Examples, exercises & solutions

August 31, 2026

Updated: 07 Jul 2026

Exercises¶

In the table below, the mass and distance from the sun of the planets in our solar system are given (in terms of the earth mass and distance from the earth to the sun). Compute for each planet-sun pair the distance from the center of mass to the center of the sun. Given: distance CM to center of sun for the earth-sun system is $450 \; \mathrm{km}$ .

planet	relative mass	relative distance to the sun
Mercurius	0.06	0.39
Venus	0.82	0.72
Earth	1.00	1.00
Mars	0.11	1.52
Jupiter	317.8	5.20
Saturnus	95.2	9.54
Uranus	14.6	19.22
Neptunus	17.2	30.06

Exercise 2 (🌶)

Two particles $m_1 = m$ and $m_2 = 2m$ are traveling both along the $x$ -axis. At $t=0$ the particles have both velocity $v_0>0$ . Their positions at $t=0$ are $x_1(0) = x_{10}$ and $x_2(0) = x_{20}$ with $x_{10}<x_{20}$ . They repel each other with a force $F_r = \frac{k}{(x_2-x_1)^2}$ . Moreover, a constant external force $F_e$ is acting on them. The problem is 1-dimensional.

Find the velocity of the center of mass for $t>0$
Find the position of the center of mass for $t>0$ .

Exercise 3 (🌶)

Two particles $m_1 = 3\mathrm{\; kg}$ and $m_2 = 2\; \mathrm{kg}$ are connected via a massless rod of length $L=50\;\mathrm{cm}$ .

Find the position of the center of mass of the system, measured from $m_1$
Calculate the reduced mass of the two-particle system.

Exercise 4 (Bumper car collision 🌶)

Two bumper cars are approaching each other in a straight line. The two cars will collide head-on. The mass of car 1 (including the driver) is $200 \mathrm{\; kg}$ , that of car 2 $300 \mathrm{\; kg}$ . Car 1 has a velocity of $8 \mathrm{\; m/s}$ ; car 2 of $-4 \mathrm{\; m/s}$ .

What is the velocity of the center of mass of the system?
What is the reduced mass of the system?
Transform the velocities of both carts to the center-of-mass frame.

Exercise 5 (🌶)

Two carts on a frictionless track move toward each other:

Cart 1: mass $m_1 = 2\mathrm{\; kg}$ , velocity $v_1 = 4 \mathrm{\; m/s}$
Cart 2: mass $m_2 = 3\mathrm{\; kg}$ , velocity $v_2 = -2 \mathrm{\; m/s}$

What is the total kinetic energy in the lab frame?
What is the velocity of the center of mass?
What is the total kinetic energy in the center-of-mass frame?
Verify that the CM frame kinetic energy equals the kinetic energy due to relative motion using the reduced mass.

Answers¶

Solution to Exercise 1

planet	relative mass	relative distance to the sun	distance CM to center of sun (km)
Mercurius	0.06	0.39	10
Venus	0.82	0.72	265
Earth	1.00	1.00	450
Mars	0.11	1.52	75
Jupiter	317.8	5.20	743 · 10³
Saturnus	095.2	9.54	409 · 10³
Uranus	14.6	19.22	126 · 10³
Neptunus	17.2	30.06	234 · 10³

Solution to Exercise 2

We set up the equation of motion for the particles:

\begin{split} m_1: m_1 \dot{v}_1 &= F_e - F_r \\ m_1: m_2 \dot{v}_2 &= F_e + F_r \end{split}

(1)

Add these two equations:

M\dot{V} = m_1 \dot{v}_1 + m_2 \dot{v}_2 = 2F_e \rightarrow \dot{V} = \frac{2F_e}{m_1+m_2} = \frac{2F_e}{3m}

(2)

As expected, we see that the repelling mutual force has no effect on the center of mass. We can solve this equation, using the initial condition the $MV(0) = m_1v_1(0) + m_2 v_2(0) \rightarrow V(0) = \frac{mv_0 + 2mv_0}{m+2m} = v_0$

V(t) = \frac{2F_e}{3m} t + C_1 = \frac{2F_e}{3m} t + v_0

(3)

As the next step we calculate $R(t)$ :

\dot{R} \equiv V = v_0 + \frac{2F_e}{3m} t \rightarrow R(t) = v_0 t + \frac{F_e}{3m} t^2 + C_2

(4)

The initial condition is: $R(0) = \frac{m_1 x_1(0) + m_2 x_2(0)}{m_1+m_2} = \frac{1}{3}x_{10} + \frac{2}{3}x_{20}$ .

This gives

R(t) = \frac{1}{3}x_{10} + \frac{2}{3}x_{20} + v_0 t + \frac{F_e}{3m} t^2

(5)

Solution to Exercise 3

The center of mass of two point masses is on the line connecting $m_1$ and $m_2$ . We denote this line as the $x$ -axis, with $m_1$ as the origin.

