February 2022 – Physicspupil

February 8, 2022October 24, 2022

202202081217 Dynamics Figures (Elementary) Q6

13. Two identical $2\,\mathrm{kg}$ trolleys are connected by a light string and pulled by a $12\,\mathrm{N}$ force as shown. Assuming the surface to be frictionless,

(a) calculate the acceleration of the two trolleys;

(b) find the tension $T$ in the string.

_{Extracted from B. Kennedy. (2001). Higher Physics Progressive Problems.}

Solution.

We draw three free-body diagrams of i. $m_a$ ; ii. $m_b$ ; and iii. $m_a+m_b$ :

In respective diagrams there are three equations of motion.

By Newton’s second law,

$\textrm{Net }\mathbf{F}=m\mathbf{a}$ :

we have

$\begin{aligned} T & = m_aa \\ 12-T & = m_ba \\ 12 & = (m_a+m_b)a \\ \end{aligned}$

Answers. (a) acceleration $a=3\,\mathrm{m\, s^{-2}}$ ; (b) tension $T=6\,\mathrm{N}$ .

February 8, 2022October 24, 2022

202202081107 Solution to 1980-AL-PHY-I-33

In an experiment, a student takes several readings of the variable $x$ and $y$ . He plots a graph of $\log y$ against $\log x$ and finds that it is a straight line of gradient $m$ , making an intercept $c$ on the positive $\log y$ axis.

What is the relation between $x$ and $y$ ?

Solution.

$\begin{aligned} \log y & = m\log x+c\\ \log y & = \log x^m + c\\ \log y-\log x^m & = c\\ \log \bigg( \frac{y}{x^m}\bigg) & = c \\ \cdots\enspace \textrm{let }c =\log a&\enspace \cdots\\ \log \bigg( \frac{y}{x^m}\bigg) & = \log a \\ \frac{y}{x^m} & = a\\ y & = ax^m \\ \end{aligned}$

February 8, 2022January 19, 2023

202202080958 Dynamics Figures (Elementary) Q5

2. A man walking with a speed $v$ constant in magnitude and direction passes under a lantern hanging at a height $H$ above the ground. Find the velocity which the edge of the shadow of the man’s head moves over the ground with if his height is $h$ .

_{Extracted from B. Bukhovtsev et al. (1978). Problems in Elementary Physics.}

Solution.

Let $x=0$ be the position of the lantern; let the man walk in the positive $x$ -direction; and let the position of the man be $x_m(t)$ and that of the shadow of his head $x_s(t)$ . So the length $s$ of his shadow is $|x_s-x_m|$ .

By comparing similar triangles, we have

$\displaystyle{\frac{H}{x_s} = \frac{h}{s}}$ .

Thus,

$\begin{aligned} \frac{H}{x_s} & = \frac{h}{x_s-x_m} \\ x_s & = \bigg(\frac{H}{H-h}\bigg) x_m \\ \dot{x}_s & = \bigg(\frac{H}{H-h}\bigg) \dot{x}_m \\ \dot{x}_s & = \bigg(\frac{H}{H-h}\bigg) v \\ \end{aligned}$

$\therefore$ The edge of the shadow of the man’s head moves with a velocity $(\frac{H}{H-h}) v\,\hat{\mathbf{i}}$ over the ground.

February 7, 2022October 24, 2022

202202071621 Dynamics Figures (Elementary) Q4

1.3.10. A $2\,\mathrm{kg}$ mass and a $3\,\mathrm{kg}$ mass are linked by a light string passed over a frictionless pulley. Calculate the acceleration of the system.

_{Extracted from B. Kennedy. (2001). Higher Physics Progressive Problems.}

Solution.

At first glance, as mass $m_2$ is heavier than mass $m_1$ , one can expect that the heavier mass will fall down and the lighter mass will move up.

