Examples, exercises and experiments - Classical Mechanics & Special Relativity for Starters

Updated: 02 feb 2026

Here are some examples and exercises that deal with forces. Make sure you practice IDEA.

Worked Examples¶

Slowing a mass down¶

In this example we will consider a 1-dimensional case: a mass, $m$ has initial velocity $v_0$ . From $t=0$ onwards a force $F$ is acting on $m$ , slowing it down. Eventually, $m$ will come to a stand still.

The final question to be answered is: what is the distance $m$ has traveled from $t=0$ until $m$ has stopped moving?

We will inspect the simplest case: the force is constant and has a magnitude $\mu mg$ (with $\mu$ a positive constant). This is one of the simplest frictional forces. It is proportional to the weight of $m$ and is a first order approximation for a mass sliding over a horizontal plane.

Interpret the problem

Develop the solution

Evaluate the solution

Assess the solution

First we make a sketch and draw what is relevant for this problem.

a point mass moving to the right with velocity v, while a force F is acting to the left. The position where the mass stops is indicated as xmax.

The problem is 1-dimensional, so we only need one coordinate. Hence we have drawn the $x$ -axis. The mass is somewhere on the axis and has at that position velocity $v$ . We also draw that, as the velocity will change. Moreover, velocity is related to momentum ( $p=mv$ ) and -as a force is acting on $m$ - we expect that we will use N2.

We also draw the force. As the force is slowing down the mass, it will have to act in the direction opposite to the velocity. That means in our case: $F$ points in the negative $x$ -direction. Finally, we indicate the position where the mass will stop moving: $x_{max}$ .

We conclude our interpret-phase with our idea on how to approach this problem:

a force is slowing down $m$ : that calls for setting up N2
at some point in time $m$ has zero velocity: we need to find that time (let’s call it $t_f$ ). We can do that via N2.
we need to find the trajectory of $m$ , i.e. $x(t)$ and than substitute $t_f$ to find $x_{max} = x(t_f)$ .

We start with solving N2:

m \frac{dv}{dt} = F \rightarrow \frac{dv}{dt} = - \frac{\mu mg}{m} = -\mu g

(5)

Note: we have now explicitly given the minus-sign as the force is acting in the negative $x$ -direction. Moreover, the friction constant $\mu >0$ . This differential equation is easy to solve, as the right hand side is a constant:

v(t) = - \mu g t + C_1

(6)

with $C_1$ an integration constant. We can find this constant by looking at the initial condition:

t=0 \rightarrow v=v_0 \Rightarrow v_0 = -\mu g 0 + C_1 \Rightarrow C_1 = v_0

(7)

Thus, the solution for $v(t)$ is:

v(t) = v_0 - \mu g t

(8)

From this equation, we can find when the mass stops moving:

v(t_f) = 0 \Rightarrow 0 = v_0 - \mu g t_f \Rightarrow t_f = \frac{v_0}{\mu g}

(9)

Now we are ready to find the trajectory $x(t)$ :

\frac{dx}{dt} = v(t) = v_0 - \mu g t \Rightarrow x(t) = v_0 t -\frac{1}{2} \mu g t^2 + C_2

(10)

and we use the initial condition to find $C_2$ :

t=0 \rightarrow x=0 \Rightarrow 0 = v_0 0 -\frac{1}{2} \mu g 0^2 + C_2 \Rightarrow C_2 = 0

(11)

Thus, the solution for the trajectory is:

x(t) = v_0 t -\frac{1}{2} \mu g t^2

(12)

Finally, we can find $x_{max}$ by substituting $t_f$ into $x(t)$ :

\begin{split} x_{max} &= x(t_f) = v_0 \frac{v_0}{\mu g} - \frac{1}{2} \mu g \left ( \frac{v_0}{\mu g} \right )^2 \\ &= \frac{v_0^2}{\mu g} - \frac{1}{2} \frac{v_0^2}{\mu g} = \frac{1}{2} \frac{v_0^2}{\mu g} \end{split}

(13)

Exercises set 1¶

Exercise 6 (Different coordinate systems 🌶 🌶)

In Classical Mechanics we often use a coordinate system to describe motion of object. In this exercise, you will look at two Cartesian coordinate systems. System S has coordinates $(x, y)$ and corresponding unit vectors $\hat{x}$ and $\hat{y}$ .
The second system, S’, uses $(x', y')$ and corresponding unit vectors. The $x'$ -axis makes an angle of $30^\circ$ with the $x$ -axis (measured counter-clockwise).

