Trial and Error – Physicspupil

February 10, 2023February 10, 2023

202302101153 Exercise 26.2

If required take $g=9.8\,\mathrm{m\,s^{-2}}$ .

1. Two trucks, masses $30\,\mathrm{kg}$ and $20\,\mathrm{kg}$ , travelling at $6\,\mathrm{m\,s^{-1}}$ and $2\,\mathrm{m\,s^{-1}}$ respectively in the same direction, collide and continue together. Find the loss of KE due to the collision, and the percentage loss of energy.
2. Two masses, of $3\,\mathrm{kg}$ and $2\,\mathrm{kg}$ , move towards each other at speeds $2\,\mathrm{m\,s^{-1}}$ and $1\,\mathrm{m\,s^{-1}}$ respectively. After colliding they move together. Find the percentage loss of energy in the collision.
3. A force of $2\,\mathrm{N}$ is applied for $5\,\mathrm{s}$ to a mass of $2\,\mathrm{kg}$ resting on a smooth horizontal surface. The mass now collides with a second mass of $3\,\mathrm{kg}$ at rest, and they continue together. Find the common velocity and the loss of KE in the impact.

_{Extracted from A. Godman & J. F. Talbert. (1975). Additional Mathematics Pure and Applied in SI Units.}

Roughwork.

1.

$\begin{aligned} m_1u_1+m_2u_2 & = (m_1+m_2)v \\ (30)(6) + (20)(2) & = (30+20)v \\ v& = 4.4\,\mathrm{m\,s^{-1}}\\ \end{aligned}$

$\begin{aligned} \textrm{KE}_\textrm{initial} & = \frac{1}{2}(30)(6)^2 + \frac{1}{2}(20)(2)^2 = 580\,\mathrm{J} \\ \textrm{KE}_\textrm{final} & = \frac{1}{2}(30+20)(4.4)^2 = 484\,\mathrm{J}\\ \Delta\textrm{KE} & = -96\,\mathrm{J}\\ \end{aligned}$

2. Not to be attempted.

3.

$\begin{aligned} F_{\textrm{net}} & = \frac{\Delta p}{\Delta t} \\ 2 & = \frac{(2)(v-0)}{5} \\ v & = 5\,\mathrm{m\,s^{-1}} \\ \end{aligned}$

$\begin{aligned} m_1u_1+m_2u_2 & = (m_1+m_2)v \\ (2)(5)+(3)(0) & = (2+3)v \\ v & = 2\,\mathrm{m\,s^{-1}} \\ \end{aligned}$

Not to be completed.

February 9, 2023February 9, 2023

202302091224 Exercise 20.2

(All accelerations are to be taken as uniform and in a straight line.)

1. A particle starts with velocity $3\,\mathrm{m\,s^{-1}}$ and accelerates at $0.5\,\mathrm{m\,s^{-2}}$ . What is its velocity after i. $3\,\mathrm{s}$ , ii. $10\,\mathrm{s}$ , iii. $t\,\mathrm{s}$ ? How far has it travelled in these times?
2. A body, decelerating at $0.8\,\mathrm{m\,s^{-2}}$ , passes a certain point with a speed of $30\,\mathrm{m\,s^{-1}}$ . Find its velocity after $10\,\mathrm{s}$ , the distance covered in that time and how much further the body will go until it stops.
3. A particle travelling with acceleration of $0.75\,\mathrm{m\,s^{-2}}$ passes a point $O$ with speed $5\,\mathrm{m\,s^{-1}}$ . How long will it take to cover a distance of $250\,\mathrm{m}$ ? What will its speed be at that time?
4. If a particle passes a certain point with speed $5\,\mathrm{m/s}$ and is accelerating at $3\,\mathrm{m/s^{2}}$ how far will it travel in the next $2\,\mathrm{s}$ ? How long will it take (from the start) to travel $44\,\mathrm{m}$ ?
5. A body falls from rest with an acceleration of $10\,\mathrm{m\,s^{-2}}$ . What is its velocity $5\,\mathrm{s}$ afterwards? How far has it fallen by then?

_{Extracted from A. Godman & J. F. Talbert. (1975). Additional Mathematics Pure and Applied in SI Units.}

Roughwork.

1.