Draw the free body diagram of each mass below:

Write the equation of motion for both masses:

$\begin{aligned} T-m_1g & = m_1a \\ m_2g-T & = m_2a \\ & \\ T-2g & = 2a \\ 3g-T & = 3a \\ & \\ a & = \frac{g}{5} \\ T & = \frac{12g}{5} \\ \end{aligned}$

$\therefore$ Both masses $m_1$ and $m_2$ will accelerate with a magnitude of $g/5=1.96\,\mathrm{m\, s^{-2}}$ .

February 7, 2022October 24, 2022

202202071437 Dynamics Figures (Elementary) Q3

1.17. A stationary object explodes into two fragments of relative mass $1:100$ . At the instant of break-up, the larger mass has a velocity of $10\,\mathrm{m\, s^{-1}}$ . Calculate i. the velocity of the smaller mass, ii. the ratio of their kinetic energies at this instant.

_{Extracted from M. Nelkon. (1971). Graded Exercises and Worked Examples in Physics.}

Solution.

Provided that $u=0$ , $m_1:m_2=1:100$ , and $v_2=10\,\mathrm{m\, s^{-1}}$ .

By the law of conservation of linear momentum,

$\begin{aligned} M\mathbf{u} & = m_1\mathbf{v}_1+m_2\mathbf{v}_2 \\ (M)(0\,\hat{\mathbf{i}}) & = \bigg(\frac{1}{101}M\bigg) (-v_1\,\hat{\mathbf{i}})+\bigg(\frac{100}{101}M\bigg) (+10\,\hat{\mathbf{i}}) \\ v_1 & = 1000 \\ \mathbf{v}_1 & = -1000\,\hat{\mathbf{i}}\\ \end{aligned}$

The ratio of their kinetic energies is given by

$\begin{aligned} &\quad \textrm{KE}_1:\textrm{KE}_2 \\ & = \frac{\frac{1}{2}m_1v_1^2}{\frac{1}{2}m_2v_2^2} \\ & = \frac{(\frac{1}{2})(\frac{1}{101}M)(1000)^2}{(\frac{1}{2})(\frac{100}{101}M)(10)^2} \\ & = 100:1 \\ \end{aligned}$

February 7, 2022October 24, 2022

202202071146 Solution to 1969-CE-AMATH-I-XX

Find $\displaystyle{\int\frac{2x+5}{(x^2+5x+1)^3}\,\mathrm{d}x}$ .

Solution.

Set-up.

$\begin{aligned} \textrm{Let }u & =x^2+5x+1 \\ \frac{\mathrm{d}u}{\mathrm{d}x} & = 2x+5 \\ \mathrm{d}x & = \frac{\mathrm{d}u}{2x+5} \\ \end{aligned}$

Then,

$\begin{aligned} &\quad \int\frac{2x+5}{(x^2+5x+1)^3}\,\mathrm{d}x \\ & = \int \frac{2x+5}{u^3}\bigg( \frac{\mathrm{d}u}{2x+5}\bigg) \\ & = \int \frac{1}{u^3} \,\mathrm{d}u \\ & = -\frac{1}{2u^2}+C\textrm{ for some constant }C \\ & = -\frac{1}{2(x^2+5x+1)^2}+C\\ \end{aligned}$

February 7, 2022October 24, 2022

202202071039 Dynamics Figures (Elementary) Q2

The ends of a rope are fastened at separate but horizontal points on a ceiling. A $10\,\mathrm{lb}$ weight is hung from the rope such that the two segments of the rope make angles of $30^\circ$ and $45^\circ$ with the ceiling. Compute the tension in each segment of the rope.

_{Extracted from R. L. Gray. (1973). Physics Problems: Mechanics and Heat.}

Solution.

Draw a free-body diagram below:

Separating the vertical and the horizontal component:

$\begin{aligned} T_1\cos 45^\circ + T_2\cos 60^\circ & = W \\ T_1\sin 45^\circ & = T_2\sin 60^\circ \\ \end{aligned}$

we have two equations with two unknowns (i.e., $T_1$ and $T_2$ ).

