<html><head><meta name="color-scheme" content="light dark"></head><body><pre style="word-wrap: break-word; white-space: pre-wrap;">% use [nosolutions] flag to hide solutions.
% use [solutions] flag to show solutions.
\documentclass[
	solutions
]{../../resources/ormc_handout}


\begin{document}

	\maketitle
		&lt;Intermediate 2&gt;
		&lt;ORMC Summer Sessions&gt;
		{Newton's Laws of Motion}
		{
			Prepared by Mark on \today \\
			Based on a handout by Oleg Gleizer
		}

	\section{Newton's First Law}
	If the net force acting on an object is zero, the velocity of that object does not change. \\
	Conversely, if the velocity of an object doesn't change, the net force acting on it is zero.

	\medskip

	In the context of vectors, the ``net force'' is the sum of all the force vectors acting on the object. ``Speed'' is the length (or \textit{magnitude}) of the velocity vector.


	\problem{}
	There are no forces acting on the object
	$A$ below. The current velocity of the object,
	in meters per second, is represented by the vector $\overrightarrow{v}$.
	Draw the position of the object two seconds later.

	\begin{center}
		\begin{normalsize}
		\begin{tikzpicture} [scale=.7]
		\draw[step=1cm, gray, very thin] (-11,-7) grid (1,1);
		\draw [line width = 1.5pt, -&gt;] (-9,0) -- (-5,-3);
		\coordinate [label = above:{$\overrightarrow{v}$}] (v) at (-6.9,-1.5);
		\filldraw (-9,0) circle (3pt);
		\coordinate [label = left:{$A$}] (a) at (-9,0);
		\end{tikzpicture}
		\end{normalsize}
	\end{center}

	The sides of the grid squares
	on the picture above are one metre long.
	What is the speed of the object?
	\vfill

	What distance would the object cover
	in two seconds? \\
	\vfill

	\pagebreak

	\problem{}
	There are no forces acting on the object
	$A$ below.
	The current velocity of this object,
	in miles per hour,
	is represented by the vector $\overrightarrow{v}$.
	Draw the position of the object half an hour later.
	\vspace{20pt}

	\begin{center}
		\begin{normalsize}
		\begin{tikzpicture} [scale=.7]
		\draw[step=1cm, gray, very thin] (-11,-7) grid (1,1);
		\draw [line width = 1.5pt, -&gt;] (-9,0) -- (-1,-6);
		\coordinate [label = above:{$\overrightarrow{v}$}] (v) at (-4.8,-3);
		\filldraw (-9,0) circle (3pt);
		\coordinate [label = left:{$A$}] (a) at (-9,0);
		\end{tikzpicture}
		\end{normalsize}
	\end{center}

	The sides of the grid squares
	on the picture above are ten miles long.
	What is the speed of this object?

	\vfill

	What distance would the object cover
	in half an hour? \\

	\vfill

	What distance would the object cover
	in three hours? \\

	\vfill

	\pagebreak

	\definition{Acceleration}
	\textit{Acceleration} is the rate at which velocity changes.
	Let's represent acceleration by the vector $\overrightarrow{a}$.

	If $\overrightarrow{a}$ does not change over time, then the speed
	of an object at time $t$ is given by the following equation:

	\begin{equation} \label{eq:ac}
		\overrightarrow{v_t} = \overrightarrow{v_0} +
		t \overrightarrow{a}.
	\end{equation}

	\problem{}
	It takes a minivan seven seconds to accelerate from 0 to 60 miles per hour. Find its acceleration in meters per second squared. \\
	\hint{1 mile $\approx$ 1600 meters}

	\vfill

	Note that in the previous problem, motion is one-dimensional (it happens on a straight line). In this case, both velocity and acceleration are one-dimensional vectors---in other words, (real) numbers! \\

	In general, velocity and acceleration are \textit{not} numbers, but vectors. You'll see this in the next few problems.

	\pagebreak


	Now, let's try a few examples with vectors: \\
	\medskip
	Consider an object, currently at $P_0$, moving along the vector $\overrightarrow{v_0}$.

	As before, let $\overrightarrow{a}$ represent the acceleration of the object. This could be caused by gravity, current, or any other constant force.