$\begin{aligned} v(t) & = u+at = 3 + 0.5t \\ v(3) & = 3+0.5(3) = 4.5\,\mathrm{m\,s^{-1}}\\ v(10) & = 3+0.5(10) = 8\,\mathrm{m\,s^{-1}} \\ \end{aligned}$

$\begin{aligned} s(t) & =ut+\frac{1}{2}at^2 = 3t+\frac{t^2}{4} \\ s(3) & = 3(3)+\frac{(3)^2}{4} = 11.25\,\mathrm{m}\\ s(10) & = 3(10)+\frac{(10)^2}{4} = 55\,\mathrm{m}\\ \end{aligned}$

2.

Not to be attempted.

3.

$\begin{aligned} s & = ut+\frac{1}{2}at^2 \\ (250) & = (5)t+\frac{1}{2}(0.75)t^2 \\ 0 & = 3t^2 + 40t - 2000 \\ 0 & = (3t+100)(t-20) \\ t & = 20\,\mathrm{s}\textrm{ \scriptsize{OR} }-\frac{100}{3}\,\mathrm{s}\textrm{ (rej.)} \\ \end{aligned}$

$\begin{aligned} v & = u+at \\ & = (5)+(0.75)(20) \\ & = 20\,\mathrm{m\,s^{-1}} \\ \end{aligned}$

or,

$\begin{aligned} v^2 & =u^2+2as \\ v & = \pm\sqrt{(5)^2+2(0.75)(250)} \\ & = 20\,\mathrm{m\,s^{-1}}\textrm{ \scriptsize{OR} }-20\,\mathrm{m\,s^{-1}}\textrm{ (rej.)} \\ \end{aligned}$

4. Not to be attempted.

5. Not to be attempted.

September 28, 2022October 25, 2022

202209281521 Problem 5.37

A bead can slide without friction on a circular hoop of radius $R$ in a vertical plane. The hoop rotates at a constant rate of $\omega$ about a vertical diameter, as shown in the figure below.

(a) Find the angle $\theta$ at which the bead is in vertical equilibrium. (Of course it has a radial acceleration toward the axis.)
(b) Is it possible for the bead to “ride” at the same elevation as the centre of the hoop?
(c) What will happen if the hoop rotates at a slower rate $\omega' = \omega /2$ ?

_{Modified from H. D. Young. (1989). University Physics.}

Roughwork.

(a)

Kinetic energy $T$ :

$\begin{aligned} T & = \frac{1}{2}mv^2 \\ \dots\enspace v & = r\omega \enspace\dots \\ T & = \frac{1}{2}mr^2\omega^2 \\ \dots\enspace r & = R\sin\theta \enspace\dots \\ T & = \frac{1}{2}mR^2\omega^2\sin^2\theta \\ \end{aligned}$

Potential energy $V$ :

$\begin{aligned} V & = mg(R-R\cos\theta ) \\ & = mgR(1-\cos\theta ) \\ \end{aligned}$

Lagrangian $\mathcal{L}=T-V$ :

$\displaystyle{\mathcal{L}=\frac{1}{2}mR^2\omega^2\sin^2\theta-mgR(1-\cos\theta )}$

Euler–Lagrange equation:

$\displaystyle{\frac{\mathrm{d}}{\mathrm{d}t}\bigg(\frac{\partial\mathcal{L}}{\partial \dot{\theta}}\bigg)-\frac{\partial\mathcal{L}}{\partial \theta}=0}$

Thereby

$\begin{aligned} \frac{\partial\mathcal{L}}{\partial\dot{\theta}} & = 0 \quad \textrm{and}\quad \frac{\mathrm{d}}{\mathrm{d}t}\bigg(\frac{\partial\mathcal{L}}{\partial\dot{\theta}}\bigg) = 0 \\ \frac{\partial\mathcal{L}}{\partial\theta} & = mR^2\omega^2\sin\theta\cos\theta -mgR\sin\theta \\ \end{aligned}$

The bead is in vertical equilibrium when:

$mR^2\omega^2\sin\theta\cos\theta -mgR\sin\theta =0$

solving for $\theta$ then :

$\begin{aligned} 0 & = (mR\sin\theta )(R\omega^2\cos\theta -g) \\ \theta & = 0\enspace\textrm{ (rej.)}\quad\textrm{\scriptsize{OR}}\quad \cos^{-1}\bigg(\frac{g}{R\omega^2}\bigg) \\ \end{aligned}$

(b)

$\theta\to 90^\circ$ iff $R\omega^2\gg g$ .

(c) This is left as an exercise to the reader.