Make a sketch.
Determine the relations between $\hat{x}'$ and $\hat{x}, \hat{y}$ as well as between $\hat{y}'$ and $\hat{x}, \hat{y}$
An object has, according to S, a velocity of $\vec{v} = 3 \hat{x} + 5 \hat{y}$ .
Determine the velocity according to S’.

Exercise 7 (Rotating unit vectors 🌶)

According to your observations, a particle is located at position (1,0) (you use a Cartesian coordinate system). The particle has no velocity and no forces are acting on it.
Another observer, S’, uses a Cartesian coordinate system described by $(x', y')$ . You notice that her unit vectors rotate at a constant speed compared to your unit vectors:

\hat{x}' = \cos(2\pi f t ) \hat{x} + \sin(2\pi f t ) \hat{y}

(15)

\hat{y}' = -\sin(2\pi f t ) \hat{x} + \cos(2\pi f t ) \hat{y}

(16)

Find the position of the particle according to the other observer, S’.
Calculate the velocity of the particle according to S’.

Exercise 9 (Parabolic trajectory with maximum area ^[1] 🌶 🌶)

A ball is thrown at speed $v$ from zero height on level ground. We want to find the angle $\theta$ at which it should be thrown so that the area under the trajectory is maximized.

Sketch of the trajectory of the ball.
Use dimensional analysis to relate the area to the initial speed $v$ and the gravitational acceleration $g$ .
Write down the $x$ and $y$ coordinates of the ball as a function of time.
Find the total time the ball is in the air.
The area under the trajectory is given by $A = \int y \mathrm{d}x$ . Make a variable transformation to express this integral as an integration over time.
Evaluate the integral. Your answer should be a function of the initial speed $v$ and angle $\theta$ .
From your answer at (6), find the angle that maximizes the area, and the value of that maximum area.

Answers set 1¶

Solution to Exercise 1

$\vec{F} = -\frac{m v_0}{\Delta t} \hat{x}$

Solution to Exercise 2

Interpret

Develop

Evaluate

Assess

Solution to Exercise 3

$a = \frac{F}{m}$ is constant
$v(t) = v_0 + at$
$x(t) = v_0 t + \frac{1}{2}at^2$

Solution to Exercise 4

$a = \frac{F}{m} = \frac{F_0}{m} \sin (2\pi f_0 t)$ is not constant
$v(t) = v_0 + \frac{F_0}{2 \pi f_0 m} \left ( 1 - \cos (2\pi f_0 t ) \right )$
$x(t) = v_0 t + \frac{F_0}{2\pi f_0 m} t - \frac{F_0}{4 \pi^2 f_0^2 m} \sin 2\pi f_0 t$

Solution to Exercise 5

Particle moves at constant velocity, thus path is a straight line:

\vec{r}(t) = \vec{r}_0 + \vec{v}_0 t = x_0 \hat{i} + y_0 \hat{j} + v_{0x} t \hat{i} + v_{0y} t \hat{j}

(20)

At $t=0: \vec{r}(0) = 0\hat{i} + y_A \hat{j} \rightarrow \vec{r}(0) = \vec{r}_0 = 0\hat{i} + y_A\hat{j} \rightarrow x_0 = 0 \text{ and } y_0 = y_A$

At $t=t_1$ :

\begin{split} \vec{r}(t_1) &= x_B\hat{i} + 0\hat(j) \rightarrow \\ \vec{r}(t_1) &= \vec{r}_0 +\vec{v}_0 t_1 \\ \; &= (0 + v_{0x}t_1 )\hat{i} + (y_A + v_{0y}t_1 ) \hat{j} \rightarrow \\ v_{0x} &= \frac{x_B}{t_1} \text{ and } v_{0y} = -\frac{y_A}{t_1} \end{split}

(21)

Thus, we find $\vec{v} = \frac{x_B}{t_1} \hat{i} -\frac{y_A}{t_1}\hat{j}$

Solution to Exercise 6

A set of x- and y-axis is drawn. In a lighter shade, another set of x- and y-axis is drawn, which is rotated theta degrees counterclockwise.

