{cm} = \frac{m_1 y_1 + m_2 y_2}{m_1 + m_2}$ $y_{cm} = \frac{\frac{Mx}{L} \cdot 0 + \frac{M(L-x)}{L} \cdot \frac{L-x}{2}}{M} = \frac{M(L-x)^2}{2LM} = \frac{(L-x)^2}{2L}$ Choice A gives $\frac{L-x}{2}$ , which is just the center of mass of the hanging portion alone, ignoring the horizontal section entirely. Choice B yields $\frac{L^2-x^2}{2L} = \frac{(L-x)(L+x)}{2L}$ , which incorrectly weights the calculation. Choice D gives $\frac{L^2+x^2}{2L}$ , suggesting the portions somehow add constructively rather than being weighted by their masses. Remember: in center of mass problems, always identify each distinct portion, find their individual centers of mass, then use the weighted average based on their masses.
A thin uniform triangular plate has vertices at $(0,0)$ , $(3,0)$ , and $(0,4)$ . If the plate has uniform mass density $\sigma$ , what are the coordinates of its center of mass?
Answer: $(1.0, 1.33)$ For a uniform triangular plate, the center of mass is located at the centroid, which is the average of the vertex coordinates: $x_{cm} = \frac{0 + 3 + 0}{3} = 1.0$ and $y_{cm} = \frac{0 + 0 + 4}{3} = \frac{4}{3} = 1.33$ . Choice B incorrectly uses the midpoint of the triangle's bounding rectangle. Choice C uses incorrect averaging for the $y$ -coordinate. Choice D represents a weighted average that doesn't apply to uniform triangular plates.
A system consists of two masses connected by a spring. Mass $m_1 = 3.0\text{ kg}$ is at rest at position $x_1 = 2.0\text{ m}$ , and mass $m_2 = 1.0\text{ kg}$ moves with velocity $v_2 = 8.0\text{ m/s}$ at position $x_2 = 6.0\text{ m}$ . What is the velocity of the center of mass?
Answer: $2.0\text{ m/s}$ The velocity of the center of mass is calculated as: $v_{cm} = \frac{m_1 v_1 + m_2 v_2}{m_1 + m_2} = \frac{3.0 \times 0 + 1.0 \times 8.0}{3.0 + 1.0} = \frac{0 + 8.0}{4.0} = 2.0\text{ m/s}$ . Choice A represents a quarter of the moving mass's velocity. Choice C uses incorrect mass ratios in the calculation. Choice D represents half of the moving mass's velocity without proper mass weighting.
A system of particles initially has its center of mass at position $(2, 3)\text{ m}$ and moves with velocity $(1, -2)\text{ m/s}$ . After $4\text{ s}$ , what is the position of the center of mass if no external forces act on the system?
Answer: $(6, -5)\text{ m}$ With no external forces, the center of mass moves with constant velocity. Using kinematic equations: $\vec{r}_{cm}(t) = \vec{r}_{cm,0} + \vec{v}_{cm} t = (2, 3) + (1, -2) \times 4 = (2, 3) + (4, -8) = (6, -5)\text{ m}$ . Choice A incorrectly uses only the $x$ -component displacement. Choice C uses incorrect time scaling. Choice D applies wrong velocity components to the calculation.
Two uniform rods, each of mass $M$ and length $L$ , are joined at right angles to form an L-shape. One rod lies along the positive $x$ -axis from the origin, and the other lies along the positive $y$ -axis from the origin. What is the position of the center of mass of the L-shaped system?
Answer: $\left(\frac{L}{4}, \frac{L}{4}\right)$ Each rod has its center of mass at $L/2$ from the origin. Rod 1 (x-axis): center at $(L/2, 0)$ . Rod 2 (y-axis): center at $(0, L/2)$ . Combined center of mass: $x_{cm} = \frac{M \cdot L/2 + M \cdot 0}{2M} = \frac{L}{4}$ and $y_{cm} = \frac{M \cdot 0 + M \cdot L/2}{2M} = \frac{L}{4}$ . Choice A represents the individual rod centers, not the system center. Choice B uses incorrect fractional positioning. Choice D overestimates the center of mass location.

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