The center of mass is than given by (with $m_1 = 3 \text{ kg}$ , $m_2 = 2 \text{ kg}$ , $x_1=0$ and $x_2 = x_1 + L = 0.5 \text{ m}$ ):

x_{cm} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2} = 0.2 \text{ m}

(6)

The reduced mass is given by:

\mu \equiv \frac{m_1 m_2}{m_1 + m_2} = \frac{6}{5} \text{ kg}

(7)

Solution to Exercise 4

This is a 1-dimensional problem.

The velocity of the center of mass is:

V_{cm} = \frac{m_1 v_1 + m_2 v_2}{m_1 + m_2} = \frac{4}{5} \text{ m/s}

(8)

The reduced mass is given by:

\mu \equiv \frac{m_1 m_2}{m_1 + m_2} = 120 \text{ kg}

(9)

In the CM frame the velocities of the cars are:

\begin{split} v_1' & = v_1 - V_{cm} = 7.2 \text{ m/s} \\ v_2' & = v_2 - V_{cm} = -4.8 \text{ m/s} \end{split}

(10)

Solution to Exercise 5

Cart 1: mass $m_1 = 2 \text{ kg}$ , velocity $v_1 = 4 \text{ m/s}$
Cart 2: mass $m_2 = 3 \text{ kg}$ , velocity $v_2 = -2 \text{ m/s}$

The total kinetic energy in the lab frame is

E_{kin} = \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2 = 22 \text{ J}

(11)

The velocity of the center of mass is

V_{cm} \equiv \frac{m_1v_1 + m_2 v_2}{m_1 + m_2} = 0.4 \text{ m/s}

(12)

The total kinetic energy in the center-of-mass frame is

E_{kin,CM} = \frac{1}{2}m_1v_1'^2 + \frac{1}{2}m_2v_2'^2

(13)

with

\begin{split} v_1' &= v_1 - V_{cm} = 3.6 \text{ m/s} \\ v_2' &= v_2 - V_{cm} = -2.4 \text{ m/s} \end{split}

(14)

Thus

E_{kin,CM} = 21.6 \text{ J}

(15)

The reduced mass is

\mu \equiv \frac{m_1 m_2}{m_1 + m_2} = 1.2 \text{ kg}

(16)

The relative velocity is

v_{rel} \equiv v_1 - v_2 = 6 \text{ m/s}

(17)

The kinetic energy associated with the motion of the reduced mass (i.e. the kinetic energy in the CM frame) is:

E_{kin, rel} \equiv \frac{1}{2} \mu v_{rel}^2 = 21.6 \text{ J}

(18)

as we expected.

Exercises¶

Exercise 6 (🌶)

Three masses are forming an equilateral triangle with sides of $2 \; \mathrm{m}$ . Mass 1 ( $10 \; \mathrm{kg}$ ) is positioned at $(x_1,y_1) = (-1m,0)$ . Mass 2 ( $6 \; \mathrm{kg}$ ) is at $(x_2,y_2) = (1m,0)$ , while mass 3 ( $2 \; \mathrm{kg}$ ) is at $(x_3,y_3) = (0,\sqrt{3})$ .

Calculate the position of the center of mass.

Exercise 7 (🌶 🌶)

Four particles are moving over the line $y=y_0$ . The particles have mass $m_1 = 4m, m_2 = 3m, m_3 = 2m, m_4 = m$ and velocity $v_1 = v, v_2 = 2v, v_3 = 3v, v_4 = 4v$ . These velocities are constant and parallel to the $x$ -axis. At $t=0$ all particles are at location $(x,y) = (0,y_0)$ .

Figure 1:Four particles moving on a line.

Calculate the velocity of the center of mass.
Calculate the position of the center of mass as a function of time.
Calculate the total angular momentum.
Calculate the angular momentum associated with the center of mass and show that in this case this is equal to the total angular momentum.

Exercise 8 (🌶 🌶)

Eight point particles (each mass $m$ ) are attached to a massless wheel of radius $R$ .

Figure 2:Eight particles on a wheel.

The wheel is moving with a velocity $V$ while it is rotating at the same time with angular velocity $\omega$ .

Calculate the total kinetic energy of this system. Hint: use the CM frame and connect that to the lab frame.

Exercise 9 (🌶 🌶 🌶)

A container of volume $V_c$ and mass $M_c$ contains Nitrogen gas. The number of molecules, $N$ , is on the order of 10²³. The container is dropped from a height $H$ . Gravity is acting on the molecules. Friction on the container is ignored.

Show that the container falls with the acceleration of gravity $g$ .

Exercise 10 (🌶 🌶 🌶)

We consider a 2-dimensional problem: $N=30$ particles move in the $xy$ -plane. Each particle has a fixed velocity $(v_x^i, v_y^i )$ with $i = 1 ..N$ . The particle velocities have a magnitude ranging from 1 to 5 (m/s) randomly chosen for each particle. The direction of each velocity vector is also randomly chosen from 0 to 2 $\pi$ . The particles move inside a box with sides $L=50 \text{ m}$ . Particles do not collide with each other, but they do collide with the walls of the container. The result of a collision is that the particle motion gets reflected.