	\medskip

	One second later, the object will be at $P_1$, and has the velocity vector $\overrightarrow{v_1} = \overrightarrow{v_0} + \overrightarrow{a}$.

	\medskip

	Two seconds later, the object will be positioned at $P_2$ and will have the velocity vector $\overrightarrow{v_2} = \overrightarrow{v_0} + 2\overrightarrow{a}$.

	\begin{center}
		\begin{normalsize}
		\begin{tikzpicture} [scale=.7]
		  \draw[step=1cm, gray, very thin] (-19,-15) grid (1,1);

		  \draw [line width = 1.5pt, -&gt;] (-18,-1) -- (-16,-4);
		  \coordinate [label = left:{$\overrightarrow{v_0}$}]
			(v0) at (-17.1,-2.6);
		  \draw [line width = 1.5pt, -&gt;] (-18,-1) -- (-15,-2);
		  \coordinate [label = above:{$\overrightarrow{a}$}]
			(a) at (-16.4,-1.5);
		  \filldraw (-18,-1) circle (3pt);
		  \coordinate [label = above:{$P_0$}] (p0) at (-18,-1);

		  \filldraw (-14.5,-4.5) circle (3pt);
		  \coordinate [label = right:{$P_1$}] (p1) at (-14.45,-4.4);
		  \draw [line width = 1.5pt, -&gt;] (-14.5,-4.5) -- (-9.5,-8.5);
		  \coordinate [label = right:{$\overrightarrow{v_1}$}]
			(v1) at (-11.75,-6.5);
		  \draw [-&gt;] (-14.5,-4.5) -- (-12.5,-7.5);
		  \coordinate [label = left:{$\overrightarrow{v_0}$}]
			(v01) at (-13.6,-6);
		  \draw (-12.5,-7.5) -- (-9.5,-8.5);
		  \coordinate [label = left:{$\overrightarrow{a}$}]
			(a1) at (-11,-8.3);

		  \filldraw (-8,-9) circle (3pt);
		  \coordinate [label = above:{$P_2$}] (b) at (-8,-9);
		  \draw [-&gt;] (-8,-9) -- (-6,-12);
		  \coordinate [label = left:{$\overrightarrow{v_0}$}]
			(v02) at (-7,-10.5);
		  \draw (-6,-12) -- (0,-14);
		  \coordinate [label = left:{$2\overrightarrow{a}$}]
			(a1) at (-3.5,-13);
		  \draw [line width = 1.5pt, -&gt;] (-8,-9) -- (0,-14);
		\end{tikzpicture}
		\end{normalsize}
	\end{center}

	$t$ seconds later, the object
	will have the velocity
	\[
		\overrightarrow{v_t} = \overrightarrow{v_0} + t \overrightarrow{a}
	\]

	Note that the word {\it acceleration} has two different meanings.
	One is a vector, as in the example above.
	The other meaning is the length
	of the vector that represents the magnitude
	of the velocity change. In this case,
	we do not put an arrow above the letter
	$a$ representing acceleration.
	In other words, $a = |\overrightarrow{a}|$.
	Similarly, speed is the length
	of the velocity vector,
	$v = |\overrightarrow{v}|$.

	\vfill
	\pagebreak


	\example{}&lt;accel_example&gt;
	Moving with a constant acceleration,
	an object moves from point $A$ to point $B$
	in one second. The velocities of the motion,
	in metres per second, are represented
	by the vectors $\overrightarrow{v_0}$ and
	$\overrightarrow{v_1}$.
	Find the acceleration vector. \\

	\begin{center}
	\begin{normalsize}
	\begin{tikzpicture} [scale=.7]
		\draw[step=1cm, gray, very thin] (-18,-8) grid (1,1);
		\draw [line width = 1.5pt, -&gt;] (-17,-1) -- (-13,-4);
		\coordinate [label = above:{$\overrightarrow{v_0}$}]
		(v0) at (-14.9,-2.5);
		\filldraw (-17,-1) circle (3pt);
		\coordinate [label = left:{$A$}] (a) at (-17,-1);
		\filldraw (-10,-3) circle (3pt);
		\coordinate [label = above right:{$B$}] (b) at (-10,-3.1);
		\draw [line width = 1.5pt, -&gt;] (-10,-3) -- (0,-4);
		\coordinate [label = above:{$\overrightarrow{v_1}$}]
		(v1) at (-5,-3.5);
	\end{tikzpicture}
	\end{normalsize}
	\end{center}

