AP Physics C Mechanics Quiz: Systems And Center Of Mass

20 questions · exam conditions

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Systems And Center Of MassQuestion 1 of 20

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?

2.3\text{ m}

2.7\text{ m}

3.3\text{ m}

4.0\text{ m}

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AP Physics C Mechanics Quiz

AP Physics C Mechanics Quiz: Systems And Center Of Mass

Practice Systems And Center Of Mass in AP Physics C Mechanics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Systems And Center Of Mass, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Physics C Mechanics.

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Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

$2.3\text{ m}$
$2.7\text{ m}$ (correct answer)
$3.3\text{ m}$
$4.0\text{ m}$

Explanation: 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.

Question 2

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?

$2.0\text{ m}$
$3.0\text{ m}$ (correct answer)
$4.0\text{ m}$
$6.0\text{ m}$

Explanation: 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.

Question 3

A uniform rod of mass $M$ and length $L$ lies along the $x$ -axis with one end at the origin. A point mass $m$ is attached to the rod at a distance $\frac{L}{4}$ from the origin. What is the $x$ -coordinate of the center of mass of the combined system?

$\frac{L(2M + m)}{4(M + m)}$ (correct answer)
$\frac{L(2M + m)}{2(M + m)}$
$\frac{L(M + 2m)}{4(M + m)}$
$\frac{L(M + 4m)}{4(M + m)}$

Explanation: The uniform rod's center of mass is at

\frac{L}{2}

. Using the center of mass formula:

x_{cm} = \frac{M \cdot \frac{L}{2} + m \cdot \frac{L}{4}}{M + m} = \frac{\frac{ML}{2} + \frac{mL}{4}}{M + m} = \frac{\frac{2ML + mL}{4}}{M + m} = \frac{L(2M + m)}{4(M + m)}

. Choice B has an incorrect denominator factor. Choice C incorrectly weights the masses. Choice D uses an incorrect coefficient for the point mass.

Question 4

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?

$(3.5, 3.25)\text{ m}$
$(3.75, 3.25)\text{ m}$ (correct answer)
$(4.25, 2.75)\text{ m}$
$(4.5, 2.5)\text{ m}$

Explanation: The original system can be treated as a single particle of mass

12\text{ kg}

(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.

Question 5

A system consists of four identical masses $m$ located at the corners of a square with side length $a$ . If one corner mass is removed, what is the distance from the geometric center of the original square to the center of mass of the remaining three-mass system?

$\frac{a}{6}$
$\frac{a}{4}$
$\frac{a\sqrt{2}}{6}$ (correct answer)
$\frac{a\sqrt{2}}{4}$

Explanation: Place the square with corners at

(\pm a/2, \pm a/2)

. Remove the mass at

(a/2, a/2)

. The remaining masses are at

(-a/2, a/2)

(-a/2, -a/2)

, and

(a/2, -a/2)

. The center of mass is at:

x_{cm} = \frac{-a/2 - a/2 + a/2}{3} = -\frac{a}{6}

and

y_{cm} = \frac{a/2 - a/2 - a/2}{3} = -\frac{a}{6}

. The distance from origin is

\sqrt{(a/6)^2 + (a/6)^2} = \frac{a\sqrt{2}}{6}

. Choice A gives only one coordinate component. Choice B uses incorrect weighting. Choice D doubles the correct result.

Question 6

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?

$\frac{2}{3}\hat{i}\text{ m/s}$
$\frac{5}{6}\hat{i}\text{ m/s}$
$\frac{7}{6}\hat{i}\text{ m/s}$ (correct answer)
$\frac{4}{3}\hat{i}\text{ m/s}$

Explanation: 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.

Question 7

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?

$\frac{R}{\pi}$
$\frac{2R}{\pi}$ (correct answer)
$\frac{R}{2}$
$\frac{R\sqrt{2}}{\pi}$

Explanation: For a uniform semicircular wire, using polar coordinates where

y = R\sin\theta

and

\theta

ranges from

0

\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.

Question 8

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?

$1.0\text{ m/s}$
$2.0\text{ m/s}$ (correct answer)
$3.0\text{ m/s}$
$4.0\text{ m/s}$

Explanation: 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.

Question 9

A uniform solid hemisphere of mass $M$ and radius $R$ sits on a flat surface with its curved surface in contact with the surface and its flat circular face pointing upward. What is the height of the center of mass above the flat surface?

$\frac{3R}{8}$ (correct answer)
$\frac{3R}{5}$
$\frac{R}{2}$
$\frac{2R}{3}$

Explanation: For a uniform solid hemisphere, the center of mass is located at a distance of

\frac{3R}{8}

from the flat face along the axis of symmetry. This can be derived using integration in spherical coordinates. When the hemisphere sits with its curved surface on the ground, the flat face is at height

R

above the ground, but the center of mass is at

\frac{3R}{8}

above the flat surface, which means it's at height

R - \frac{3R}{8} = \frac{5R}{8}

above the ground. However, the question asks for the height above the flat surface, which is

\frac{3R}{8}

. Choice B gives an incorrect fraction. Choice C would be correct for a uniform semicircular disk. Choice D overestimates the center of mass location.

