February 7, 2022 – Physicspupil

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.