$\begin{split} \hat{x}' = \cos \theta \hat{x} + \sin \theta \hat{y} = \frac{1}{2}\sqrt{3}\hat{x} + \frac{1}{2} \hat{y}\\ \hat{y}' = -\sin \theta \hat{x} + \cos \theta \hat{y} = -\frac{1}{2} \hat{x} + \frac{1}{2}\sqrt{3} \hat{y}\end{split}$
(22)
Invert:
$\begin{split} \hat{x} = \cos \theta \hat{x}' - \sin \theta \hat{y}' = \frac{1}{2}\sqrt{3}\hat{x}' - \frac{1}{2} \hat{y}'\\ \hat{y} = \sin \theta \hat{x}' + \cos \theta \hat{y}' = \frac{1}{2} \hat{x}' + \frac{1}{2}\sqrt{3} \hat{y}' \end{split}$
(23)

velocity:

\begin{split} \vec{v} &= v_x \hat{x} + v_y \hat{y} \\ &= v_x \left ( \cos \theta \hat{x}' - \sin \theta \hat{y}' \right ) +v_y \left ( \sin \theta \hat{x}' + \cos \theta \hat{y}' \right ) \\ &= \left ( v_x \cos \theta + v_y \sin \theta \right ) \hat{x}' + \left ( -v_x \sin \theta + v_y \cos \theta \right ) \hat{y}' \end{split}

(24)

from which we find

\vec{v} = \left ( \frac{3}{2}\sqrt{3} + \frac{5}{2} \right ) \hat{x}' + \left ( -\frac{3}{2} + \frac{5}{2}\sqrt{3} \right ) \hat{y}'

(25)

Solution to Exercise 7

\begin{split} \hat{x}' = \cos (2\pi ft) \hat{x} + \sin (2\pi ft) \hat{y}\\ \hat{y}' = -\sin (2\pi ft) \hat{x} + \cos (2\pi ft) \hat{y} \end{split}

(26)

The unit vectors of S’ rotate with a frequency $f$ with respect to the unit vectors of S. This means, that the coordinate system of S’ rotates: the rotation angle is a function of time, i.e. $\theta (t) = 2 \pi f t$

From the figure we see, that the coordinates of a point P, $(x_p, y_p )$ according to S, are related to those used by S’, $(x'_p, y'_p )$ via:

\begin{split} x_p = OP \cos ( \alpha + \theta ) = OP \left ( \cos \alpha \cos \theta - \sin \alpha \sin \theta \right) = x'_p \cos \theta - y'_p \sin \theta \\ y_p = OP \sin ( \alpha + \theta ) = OP \left ( \cos \alpha \sin \theta + \sin \alpha \cos \theta \right) = x'_p \sin \theta + y'_p \cos \theta \end{split}

(27)

or written as the coordinate transformation:

\begin{split} x = x' \cos \theta - y' \sin \theta \\ y = x' \sin \theta + y' \cos \theta \end{split}

(28)

with its inverse

\begin{split} x' = x \cos \theta + y \sin \theta \\ y' = -x \sin \theta + y \cos \theta \end{split}

(29)

Note that in this case $\theta = 2 \pi ft$ , that is: it is a function of $t$ .

a) From the above relation we find that the point (1,0) in S will be denoted by S’ as $(x'(t), y'(t)) = (\cos (2 \pi ft), -\sin ( 2\pi ft))$

b) the velocity of the point (1,0) in S is according to S of course zero: $\frac{dx}{dt} = 0, \frac{dy}{dt} = 0$ S’ will say:

\begin{split} x'(t) = \cos ( 2\pi ft) \rightarrow \frac{dx'}{dt} = -2 \pi f \sin (2\pi ft) \\ y'(t) = -\sin ( 2\pi ft) \rightarrow \frac{dy'}{dt} = 2 \pi f \cos (2\pi ft) \end{split}