Question 10

Three particles are arranged in a straight line. Particle A has mass $2m$ at position $x = 0$ , particle B has mass $m$ at position $x = L$ , and particle C has mass $3m$ at position $x = 2L$ . What is the distance from particle A to the center of mass of the system?

$\frac{4L}{3}$
$\frac{5L}{3}$
$\frac{7L}{6}$ (correct answer)
$\frac{4L}{3}$

Explanation: The center of mass position is:

x_{cm} = \frac{2m \cdot 0 + m \cdot L + 3m \cdot 2L}{2m + m + 3m} = \frac{0 + L + 6L}{6m} = \frac{7L}{6}

. The distance from particle A (at

x = 0

) to the center of mass is

\frac{7L}{6} - 0 = \frac{7L}{6}

. Choice A represents an incorrect average. Choice B uses wrong mass weighting. Choice D duplicates choice A with calculation error.

Question 11

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 $x = R$ ?

$m = \frac{M}{2}$ (correct answer)
$m = \frac{M}{3}$
$m = \frac{2M}{3}$
$m = M$

Explanation: 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

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.

Question 12

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?

$( 2.5, 0.5, 2.5)\text{ m}$
$(3.0, 0, 2.17)\text{ m}$
$(3.0, 0, 2.33)\text{ m}$ (correct answer)
$(2.67, 0.67, 2.67)\text{ m}$

Explanation: 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.

Question 13

A thin uniform rod of mass $M$ and length $L$ has a linear mass density that varies as $\lambda(x) = \lambda_0\left(1 + \frac{x}{L}\right)$ , where $x$ is measured from one end. What is the $x$ -coordinate of the center of mass?

$\frac{5L}{9}$
$\frac{7L}{12}$ (correct answer)
$\frac{2L}{3}$
$\frac{3L}{4}$

Explanation: The center of mass is

x_{cm} = \frac{\int_0^L x \lambda(x) dx}{\int_0^L \lambda(x) dx}

. First,

\int_0^L \lambda_0(1 + x/L) dx = \lambda_0[x + x^2/(2L)]_0^L = \lambda_0(L + L/2) = \frac{3\lambda_0 L}{2}

. Then,

\int_0^L x \lambda_0(1 + x/L) dx = \lambda_0 \int_0^L (x + x^2/L) dx = \lambda_0[x^2/2 + x^3/(3L)]_0^L = \lambda_0(L^2/2 + L^2/3) = \frac{5\lambda_0 L^2}{6}

. Therefore,

x_{cm} = \frac{5\lambda_0 L^2/6}{3\lambda_0 L/2} = \frac{5L}{9} \cdot \frac{2}{3} = \frac{7L}{12}

. The other choices result from integration errors or incorrect density function applications.

Question 14

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.

The CM accelerates at $a_{\text{CM}}=8.0\,\text{m/s}^2$ to the right.
The CM accelerates at $a_{\text{CM}}=2.0\,\text{m/s}^2$ to the right. (correct answer)
The CM accelerates at $a_{\text{CM}}=0\,\text{m/s}^2$ because tension cancels the force.
The CM accelerates at $a_{\text{CM}}=2.0\,\text{m/s}^2$ to the left.

Explanation: 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.

Question 15

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?

$(1.0, 1.5)$
$(2.0, 1.5)$
$(3.0, 1.5)$ (correct answer)
$(2.5, 3.0)$

Explanation: 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.

Question 16

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?

$(4, -2)\text{ m}$
$(6, -5)\text{ m}$ (correct answer)
$(8, -3)\text{ m}$
$(10, -1)\text{ m}$

Explanation: 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.

Question 17

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?

$\frac{a}{3}$
$\frac{a\sqrt{3}}{6}$
$\frac{a\sqrt{3}}{3}$ (correct answer)
$\frac{a}{2}$

Explanation: 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

\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.

Question 18

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?

$\left(\frac{L}{2}, \frac{L}{2}\right)$
$\left(\frac{L}{3}, \frac{L}{3}\right)$
$\left(\frac{L}{4}, \frac{L}{4}\right)$ (correct answer)
$\left(\frac{2L}{3}, \frac{2L}{3}\right)$

Explanation: 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.

Question 19

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?

$\frac{L}{2}$
$\frac{L}{3}$
$\frac{L}{4}$ (correct answer)
$\frac{L}{6}$

Explanation: 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.

Question 20

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?

$12 \text{ N}$ (correct answer)
$15 \text{ N}$
$18 \text{ N}$
$20 \text{ N}$

Explanation: 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.