March 2024 – Physicspupil

March 19, 2024March 19, 2024

202403191813 饔飧

中國人：日出而作，朝飯嘗飽，過中不食，日入而息，是上古先民的生活寫照。

英國人：Breakfast (to break one’s fast) 者，一吞一嚥；lunch (= noon + shench) 者，一啖一啜；supper (to sup/soup) 者，一咀一嚼。水穀精微，以固體難消化、流質易吸收。

法國人：Le déjeuner (= dé- +‎ jeûner) 本與 to break fast 同義；不過我輩晏起，又忌空腹食大餐，故襲以早餐為 le petit déjeuner，午餐為 le déjeuner。（注：petit = little）

March 5, 2024March 5, 2024

202403051417 Revision Paper XIII Q10

To a man walking along a horizontal road at $1.5\,\mathrm{m/s}$ the rain is coming towards him and appears to be falling at $3.5\,\mathrm{m/s}$ at an angle of $30^\circ$ to the vertical. Find, by calculation or drawing, the true speed of the rain and the angle this makes with the vertical.

_{Extracted from A. Godman & J. F. Talbert. (1973). Additional Mathematics Pure and Applied in SI Units.}

Setup.

As always, visualise the scene.

Two observers in different reference frames $S$ and $S'$ can give different descriptions of the same physical event $(x,y,z,t)$ . Where something is depends on when you check on it and on the movement of your own reference frame. Time and space are not independent quantities; they are related by relative velocity.

If $S'$ is moving with speed $v$ in the positive $x$ -direction relative to $S$ , then its coordinates in $S'$ are

$\begin{aligned} x' & = x-vt \\ y' & = y \\ z' & = z \\ t' & = t \\ \end{aligned}$

and if an object has velocity $\mathbf{u}$ in frame $S$ , then velocity $\mathbf{u}'$ of the object in frame $S'$ is

$\begin{aligned} u' & = \frac{\mathrm{d}x'(t)}{\mathrm{d}t} \\ & = \frac{\mathrm{d}}{\mathrm{d}t}(x(t)-vt) \\ & = \frac{\mathrm{d}x(t)}{\mathrm{d}t}-v \\ & = u-v \\ \end{aligned}$

in magnitude.

_{Cf. Ming-chang Chen. (2017). NTHU EE211000 Modern Physics.}

Roughwork.

Recall that a triangle is uniquely determined by not all but some of its three sides and three angles:

$\begin{aligned} \textrm{SSS} & \quad \textrm{(Side-Side-Side)} \\ \textrm{SAS} & \quad \textrm{(Side-Angle-Side)} \\ \textrm{ASA} & \quad \textrm{(Angle-Side-Angle)} \\ \textrm{AAS} & \quad \textrm{(Angle-Angle-Side)} \\ \textrm{RHS} & \quad \textrm{(Right angle-Hypotenus-Side)} \\ \end{aligned}$

Hence, provided in part

$\begin{aligned} u' & = 3.5 \\ v & = 1.5 \\ \measuredangle (u',v) & = 60^\circ \\ \end{aligned}$

we can supply in whole

$\begin{aligned} u & = \cdots \\ \measuredangle (v,u) & = \cdots \\ \measuredangle (u,u') & = \cdots \\ \end{aligned}$

This problem is not to be attempted.

March 4, 2024March 4, 2024

202403041703 Revision Paper II Q10

A body travels in a straight line but its speed at any time does not exceed $5\,\mathrm{m\,s^{-1}}$ . If it accelerates and decelerates at $2\,\mathrm{m\,s^{-2}}$ , find the shortest time needed to cover a distance of $30\,\mathrm{m}$ from rest to rest.

_{Extracted from A. Godman & J. F. Talbert. (1973). Additional Mathematics Pure and Applied in SI Units.}

Roughwork.

For avoidance of physics is one by mathematical formulation.