(30)

Solution to Exercise 8

Since $\vec{v}_0$ and $\vec{F}$ are parallel, the particle will not deviate from the line x=y. Hence, we are dealing with a 1-dimensional problem. The original coordinate system, $(x,y)$ , is not wrong: it is just not handy as it makes the problem look like 2D. Thus, we change our coordinate system, such that the new $x$ -axis coincides with the original x=y line.
N2: $m\frac{dv}{dt} = -b v$ with initial conditions: $t= 0 \rightarrow x=0$ and $t=0 \rightarrow v = v_0$
$\frac{dv}{dt} - \frac{b}{m} v = 0 \rightarrow v = A e^{-\frac{b}{m}t}$ initial condition: $t= 0 \rightarrow v=v_0 \Rightarrow A = v_0$ Thus: $v(t) = v_0 e^{-\frac{b}{m}t}$
for $t \rightarrow \infty: v \rightarrow 0$ . The particle comes to rest and then, obviously, the friction force is zero.

Solution to Exercise 9

We expect that the area, $A$ , under the trajectory of the ball is a function of $v$ , $g$ , and $\theta$ . In a dimensional analysis we write this as ‘product of powers’:

A = v^a \cdot g^b \cdot \theta^c

(31)

and we make this expression dimensional correct. (Note: we don’t mean that the final outcome of a full analysis is a product of powers, it can be any function but the units should be related in the right way and that is what this ‘trick’ with powers ensures.)

The area has units m $^2$ , velocity m/s, g m/s $^2$ and $\theta$ is dimensionless (radians don’t count as a dimension or unit). Thus:

\begin{split} m: 2 &= a + b \\ s: 0 &= -a -2b \end{split}

(32)

This yields: $a=4, b=-2$ . Thus on dimensional grounds we may expect: $A \sim \frac{v^4}{g^2}$ .

In the x-direction: no forces, hence $m\frac{v_x}{dt} = 0 \rightarrow x(t) = v \cos \theta t$

In the y-direction: $m\frac{dv_y}{dt} = -mg \rightarrow y(t) = v \sin \theta t -\frac{1}{2}gt^2$ . Where we have used the initial conditions: $x(o)=0, y(0)=0, v_x(0)=v\cos \theta, v_y(0) = v \sin \theta$

Total time in the air: $v_y(t^*)=0 \rightarrow t^* = \frac{2v}{g} \sin \theta$

5+6. Evaluate the area under the trajectory:

\begin{split} A &= \int_0^{x_{max}} ydx \\ &= \int_0^{t^*} \left ( v \sin \theta t -\frac{1}{2}gt^2 \right ) v \cos \theta dt \\ &= v^2 \sin \theta \cos\theta \frac{1}{2} (t^*)^2 -/frac{1}{6} gv \cos \theta (t^*)^3 \\ &= \frac{2}{3} \frac{v^4}{g^2} \cos \theta \sin^3 \theta \end{split}

(33)

We maximize the function $f(\theta) = \cos \theta \sin^3 \theta$ :

\frac{df}{d\theta} = \sin^2 \theta (-\sin^2 \theta + 3 \cos^2 \theta)

(34)

\frac{df}{d\theta} = 0 \rightarrow \sin \theta = 0 \text{ or } \sin^2 \theta = 3 \cos^2 \theta

(35)

The first solution give a minimum for the area ( $A=0$ ). So we need the second solution:

\frac{\sin^2 \theta}{\cos^2 \theta} = \tan^2 \theta =3 \rightarrow \tan \theta =\sqrt{3} \rightarrow \theta = \frac{\pi}{3}

(36)

Solution to Exercise 10

Interpret

Develop

Evaluate

Assess

We start with a sketch.

This is a 1-dimensional problem. We will use $r$ as the coordinate. Moreover, it is a problem involving two particles, that both can move. This makes it more difficult than 1-dimensional cases with only one particle.