	According to formula on the previous page,
	$\overrightarrow{v_1} = \overrightarrow{v_0} + (1\text{ sec}) \times \overrightarrow{a}$. \\

	\begin{center}
	\begin{normalsize}
	\begin{tikzpicture} [scale=.7]
		\draw[step=1cm, gray, very thin] (-18,-8) grid (1,1);
		\draw [line width = 1.5pt, -&gt;] (-17,-1) -- (-13,-4);
		\coordinate [label = above:{$\overrightarrow{v_0}$}]
		(v0) at (-14.9,-2.5);
		\filldraw (-17,-1) circle (3pt);
		\coordinate [label = left:{$A$}] (a) at (-17,-1);
		\filldraw (-10,-3) circle (3pt);
		\coordinate [label = above right:{$B$}] (b) at (-10,-3.1);
		\draw [line width = 1.5pt, -&gt;] (-10,-3) -- (0,-4);
		\coordinate [label = above:{$\overrightarrow{v_1}$}]
		(v1) at (-5,-3.5);
		\draw [line width = 1.5pt, -&gt;] (-10,-3) -- (-6,-6);
		\coordinate [label = below:{$\overrightarrow{v_0}$}]
		(v01) at (-8.2,-4.5);
	\end{tikzpicture}
	\end{normalsize}
	\end{center}

	\problem{}
	Complete \ref{accel_example}: \\
	Draw the acceleration vector on the above diagram.


	\vfill
	\pagebreak

	\problem{}
	At an initial moment in time,
	the object at point $A$
	is moving with the velocity vector
	$\overrightarrow{v_0}$,
	measured in metres per second.
	The constant acceleration
	acting on the object
	is represented by the vector
	$\overrightarrow{a}$,
	measured in meters per second squared.
	Two seconds later, the object is located
	at point $B$. Draw its velocity vector
	at that moment.

	\begin{center}
	\begin{normalsize}
	\begin{tikzpicture} [scale=.7]
		\draw[step=1cm, gray, very thin] (-19,-15) grid (1,1);
		\draw [line width = 1.5pt, -&gt;] (-18,-1) -- (-16,-4);
		\coordinate [label = left:{$\overrightarrow{v_0}$}]
		(v0) at (-17.1,-2.6);
		\draw [line width = 1.5pt, -&gt;] (-18,-1) -- (-15,-2);
		\coordinate [label = above:{$\overrightarrow{a}$}]
		(a) at (-16.4,-1.5);
		\filldraw (-18,-1) circle (3pt);
		\coordinate [label = left:{$A$}] (a) at (-18,-1);
		\filldraw (-8,-9) circle (3pt);
		\coordinate [label = left:{$B$}] (b) at (-8,-9);
		%\draw [line width = 1.5pt, -&gt;] (-8,-9) -- (0,-14);
	\end{tikzpicture}
	\end{normalsize}
	\end{center}
	\vspace{12pt}


	The side length of the grid squares
	on the picture above is one metre.
	Find the speed of the object when it is
	at point $B$.

	\vfill
	\pagebreak

	\section{Newton's Second Law}

	The second law is simple:
	\[
		\overrightarrow{F} = m \times \overrightarrow{a}
	\]

	Here, $\overrightarrow{F}$ is the net force acting on an object with mass $m$, and $\overrightarrow{a}$ is the acceleration the object experiences as a result of this action. Mass is a measure of an object's \textit{inertia}: the heavier an object is, the more effort it takes to change its velocity. \\

	In civilized countries, mass is measured in grams and force is measured in \textit{newtons}. One newton is the force it takes to accelerate 1 kg of mass to 1 meter per second. In other words,

	\[
		1\ N = (1\ kg) (1\ \frac{m}{s})
	\]