$\begin{aligned} v & = u+at \\ 5 & = 0 + 2t_1 \\ \textrm{Eq. (1):}\qquad\qquad t_1 & = 2.5\,\mathrm{s} \\ & \\ v & = u+at \\ 0 & = 5-2(t_3-t_2) \\ \textrm{Eq. (2):}\qquad t_3-t_2 & = 2.5\,\mathrm{s} \\ & \\ s & = \frac{((t_2-t_1)+(t_3-0))(v_{\textrm{max}})}{2} \\ 30 & = \frac{(t_2-2.5+t_3)(5)}{2} \\ \textrm{Eq. (3):}\qquad t_2+t_3 & = 14.5\,\mathrm{s} \\ \end{aligned}$

What is $t_3$ then, in unit $\textrm{sec}$ ?

This problem is not to be attempted.

March 1, 2024March 1, 2024

202403011713 Revision Paper IV Q8

Ball $A$ (mass $3\,\mathrm{kg}$ ) is travelling due east in a straight line with speed $5\,\mathrm{m\,s^{-1}}$ when it collides with ball $B$ (mass $4\,\mathrm{kg}$ ) which was at rest. After the collision both balls move east with relative speed $2.5\,\mathrm{m\,s^{-1}}$ . Find their separate speeds.

_{Extracted from A. Godman & J. F. Talbert. (1973). Additional Mathematics Pure and Applied in SI Units.}

Roughwork.

Take eastward positive.

$\begin{aligned} \textrm{CoM:}\qquad p_{\textrm{initial}} & = p_{\textrm{final}} \\ \sum_{i=A}^{B}m_iu_i & = \sum_{i=A}^{B}m_iv_i \\ m_Au_A+m_Bu_B & = m_Av_A+m_Bv_B \\ (3)(5)+(4)(0) & = (3)(v_A)+(4)(v_B) \\ \textrm{Eq. (1):}\qquad\quad 15 & = 3v_A+4v_B \\ \textrm{Eq. (1)':}\qquad\quad v_A & = 5-\frac{4}{3}v_B \\ \end{aligned}$

$\begin{aligned} \textrm{CoE:}\qquad\textrm{KE}_{\textrm{initial}} & = \textrm{KE}_{\textrm{final}} \\ \sum_{i=A}^{B}\frac{1}{2}m_iu_i^2 & = \sum_{i=A}^{B}\frac{1}{2}m_iv_i^2 \\ m_Au_A^2+m_Bu_B^2 & = m_Av_A^2+m_Bv_B^2 \\ (3)(5)^2 + (4)(0)^2 & = (3)(v_A)^2 + (4)(v_B)^2 \\ \textrm{Eq. (2):}\qquad\qquad 75 & = 3v_A^2 + 4v_B^2 \\ \end{aligned}$

$\begin{aligned} 75 & = 3\bigg( 5-\frac{4}{3}v_B\bigg)^2 + 4v_B^2 \\ 25 & = \bigg( 25 - \frac{40}{3}v_B + \frac{16}{9}v_B^2\bigg) + \frac{4}{3}v_B^2 \\ 0 & = v_B\bigg( -\frac{40}{3} + \frac{28}{9}v_B \bigg) \\ v_B & = 0\textrm{ \scriptsize{OR} }\frac{30}{7} \\ v_A & = \bigg[ 5-\frac{4}{3}(0)\bigg] \textrm{ \scriptsize{OR} }\bigg[ 5-\frac{4}{3}\bigg(\frac{30}{7}\bigg)\bigg] \\ & = 5\textrm{ \scriptsize{OR} }-\frac{5}{7} \\ \end{aligned}$

$(v_A,v_B) = \bigg( \displaystyle{-\frac{5}{7} , \frac{30}{7}}\bigg) \qquad \text{}^*\textrm{(5,0) is rejected.}$

This contradicts with $v_B-2.5=v_A>0$ . As the law of conservation of momentum is always observed, herein has energy $\textrm{\scriptsize{NOT}}$ been conserved.

This problem is not to be attempted.