The rest is left the reader as an exercise.

February 7, 2022October 24, 2022

202202070828 Dynamics Figures (Elementary) Q1

As shown in the sketch above, a mass $m$ is suspended from a string of length $L$ . Gravity $\mathbf{g}$ is acting downward in the diagram. A second string exerts a horizontal force $\mathbf{F}$ such that $m$ is in equilibrium at a horizontal distance $x$ .

(a) Draw a diagram showing all forces acting on $m$ .

(b) What is the tension in the suspension string? Give both direction and magnitude.

_{Extracted from R. L. Gray. (1973). Physics Problems: Mechanics and Heat.}

Solution.

(a) The free body diagram is shown below:

(b) Obviously,

$\begin{aligned} \mathbf{T} & =F\,\hat{\mathbf{i}}+mg\,\hat{\mathbf{j}} \\ T & =|\mathbf{T}|=\sqrt{(F)^2+(mg)^2} \\ \end{aligned}$ .

Or, in an old-school way, we have horizontally

$F=T\cos\theta$

and vertically

$mg=T\sin\theta$ ,

squaring and summing up the equations,

$\begin{aligned} (F)^2+(mg)^2 & =(T\cos\theta )^2+(T\sin\theta )^2 \\ F^2+m^2g^2 & = T^2(\cos^2\theta +\sin^2\theta ) \\ F^2+m^2g^2 & = T^2 \\ T & = \sqrt{F^2+m^2g^2} \\ \end{aligned}$

Dividing the equation of the vertical by the horizontal,

$\begin{aligned} \frac{mg}{F} & = \frac{T\sin\theta}{T\cos\theta} \\ \frac{mg}{F} & = \tan\theta \\ \theta & = \tan^{-1}\bigg(\frac{mg}{F}\bigg) \\ \end{aligned}$

the direction of tension makes an angle $\theta =\tan^{-1}(\frac{mg}{F})$ with the level.

February 6, 2022October 24, 2022

202202070744 Solution to 1997-CE-AMATH-II-2

Find $\displaystyle{\int x\sqrt{x-1}\,\mathrm{d}x}$ .

[Hint: Let $u=x-1$ .]

Solution.

$\begin{aligned} \textrm{Let } & u = x-1 \\ \textrm{then } & x = u+1 \\ \textrm{and }&\mathrm{d}x = \mathrm{d}u\\ \end{aligned}$

$\begin{aligned} &\quad \int x\sqrt{x-1}\,\mathrm{d}x\\ & = \int (u+1)(u^{\frac{1}{2}})\,\mathrm{d}u \\ & = \int (u^{\frac{3}{2}}+u^{\frac{1}{2}})\,\mathrm{d}u \\ & = \frac{2u^{\frac{5}{2}}}{5} + \frac{2u^{\frac{3}{2}}}{3} + C\textrm{ for some constant }C \\ & = \frac{2(x-1)^{\frac{5}{2}}}{5} + \frac{2(x-1)^{\frac{3}{2}}}{3} + C \\ \end{aligned}$

February 4, 2022October 24, 2022

202204021539 Example 3, Chapter 1.3, Methods in Physics II (2015-2016 Lectures)

If $\mathbf{f}(t)=t\,\hat{\mathbf{i}}+t^3\,\hat{\mathbf{j}}$ , $\mathbf{g}(t)=\cos t\,\hat{\mathbf{i}}+\sin t\,\hat{\mathbf{j}}$ , and $\mathbf{v}=2\,\hat{\mathbf{i}}-3\,\hat{\mathbf{j}}$ . Calculate

(a) $(\mathbf{f}+\mathbf{g})'$ ;

(b) $(\mathbf{v}\cdot\mathbf{f})'$ ;

(c) $(\mathbf{f}\cdot\mathbf{g})'$ .