May 31, 2022October 24, 2022

202205311107 Problem 4.23

Boxes $A$ and $B$ are in contact on a horizontal, frictionless (i.e. $f=0$ ) surface, as shown in the Figure below. Box $A$ has mass $20.0\,\mathrm{kg}$ and box $B$ has mass $5.0\,\mathrm{kg}$ . A horizontal force of $100\,\mathrm{N}$ is exerted on box $A$ . What is the magnitude of the force that box $A$ exerts on box $B$ ?

_{extracted from Problem 4.23, Sears and Zemansky’s University Physics}

Steps.

Draw the free-body diagrams of $A$ , $B$ , and $A+B$ .

Apply Newton’s $2^\textrm{nd}$ law $\textrm{Net }F=ma$ :

$\begin{aligned} F_A- {}_{B}F_A - f_A & = m_Aa_A \\ {}_{A}F_B - f_B & = m_Ba_B \\ F_A-f_{A+B} & = m_{A+B}a_{A+B} \\ \end{aligned}$

Conditioning the equations of motion:

$\begin{aligned} a_A=a_B & =a_{A+B} \\ f_A=f_B=f_{A+B} & =0 \\ {}_{B}F_A & ={}_{A}F_{B} \\ m_A+m_B & =m_{A+B} \\ \end{aligned}$

and substituting numbers for symbols, write:

$\begin{aligned} 100 - {}_{A}F_{B} & = 20a \\ {}_{A}F_{B} & = 5a \\ 100 & =25a \\ \end{aligned}$

$\therefore$ The magnitude ${}_{A}F_{B}$ of the force that box $A$ exerts on box $B$ is $20\,\mathrm{N}$ .

January 21, 2022October 24, 2022

202201211321 Problem 1.1

Two particles move along the $x$ -axis uniformly with speeds $v_1=8\,\mathrm{m/s}$ and $v_2=4\,\mathrm{m/s}$ . At the initial moment the first point was $21\,\mathrm{m}$ to the left of the origin and the second $7\,\mathrm{m}$ to the right of the origin. When will the first point catch up with the second? Where will this take place? Plot the graph of the motion.

_{Extracted from A. A. Pinsky. (1980). Problems in Physics.}

Set-up.

Rename the two particles by $a$ and $b$ . The velocity of particle $a$ is $\mathbf{v}_a=+8\,\mathrm{(m\, s^{-1})}\enspace\hat{\mathbf{i}}$ and that of particle $b$ is $\mathbf{v}_b=+4\,\mathrm{(m\, s^{-1})}\enspace\hat{\mathbf{i}}$ . The particles at time $t=0$ are located on the $x$ -axis with $x$ -coordinates $x_a=-21$ and $x_b=+7$ respectively.

Roughwork.

When $t=0$ :

when $t=1$ :

when $t=2$ :

when $t=3$ :

when $t=4$ :

when $t=5$ :

when $t=6$ :

when $t=7$ :

Solution.

The positions $x_a(t)$ , $x_b(t)$ of particle $a$ , $b$ can be expressed in a function of discrete time interval

$t=\{ t_i\in\mathbb{Z^{+}}\textrm{ s.t. } t_{i+1}-t_{i}=t_{i}-t_{i-1}=1\}$ ,

i.e.,

$\begin{aligned} x_a(t_{i+1}) & =x_a(t_{i})+8 \\ x_b(t_{i+1}) & =x_b(t_{i})+4 \\ \end{aligned}$

or simply, in continuous time intervals,

$\begin{aligned} x_a(t) & =-21+8t \\ x_b(t) & = 7+4t \\ \end{aligned}$

Particle $a$ will meet particle $b$ when $x_a(t)=x_b(t)$ at some time $t'$ , as

$\begin{aligned} x_a(t') & = x_b(t') \\ -21+8t' & = 7+4t' \\ 4t' & = 28 \\ t' & = 7 \\ \end{aligned}$

so the place of meeting is

$x_a(7)=-21+8(7)=\boxed{35}=7+4(7)=x_b(7)$ .

December 2, 2021December 12, 2022

202112021037 Exercise 1.4 Lagrangian actions

For a free particle an appropriate Lagrangian is

Eq. (1.8):

$\mathcal{L}(t,x,v)=\displaystyle{\frac{1}{2}mv^2}$ .

Suppose that $x$ is the constant-velocity straight-line path of a free particle, such that $x_a=x(t_a)$ and $x_b=x(t_b)$ . Show that the action on the solution path is

Eq. (1.9):

$\displaystyle{\frac{m}{2}\frac{(x_b-x_a)^2}{t_b-t_a}}$ .