Exercises set 2¶


import numpy as np
import matplotlib.pyplot as plt
from matplotlib.patches import Rectangle
from matplotlib.transforms import Affine2D
from ipywidgets import interact
import ipywidgets as widgets

# Constants
def update(theta,F_girl):
    # Compute arrow end coordinates
    F_boys_y = F_girl / 2
    F_boys_x = F_girl*np.tan(theta)/2
    
    # Clear figure
    plt.clf()
    fig, ax = plt.subplots()
    ax.set_xlim(-20, 20)
    ax.set_ylim(-20, 20)
    ax.set_aspect('equal')
    ax.grid(True)

    # Draw arrow
    
    # rope
    ax.arrow(0, 0, -20*np.sin(theta), 20*np.cos(theta),
             head_width=0, head_length=0, fc='black', ec='black')
    
    ax.arrow(0, 0, 20*np.sin(theta), 20*np.cos(theta),
             head_width=0, head_length=0, fc='black', ec='black')
    
    #Forces
    ax.arrow(0, 0, 0, -F_girl,
             head_width=1, head_length=1, fc='green', ec='green')

    ax.arrow(0, 0, F_boys_x, F_boys_y,
             head_width=1, head_length=1, fc='red', ec='red')
    
    ax.arrow(0, 0, -F_boys_x, F_boys_y,
             head_width=1, head_length=1, fc='red', ec='red')

    
    plt.show()

# Use FloatSlider for smooth interaction
interact(update, theta=widgets.FloatSlider(min=0.1, max=np.pi/2, step=np.pi/16, value=np.pi/4),
         F_girl=widgets.FloatSlider(min=3, max=10, step=1, value=3))

<function __main__.update(theta, F_girl)>

import numpy as np
import matplotlib.pyplot as plt
from matplotlib.patches import Rectangle
from matplotlib.transforms import Affine2D
from ipywidgets import interact
import ipywidgets as widgets

# Constants
start = [0, 0]

L = 2
m = 10 #kg
g = 9.81 #kgm/s^2
t = np.linspace(0,5,100)

def Fw(theta,mu):
    return mu*m*g*np.cos(theta)

def Fzx(theta):
    return m*g*np.sin(theta)

def update(theta,mu):
    # Compute arrow end coordinates
    end = np.array([L * np.cos(theta), L * np.sin(theta)])

    # Clear figure
    plt.clf()
    fig, axs = plt.subplots(1,2, figsize=(10, 5), gridspec_kw={'wspace': 0.4})
    
    # first plot showing angle and box
    ax = axs[0]
    ax.set_xlim(0, 2)
    ax.set_ylim(0, 2)
    ax.set_aspect('equal')
    ax.grid(True)

    if Fw(theta,mu)>Fzx(theta):
        ax.text(1.5, 1.5, 'Stable', fontsize=12)
    else:
        ax.text(1.5, 1.5, 'Sliding', fontsize=12)
    # Draw arrow
    ax.arrow(start[0], start[1], end[0], end[1],
             head_width=0, head_length=0, fc='black', ec='black')

    # Box properties
    box_width = 0.4
    box_height = 0.2

    # Create unrotated rectangle at (0,0)
    rect = Rectangle((-box_width / 2, 0 ),
                     box_width, box_height,
                     linewidth=1, edgecolor='red', facecolor='lightgray')

    # Compute arrow end coordinates
    end = start + np.array([L * np.cos(theta), L * np.sin(theta)])

    # Compute midpoint of arrow
    mid = start + np.array([L/2 * np.cos(theta), L/2 * np.sin(theta)])
    
    # Transformation: rotate around origin, then translate to arrow tip
    trans = (Affine2D()
             .rotate(theta)
             .translate(mid[0], mid[1]) + ax.transData)

    rect.set_transform(trans)
    ax.add_patch(rect)

    # second plot showing motion with and without friction
    ax2 = axs[1]
    
    ax2.set_xlabel('$t$(s)')
    ax2.set_ylabel('$s$(m)')
    
    
    ax2.plot(t,1/2*g*np.sin(theta)*t**2,'k-',label='without friction')
    
    if Fw(theta,mu)>Fzx(theta):
        ax2.plot(t,np.zeros(len(t)),'r-',label='with friction')
    else:
        ax2.plot(t,1/2*g*(np.sin(theta)-mu*np.cos(theta))*t**2,'r-',label='with friction')
        
  
    ax2.set_ylim(0,120)
    ax2.legend()
    