Settings. (Some properties of vector differentiation)

i. If $\mathbf{f}$ and $\mathbf{g}$ are differentiable vector functions,

$\displaystyle{\frac{\mathrm{d}}{\mathrm{d}t}(\mathbf{f}+\mathbf{g})=\frac{\mathrm{d}\mathbf{f}}{\mathrm{d}t}+\frac{\mathrm{d}\mathbf{g}}{\mathrm{d}t}}$

ii. If $\mathbf{f}$ is a differentiable vector function and $\alpha$ a constant scalar,

$\displaystyle{\frac{\mathrm{d}}{\mathrm{d}t}(\alpha\mathbf{f})=\alpha\frac{\mathrm{d}\mathbf{f}}{\mathrm{d}t}}$

iii. If $\mathbf{f}(t)$ is a vector function, $\mathbf{v}$ a constant vector, and $\mathbf{v}\cdot\mathbf{f}$ differentiable,

$\displaystyle{\frac{\mathrm{d}}{\mathrm{d}t}(\mathbf{v}\cdot\mathbf{f})=\mathbf{v}\cdot\frac{\mathrm{d}\mathbf{f}}{\mathrm{d}t}}$

iv. If $h(t)$ is a scalar function, $\mathbf{f}(t)$ a vector function and $h\mathbf{f}$ differentiable,

$\displaystyle{\frac{\mathrm{d}}{\mathrm{d}t}(h\mathbf{f})=h\frac{\mathrm{d}\mathbf{f}}{\mathrm{d}t}+\mathbf{f}\frac{\mathrm{d}h}{\mathrm{d}t}}$

v. If $\mathbf{f}$ and $\mathbf{g}$ are vector functions and $\mathbf{f}\cdot\mathbf{g}$ differentiable,

$\displaystyle{\frac{\mathrm{d}}{\mathrm{d}t}(\mathbf{f}\cdot\mathbf{g})=\mathbf{f}\cdot\frac{\mathrm{d}\mathbf{g}}{\mathrm{d}t}+\mathbf{g}\cdot\frac{\mathrm{d}\mathbf{f}}{\mathrm{d}t}}$

vi. If $\mathbf{f}$ and $\mathbf{g}$ are vector functions and $\mathbf{f}\times\mathbf{g}$ differentiable,

$\displaystyle{\frac{\mathrm{d}}{\mathrm{d}t}}(\mathbf{f}\times\mathbf{g})=\displaystyle{\bigg( \frac{\mathrm{d}\mathbf{f}}{\mathrm{d}t}\times\mathbf{g}\bigg) + \bigg( \mathbf{f}\times \frac{\mathrm{d}\mathbf{g}}{\mathrm{d}t}\bigg)}$ .

Solution.

(a)

$\begin{aligned} \mathbf{f}+\mathbf{g} & = (t+\cos t)\,\hat{\mathbf{i}}+(t^3+\sin t)\,\hat{\mathbf{j}} \\ (\mathbf{f}+\mathbf{g})' & = (t+\cos t)'\,\hat{\mathbf{i}}+(t^3+\sin t)'\,\hat{\mathbf{j}} \\ & = (1-\sin t)\,\hat{\mathbf{i}}+(3t^2+\cos t)\,\hat{\mathbf{j}} \\ \end{aligned}$

(b)

$\begin{aligned} \mathbf{v}\cdot\mathbf{f} & = (2,-3)\cdot (t,t^3) \\ & = 2t-3t^3 \\ (\mathbf{v}\cdot\mathbf{f})' & = (2t-3t^3)' \\ & = 2-9t^2 \\ \end{aligned}$

(c)

$\begin{aligned} \mathbf{f}\cdot\mathbf{g} & = (t,t^3) \cdot (\cos t,\sin t) \\ & = t\cos t+t^3\sin t \\ (\mathbf{f}\cdot\mathbf{g})' & = \cos t-t\sin t+t^3\cos t + 3t^2\sin t\\ \end{aligned}$