_{Extracted from Structure and Interpretation of Classical Mechanics, SICP, 2e}

Background. (Lagrangians and Lagrangian actions)

The function $\mathcal{L}$ is called a Lagrangian for the system, and the resulting action,

Eq. (1.4):

$S[q](t_1,t_2)=\displaystyle{\int_{t_1}^{t_2}\mathcal{L}\circ\Gamma [q]}$ ,

is called the Lagrangian action. For Lagrangians that depend only on time, positions, and velocities the action can also be written

Eq. (1.5):

$S[q](t_1,t_2)=\displaystyle{\int_{t_1}^{t_2}\mathcal{L}(t,q(t),\mathrm{D}q(t))\,\mathrm{d}t}$ .

_{(Section 1.3 The Principle of Stationary Action)}

Working. (roughly)

$\begin{aligned} S & = \int_{t_a}^{t_b}\frac{1}{2}mv^2\,\mathrm{d}t \\ & = \frac{m}{2}\int_{t_a}^{t_b}(\mathbf{v}\cdot\mathbf{v})\,\mathrm{d}t \\ & = \frac{m}{2}\int_{t_a}^{t_b}\bigg( \frac{\mathrm{d}\mathbf{x}}{\mathrm{d}t}\cdot\frac{\mathrm{d}\mathbf{x}}{\mathrm{d}t}\bigg) \,\mathrm{d}t \\ & = \frac{m}{2}\int_{x_a,t_a}^{x_b,t_b}\frac{(\mathrm{d}x)^2}{(\mathrm{d}t)} \\ & = \frac{m}{2}\frac{(x_b-x_a)^2}{t_b-t_a} \\ \end{aligned}$

April 16, 2021October 24, 2022

202104160754 Homework 1 (Q3)

A ship $A$ , which can sail at a constant speed $60\,\mathrm{km/hr}$ to meet a second ship $B$ which is $100\,\mathrm{km}$ away in the direction of $\mathrm{S60^\circ W}$ and is sailing due east at constant speed $30\,\mathrm{km/hr}$ . Find the sailing direction of $A$ and the time required to meet $B$ .

Solution.

(The solution below is based on the manuscript of 2014-2015 PHYS1250 Fundamental Physics Homework Solutions.)

Draw a diagram as follows:

Setup.

By the law of sines,

$\begin{aligned} \frac{V_A}{\sin 30^\circ} & = \frac{V_B}{\sin\theta} \\ \frac{60}{\sin 30^\circ} & = \frac{30}{\sin\theta} \\ \sin\theta & = 0.25 \\ \theta & = 14.5^\circ \end{aligned}$

Direction of $\mathbf{v}_A$ : $\mathrm{S45.5^\circ W}$

$\because 180^\circ -30^\circ -14.5^\circ - 90^\circ = 45.5^\circ$

Calculating $v_{AB}$ :

$\begin{aligned} |\mathbf{v}_{AB}| & = |\mathbf{v}_{A}|\cos\theta + |\mathbf{v}_{B}|\cos 30^\circ \\ & = 60 \cos 14.5^\circ + 30\cos 30^\circ \\ & = 84.1\,\mathrm{km/hr} \end{aligned}$

The time needed to meet ship $B$ is

$\begin{aligned} t & = \frac{100\,\mathrm{km}}{|\mathbf{v}_{AB}|} \\ & = \frac{100\,\mathrm{km}}{84.1\,\mathrm{km/hr}} \\ & = 1.19\,\mathrm{hr} \end{aligned}$

April 16, 2021October 24, 2022

202104160620 Homework 1 (Q4)

A particle is projected from a point $O$ on the horizontal floor. The range of the projectile is $R$ and the maximum height that the particle can reach is $h$ . Show that the equation of trajectory of the particle is

$\displaystyle{\frac{y}{h}=\frac{4x}{R}\bigg( 1-\frac{x}{R}\bigg)}$ .

Solution.

(The solution below is based on the manuscript of 2014-2015 PHYS1250 Fundamental Physics Homework Solutions.)

The trajectory of projectile motion must be a parabola, which can be expressed in the form of a quadratic equation:

$y=ax^2+bx+c$ ;

And since the particle passes through the points $(0,0)$ and $(R,0)$ , the equation of trajectory can be expressed in the form:

$y=A(x-0)(x-R)$ .

When the particle has traveled a horizontal distance $x=\displaystyle{\frac{R}{2}}$ , it reaches the maximum height $y=h$ .