Write a python program that generates N particles starting all at $(x,y)=(0,0)$ .
Compute the position of all particles after 1 second and compute the velocity and position of the center of mass.
Write a loop that updates the particle velocities after a time step $dt$ and recompute the velocity and position of the center of mass.
Run the loop $M$ times and plot the position of the center of mass in the $xy$ -plane as a function of time.
What happens if you change the number of particles from 30 to 3 or to 300?

Exercise 11 (Deterministic nature of physics 🌶)

Write down whether your deterministic view on nature (everything in nature can be described by physics) has changed given the three-body problem.

Answers¶

Solution to Exercise 6

The position of the center of mass is

\vec{R} \equiv \frac{\sum_i m_i \vec{x}_i}{\sum_i m_i} = \frac{(m_1 x_1) \hat{x} + (m_2 x_2) \hat{x} + (m_3 y_3) \hat{y}}{m_1 + m_2 + m_3} = -\frac{2}{9}[\mathrm{m}] \hat{x} + \frac{1}{9}[\mathrm{m}]\hat{y}

(19)

where $[\mathrm{m}]$ indicates that the unit is meters.

Note: $\hat{x}$ and $\hat{y}$ do not carry units!

Solution to Exercise 7

Velocity of the center of mass:
$\vec{V} = \frac{\sum_i m_i \vec{v}_i}{\sum_i m_i}$
(20)
Since the velocities are all parallel to the $x$ -axis, we can drop the vector notation. Substituting the data for mass and velocity, gives:

V_x = \frac{4mv + 6mv + 6mv + 4mv}{4m + 3m + 2m + m} = 2v

(21)

Position of the center of mass:
$\vec{V} = \frac{d\vec{R}}{dt} \rightarrow \vec{R}(t) = 2vt \hat{x} +\vec{c}$
(22)
At $t=0$ all particles at location $(0,y_0)$ . Thus, we find

\vec{R}(t) = 2vt \hat{x} + y_0\hat{y}

(23)

Total angular momentum:

\begin{split} \vec{L_{tot}} &= \sum_i \vec{l}_i \\ &= y_0 \cdot 4mv \hat{z} +y_0 \cdot 3m\cdot 2v \hat{z} + y_0 \cdot 2m\cdot 3v \hat{z} + y_0 \cdot m \cdot 4v \hat{z} \\ & = 20mvy_0 \hat{z} \end{split}

(24)

Angular momentum associated with the center of mass:

\vec{L} = \vec{R} \times M\vec{V} = y_0 10m \cdot 2v \hat{z} = 20mvy_0 \hat{z}

(25)

which is indeed the same as the total angular momentum. This is in this case to be expected as the angular momentum seen from the CM frame is $\vec{L}' = 0$ as in the CM frame the position vector and momentum vector are parallel for all four particles.

Solution to Exercise 8

We split the kinetic energy in the kinetic energy associated with the center of mass and the kinetic energy as seen from the CM frame:

E_{kin} = \frac{1}{2}MV^2 + E'_{kin}

(26)

Due to symmetry, the center of mass velocity is $V$ .

In the CM frame, all particles rotate with $\omega$ and thus have a velocity of magnitude $v' = \omega R$ . As all particles have the same mass, we have $M = 8m$ . The kinetic energy is:

E_{kin} = \frac{1}{2}8V^2 + 8 \cdot \frac{1}{2}m \omega^2 R^2 = 4mV^2 +4mR^2 \omega^2

(27)

Solution to Exercise 9

All nitrogen molecules feel gravity and have interaction with each other and with the wall of the container. If we write down the equation of motion for all molecules (labelled $i$ ) and the container we get:

\begin{split} M_c \ddot{\vec{x}_c} &= M_c \vec{g} + \sum_i \vec{F}_{molecule \; i \; on \; vessel \; wall} \\ m_i \ddot{\vec{x}_i} & = -m_i \vec{g} + \vec{F}_{ vessel \; wall \;on \; molecule \; i} +\sum_{j\neq i} \vec{F}_{ji} \end{split}

(28)

with $\vec{F}_{molecule \; i \; on \; vessel \; wall}$ the force of molecule $i$ on the vessel wall and $\vec{F}_{ji}$ the force from molecule $j$ on molecule $i$ . All these forces are internal forces and when summing over all particles (including the vessel) will cancel each other as they all obey N3.

Thus is we add the equations, we find:

\frac{d}{dt} \left ( M_c \dot{\vec{x}_c} + \sum_i m_i \dot{\vec{x}_i} \right ) = \left ( M_c + \sum m_i\right ) \vec{g}

(29)

On the left side, we recognize the total momentum which we can write in terms of the center of mass: $M_c \dot{\vec{x}_c} + \sum_i m_i \dot{\vec{x}_i} = M\vec{V}$ .

And on the right hand side we see the total mass $M = M_c + \sum m_i$ .

Thus, we conclude:

M\dot{\vec{V}} = M\vec{g} \rightarrow \dot{V} = -g

(30)

The entire container drops with acceleration $-g$ .

Solution to Exercise 10

Figure 4:30 particles: left motion of the center of mass, right motion of all particles.

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