   
    
    plt.show()

# Use FloatSlider for smooth interaction
interact(update, theta=widgets.FloatSlider(min=0, max=np.pi/2, step=np.pi/16, value=np.pi/4),
         mu=widgets.FloatSlider(min=0, max=1, step=0.05, value=1))

<function __main__.update(theta, mu)>

import numpy as np
import matplotlib.pyplot as plt
from ipywidgets import interact
import ipywidgets as widgets

# Parameters
t_stop = 5
dt = 0.01
t = np.arange(0, t_stop + dt, dt)

# Force functions definitions
def force1(t):
    a = 10
    x = 0.5 * a * t**2
    v = a * t
    return x, v

def force2(t):
    x = (1/6) * 10 * t**3
    v = 0.5 * 10 * t**2
    return x, v

def force3(t):
    x = 125 * (1 - np.cos(np.pi * t / 2))
    v = 125 * (np.pi / 2) * np.sin(np.pi * t / 2)
    return x, v

def update(force_num):
    fig, axs = plt.subplots(2, 1, figsize=(8, 6), gridspec_kw={'hspace': 0.4})

    if force_num == 1:
        selected_force = force1
    elif force_num == 2:
        selected_force = force2
    else:
        selected_force = force3

    x_vals, v_vals = selected_force(t)

    # Velocity vs Time
    ax1 = axs[0]
    ax1.plot(t, v_vals, 'b-')
    ax1.set_xlim(0, t_stop)
    ax1.set_ylim(-np.max(v_vals) * 1.2, np.max(v_vals) * 1.2)
    ax1.set_xlabel('Time (s)')
    ax1.set_ylabel('Velocity (cm/s)')
    ax1.set_title('Velocity vs Time')
    ax1.grid(True)

    # Position vs Time
    ax2 = axs[1]
    ax2.plot(t, x_vals, 'r-')
    ax2.set_xlim(0, t_stop)
    ax2.set_ylim(0, np.max(x_vals) * 1.2)
    ax2.set_xlabel('Time (s)')
    ax2.set_ylabel('Position (cm)')
    ax2.set_title('Position vs Time')
    ax2.grid(True)

    plt.show()

interact(update, force_num=widgets.IntSlider(min=1, max=3, step=1, value=1))

<function __main__.update(force_num)>

import numpy as np
import matplotlib.pyplot as plt
from matplotlib.patches import Rectangle
from matplotlib.animation import FuncAnimation
from IPython.display import HTML, display
from ipywidgets import interact, FloatSlider, IntSlider

def run_animation(g=9.81, M=1):
    m1 = 1.0  # red block mass
    acc = M / (m1 + M) * g  # Net acceleration

    t_stop = 1.5
    dt = 0.02
    t_vals = np.arange(0, t_stop, dt)

    scale_factor = 25
    x0 = 40
    y0 = 210

    fig, ax = plt.subplots(figsize=(9, 5))
    ax.set_xlim(0, 900)
    ax.set_ylim(500, 0)
    ax.axis('off')

    # Static scene
    ax.fill_between([20, 320], 180, 190, color='black')  # floor
    pulley = plt.Circle((320, 180), 15, color='grey')
    ax.add_patch(pulley)

    ax.plot([450, 450], [20, 470], color='black')   # y-axis
    ax.plot([450, 800], [420, 420], color='black')  # x-axis
    ax.text(408, 170, "x (m)", fontsize=12)
    ax.text(780, 440, "t (s)", fontsize=12)

    for i, y in enumerate(range(125, 426, 75)):
        ax.text(423, y, f"{(4 - i):.1f}", fontsize=10)
    for i, x in enumerate([540, 640, 740]):
        ax.text(x, 440, f"{0.5 * (i + 1):.1f}", fontsize=10)

    for i in range(24):
        ax.plot([450, 800], [420 - 15 * (i + 1)] * 2, color='grey', linewidth=0.5)
    for i in range(17):
        ax.plot([450 + 20 * (i + 1)] * 2, [60, 420], color='grey', linewidth=0.5)