Inserting the point $(\frac{R}{2},h)$ into the trajectory equation, we solve for the unknown $A$ :

$\begin{aligned} h & = A\bigg(\frac{R}{2}-0\bigg)\bigg(\frac{R}{2}-R\bigg) \\ h & = -\frac{AR^2}{4} \\ \Rightarrow \qquad A & = -\frac{4h}{R^2} \end{aligned}$

Thus,

$y = -\displaystyle{\frac{4h}{R^2}}x(x-R)$ ,

or,

$\boxed{\frac{y}{h} = \frac{4x}{R}\bigg( 1-\frac{x}{R}\bigg)}$

April 15, 2021October 24, 2022

202104150814 Homework 1 (Q2)

A particle is thrown with speed $v_0$ and an elevated angle $\theta$ on the floor. The air resistance is negligible.

(a) During the flight, the following quantities are investigated. Determine whether the following items are constants.

i. $\displaystyle{\frac{\mathrm{d}v}{\mathrm{d}t}}$ , where $v$ is the speed of the particle.

ii. $\displaystyle{\frac{\mathrm{d}\mathbf{v}}{\mathrm{d}t}}$ , where $\mathbf{v}$ is the velocity of the particle.

(b) What is the radius of curvature of the path when the particle reaches the highest point?

Answer.

(The solution below is based on the manuscript of 2014-2015 PHYS1250 Fundamental Physics Homework Solutions.)

(a)

i. $\displaystyle{\frac{\mathrm{d}v}{\mathrm{d}t}}$ varies with time.

ii. $\displaystyle{\frac{\mathrm{d}\mathbf{v}}{\mathrm{d}t}}=\mathbf{a}=\mathbf{g}=\textrm{Const.}$

Explanation.

(b)

$\begin{aligned} a=\frac{v^2}{r} & \Rightarrow g=\frac{v_0^2\cos^2\theta}{r} \\ & \Rightarrow r=\frac{v_0^2\cos^2\theta}{g} \end{aligned}$

April 15, 2021October 24, 2022

202104150729 Homework 1 (Q1)

A man starts from the origin and walks $30\,\mathrm{m}$ due east in $25\,\mathrm{s}$ . Then, he walks $10\,\mathrm{m}$ due south in $10\,\mathrm{s}$ and $18\,\mathrm{m}$ due northwest in $15\,\mathrm{s}$ . Please take the paths due east and due north as the positive $x$ and $y$ directions respectively in the Cartesian plane.

(a) Sketch the man’s path on the Cartesian plane.

(b) What is the average velocity of the man?

(c) What is the average speed of the man?

Solution.

(The solution below is based on the manuscript of 2014-2015 PHYS1250 Fundamental Physics Homework Solutions.)

(a)

(b) The average velocity of the man is

$\begin{aligned} \mathbf{OC} & = \mathbf{OA} + \mathbf{AB} + \mathbf{BC} \\ & = 30\,\hat{\mathbf{i}} - 10\,\hat{\mathbf{j}} + 18\bigg( -\frac{1}{\sqrt{2}}\,\hat{\mathbf{i}} + \frac{1}{\sqrt{2}}\,\hat{\mathbf{j}} \bigg) \\ & = \bigg( 30-\frac{18}{\sqrt{2}} \bigg) \hat{\mathbf{i}} + \bigg( \frac{18}{\sqrt{2}}-10 \bigg) \,\hat{\mathbf{j}}\\ & = 17.27\,\hat{\mathbf{i}} + 2.73\,\hat{\mathbf{j}} \end{aligned}$

$\begin{aligned} \therefore \mathbf{v}_{\textrm{avg}}& =\frac{\mathbf{OC}}{\Delta t} \\ & = \frac{17.27\,\hat{\mathbf{i}} + 2.73\,\hat{\mathbf{j}}}{50} \\ & = 0.35\,\hat{\mathbf{i}} + 0.055\,\hat{\mathbf{j}} \\ \therefore\quad |\mathbf{v}_{\textrm{avg}}| & = \sqrt{0.35^2+0.055^2} = 0.35\,\mathrm{m\, s^{-1}} \end{aligned}$

Direction of $\mathbf{v}_{\textrm{avg}}$ :

$\begin{aligned} \tan\theta & = \frac{0.055}{0.35}=0.16 \\ \theta & = 8.9^\circ \end{aligned}$

(c) The average speed of the man is

$\begin{aligned} v_{\textrm{avg}} & = \frac{OA+AB+BC}{\Delta t} \\ & = \frac{30+10+18}{50} \\ & = 1.16\,\mathrm{m\,s^{-1}} \end{aligned}$