	\problem{}
	The {\it Millennium Falcon}, at point $A$ at the moment,
	is trying to escape from the Death Star, which is trying
	to arrest the ship using its attracting beam.
	The thrust of the Falcon's engines,
	200,000 kN in total, is represented
	by the vector $\overrightarrow{T}$.
	The force of the beam is represented
	by  the vector $\overrightarrow{B}$.
	In addition, a nearby star
	exerts a gravitational force of
	$\overrightarrow{G}$ on the ship.
	Draw the vector of the net force
	acting on the vessel. \\

	\begin{center}
	\begin{normalsize}
	\begin{tikzpicture} [scale=.7]
		\draw[step=1cm, gray, very thin] (-16,-9) grid (1,1);
		\draw [line width = 1.5pt, -&gt;] (-8,-5) -- (-12,-5);
		\coordinate [label = below:{$\overrightarrow{T}$}] (v) at (-10,-5.1);
		\filldraw (-8,-5) circle (3pt);
		\coordinate [label = below:{$A$}] (a) at (-8,-5.1);
		\draw [line width = 1.5pt, -&gt;] (-8,-5) -- (-3,-6);
		\coordinate [label = above:{$\overrightarrow{B}$}] (b) at (-5.4,-5.5);
		\draw [line width = 1.5pt, -&gt;] (-8,-5) -- (-9,-2);
		\coordinate [label = left:{$\overrightarrow{G}$}] (g) at (-8.6,-3.5);
	\end{tikzpicture}
	\end{normalsize}
	\end{center}

	The mass of the ship
	is 400,000 kg. What acceleration,
	in metres per second squared,
	does the ship experience?
	\vfill



	The purpose of this daring mission
	was to find out the force
	that the the attracting beam exerts.
	However, Han Solo is not particularly
	good with vectors. Please help him complete the mission.
	\vfill
	\pagebreak


	\section{Newton's Third Law}

	Newton's third law also concerns forces. It states that \textit{every action has an equal and opposite reaction}.

	In other words, this means that when one object exerts force on another, the second simultaneously exerts a force equal in magnitude and opposite in direction to the force exerted on it by the first.

	\medskip

	In fact, we saw this in our first handout!

	\begin{tcolorbox}[
		colback=white,
		colframe=gray!75!black,
		title={ Handout 1, Page 7 }
	]

	Here is an important example of an inverse vector. When you stand still, the floor pushes you up with the force opposite to the force of the gravitational pull, a.k.a. \textit{weight}.

	\begin{center}
		\begin{tikzpicture}
			\begin{normalsize}
				\draw (0,0) -- (6,0);
				\draw (2.5,0) -- (3,1);
				\draw (3.5,0) -- (3,1);
				\draw (3,1) -- (3,2);
				\draw (3,2.3) circle (.3);
				\draw (2.6,1.7) -- (3.4,1.7);
				\draw [-&gt;, line width = 2pt] (2.5,0) -- (2.5,-1);
				\coordinate [label = below: {weight}]
					(g) at (2.5,-1.1);
				\draw [-&gt;, line width = 2pt] (3.5,0) -- (3.5,1);
				\coordinate [label = right: {floor reaction}]
					(w) at (3.7,1);
			\end{normalsize}
		\end{tikzpicture}
	\end{center}
	The two opposing vectors add up to the zero vector, and therefore you don't move.

	\end{tcolorbox}

	\problem{}
	In a second, the truck and car on the picture below will collide in a crash test.
	The weight of the truck is 20 metric tons. The weight of the car is 2 metric tons.
	Find the ratio of the accelerations,
	$a_t$ (acceleration of the truck) and
	$a_c$ (that of the car),
	the vehicles will undergo. \\

	\begin{center}
	\begin{tikzpicture}
		\draw [gray!50!black] (-7,0) -- (7,0);

		\filldraw [gray!80!blue] (-3.6,1.5) -- (-4.4,1.5) --
		(-4.4,2) -- (-6.8,2) -- (-6.8,.3) -- (-3.6,.3) --
		(-3.6,1.5);
		\filldraw (-6,.3) circle (.3);
		\filldraw [gray] (-6,.3) circle (.15);
		\filldraw (-4,.3) circle (.3);
		\filldraw [gray] (-4,.3) circle (.15);