    # Dynamic elements
    red_block = Rectangle((x0, 150), 30, 30, color='red')
    grey_block = Rectangle((320, y0), 30, 30, color='grey')
    ax.add_patch(red_block)
    ax.add_patch(grey_block)
    cord1, = ax.plot([], [], color='black')
    cord2, = ax.plot([], [], color='black')
    trace, = ax.plot([], [], color='blue')
    x_trace, y_trace = [], []

    def update(frame):
        t = frame * dt
        disp = scale_factor * 0.5 * acc * t**2

        x = min(x0 + disp, 270)
        y = min(y0 + disp, 440)

        red_block.set_xy((x, 150))
        grey_block.set_xy((320, y))
        cord1.set_data([x + 30, 320], [165, 165])
        cord2.set_data([335, 335], [180, y])

        if x < 270:
            x_trace.append(450 + (200 * t))
            y_trace.append(420 - 1.53 * (x - x0))
            trace.set_data(x_trace, y_trace)

        return red_block, grey_block, cord1, cord2, trace

    ani = FuncAnimation(fig, update, frames=len(t_vals), blit=True, interval=dt * 1000)

    plt.close(fig)  # Prevent duplicate static figure
    display(HTML(ani.to_jshtml()))

# Interact widget with sliders
interact(run_animation,
         g=FloatSlider(min=1.5, max=15.0, step=0.1, value=9.81, description='g (m/s²)'),
         M=IntSlider(min=1, max=5, step=1, value=1, description='Mass (kg)'))

<function __main__.run_animation(g=9.81, M=1)>

Exercise 16 (🌶 🌶 🌶)

A point particle (mass $m$ ) is from position $z=0$ shot with a velocity $v_0$ straight upwards into the air. On this particle only gravity acts, i.e. friction with the air can be ignored. The acceleration of gravity, $g$ , may be taken as a constant.

The following questions should be answered.

What is the maximum height that the particle reaches?
How long doe it take to reach that highest point?

Solve this exercise using IDEA.

Sketch the situation and draw the relevant quantities.
Reason that this exercise can be solved using $\vec{F}=m\vec{a}$ (or $d\vec{p}/dt = \vec{F}$ ).
Formulate the equation of motion (N2) for m.
Classify what kind of mathematical equation this is and provide initial or boundary conditions that are needed to solve the equation.
Solve the equation of motion and answer the two questions.
Check your math and the result for dimensional correctness. Inspect the limit: $F_{zw} \rightarrow 0$ .

Solution to Exercise 19

When you push with your foot on the pedal, that force is transferred to the chain of your bike. That chain exerts a force on the gear of your bike’s rear wheel, trying to get it to rotate. Your wheel touches the ground and, because of the force on the gear, the wheel exerts a force in the ground, trying to push the ground backwards. Due to action=-reaction, the ground exerts a forward force on your wheel. So actually, biking means “making the ground push you forward”!

Solution to Exercise 20

When you try to step on the jetty, a force needs to be exerted on you, otherwise you can’t move forward. The way you achieve that: you push with your back foot on the boat. And as a result of Newton 3, the boat will push back, but the force from the boat on you is forward directed. That is exactly what you need!

However, while you push, the boat will move backwards due to the force you exert on it. Consequently, your point of contact with the boat shifts away from the jetty. Either you let the boat go and no force from the boat is acting on you. Now gravity will do its work and if your forward velocity is not sufficient, you will not reach the jetty. Or your foot will try to follow the boat and that requires a force to the wrong direction acting on you.

Pushing harder seems an option: your forward velocity might increase more. However, the boat will also be pushed harder and moves quicker away from you. Consequently, the time interval of contact with the boat decreases. Thus, with Newton 2: dp = Fdt your increase in velocity due to the larger force might be compensated by a smaller duration that the force can do so. And you may still end up in the water.

Experiments¶

Footnotes¶

Exercise from Idema, T. (2023). Introduction to particle and continuum mechanics. Idema (2023)
↩↩

References¶

Idema, T. (2023). Introduction to particle and continuum mechanics. TU Delft OPEN Publishing. 10.59490/tb.81