February 8, 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.