		\filldraw [red!80!white] (4,.2) -- (6.6,.2) -- (6.55,.6) --
		(6.2,.6) -- (6.1,.8) -- (4.9,.8) -- (4.7,.6)--
		(4.2,.6) -- (4,.45) -- (4,.2);
		\filldraw (6,.2) circle (.2);
		\filldraw [gray] (6,.2) circle (.1);
		\filldraw (4.6,.2) circle (.2);
		\filldraw [gray] (4.6,.2) circle (.1);
	\end{tikzpicture}
	\end{center}

	\medskip

	\begin{huge}
	$a_t \div a_c =$
	\end{huge}

	\vfill
	\newpage

	\section{Bonus Problems}

	\problem{}
	Use vector algebra to prove
	that diagonals of a parallelogram
	in the Euclidean plane
	split each other in halves. \\

	\begin{center}
	\begin{tikzpicture}
	\begin{normalsize}
		\draw (0,0) -- (4,-2);
		\coordinate [label = left: {$A$}] (a) at (0,0);
		\coordinate [label = right: {$B$}] (b) at (4,-2);
		\coordinate [label = above: {$\overrightarrow{v}$}] (v1) at (2,-.9);
		\draw (-1,-3) -- (3,-5);
		\coordinate [label = left: {$C$}] (c) at (-1,-3);
		\coordinate [label = right: {$D$}] (d) at (3,-5);
		\coordinate [label = below: {$\overrightarrow{v}$}] (v2) at (1,-4.1);
		\draw (-1,-3) -- (0,0);
		\draw (3,-5) -- (4,-2);
		\coordinate [label = left: {$\overrightarrow{w}$}] (w1) at (-.55,-1.5);
		\coordinate [label = right: {$\overrightarrow{w}$}] (w2) at (3.55,-3.5);
		\draw (0,0) -- (3,-5);
		\draw (-1,-3) -- (4,-2);
		\coordinate [label = above: {$M$}] (m) at (1.7,-2.5);
	\end{normalsize}
	\end{tikzpicture}
	\end{center}

	\vfill
	\newpage

	\problem{}
	Use vector algebra to prove that
	all the three medians of a triangle
	in the Euclidean plane intersect
	at one point that splits each of the medians
	in the ratio 2:1, counting from the vertices. \\

	\begin{center}
	\begin{normalsize}
	\begin{tikzpicture}
		\draw (0,0) -- (6,-1);
		\draw (0,0) -- (3,-5);
		\draw (3,-5) -- (6,-1);
		\draw [color = white] (3,-5) -- (9,-6);
		\draw [color = white] (6,-1) -- (9,-6);
		\coordinate [label = left: {$A$}] (a) at (0,0);
		\coordinate [label = right: {$B$}] (b) at (6,-1);
		\coordinate [label = below: {$C$}] (c) at (3,-5);
		\coordinate [label = above: {$\overrightarrow{v}$}]
		(v) at (3,-.45);
		\coordinate [label = left: {$\overrightarrow{w}$}]
		(w) at (1.45,-2.5);
		\filldraw (4.5,-3) circle (1.5pt);
		\coordinate [label = right: {$M_A$}] (ma) at (4.65,-2.85);
		\draw (0,0) -- (4.5,-3);
	\end{tikzpicture}
	\end{normalsize}
	\end{center}

	\vfill
	\newpage

	\problem{}
	Find the area of an equilateral triangle
	with the side length $a$.
	\vfill

	\problem{}
	Does there exist an equilateral triangle
	in the Euclidean plane
	such that all of its vertices
	have integral coordinates?
	Why or why not? \\

	\begin{center}
	\begin{normalsize}
	\begin{tikzpicture} [scale=.7]
		\draw[step=1cm, gray, very thin] (-15,-9) grid (1,1);
		\draw [line width = 1.5pt, -&gt;] (-7,-9) -- (-7,1);
		\coordinate [label = above:{$y$}] (y) at (-7,1);
		\draw [line width = 1.5pt, -&gt;] (-15,-4) -- (1,-4);
		\coordinate [label = right:{$x$}] (x) at (1,-4);
		\coordinate [label = below:{$1$}] (ox) at (-6,-4);
		\coordinate [label = left:{$1$}] (oy) at (-7,-3);
	\end{tikzpicture}
	\end{normalsize}
	\end{center}
	\vfill

\end{document}</pre></body></html>