PAT 2022

1. What is the total resistance of the circuit ?

2. For which values of $x$ is $\qty(24-14x -3x^2)^{-1}$ positive?

The $-1$ exponent does not change the sign. We can factorise $24-14x -3x^2$ as $(-3x+4)(x+6)$ which is positive either if $x < \frac{4}{3}, x > -6$ or if $x > \frac{4}{3}$, $x < -6$. The second condition cannot hold so we have to go with (D).

3. Molecules of oxygen in the atmosphere absorb solar radiation in bands centred at about 80 nm, 650 nm and 1000 nm. In which parts of the electromagnetic spectrum are these absorption bands?

Picture from tikz.net

The visible spectrum is in the very narrow band from 380 nm (violet) to 750 nm (red) so 650 nm is in the visible spectrum in the red zone. $1000 \ {\rm nm}=10^{-6} \ {\rm m}$ which is right at the edge of infrared. Finally $80 \ {\rm nm}= 0.8 \times 10^{-7} \ {\rm m}$ which is close to the middle of the ultraviolet spectrum.

4. Which of these polynomial functions has the largest second derivative at $x = 0$?

$$ \begin{array}{|c|c|c|c|c|} \hline \textbf{A} & \textbf{B} & \textbf{C} & \textbf{D} & \textbf{E} \\ \hline 5x^5-x^3+4x & 3x^4+x^2+16 & 4x^6+x^2-1 & x^3+2x^2-5x+10 & 10x^5+3x^3-7x +2 \\ \hline \end{array} $$

For the second derivative at $x=0$ only the coefficient of the $x^2$ matters. So the answer is (D).

5. An asteroid of mass $10^3 \ {\rm kg}$ is moving towards a space station at $1\ {\rm m} \ {\rm s}^{-1}$. It is proposed to stop it by firing a $1 \ {\rm MW}$ laser at it. For how long must the laser be fired? You may assume that the surface of the asteroid is perfectly reflective, all photons are incident perpendicular to the surface of the asteroid, and a photon’s momentum is related to its energy by $p = E/c$ , where $c = 3 \times 10^{-8} \ {\rm m} \ {\rm s}^{-1}$ is the speed of light.

The momentum of the asteroid is $10^3 \ {\rm kg} \ {\rm m} \ {\rm s}^{-1}$. If we fire the laser for some time $\tau$ we generate $E=\tau \ 10^6\ {\rm J}$ of energy which corresponds to $(\tau / 3) \times 10^{-2} \ {\rm kg} \ {\rm m} \ {\rm s}^{-1}$ of momentum. After the asteroid has ceased to move, the photons are travelling in the opposite direction since the asteroid is perfectly reflective. Therefore we have $$ (\tau / 3) \times 10^{-2} - 10^3 =- (\tau / 3) \times 10^{3} $$ using the law of conservation of momentum. So we must fire the laser for $\tau = 1.5 \times 10^{5} \ {\rm s}$ (C).

6. Which expression correctly represents the sum $\sum_{k=0}^n ar^{2k}$ ?

The formula for a geometric series is $$ \sum_{k=0}^n r^k = \frac{1-r^{n+1}}{1-r}. $$ In the problem we must substitute $r^2$ for $r$ to obtain $$ \sum_{k=0}^n ar^{2k} = a \frac{1-r^{2(n+1)}}{1-r^2}. $$ The answer is (E).

7. In a cathode ray tube, an electron (mass $9.1 \times 10^{-31} \ {\rm kg}$, charge $-1.6 \times 10^{-19} \ {\rm C}$) is accelerated from rest by a uniform electric field of strength $20 \ {\rm kV} \ {\rm m}^{-1}$. How much time does it take to travel $50 \ {\rm cm}$?

The force acting on the charge is $$ F = 2 \times 10^4 \ {\rm V} \ {\rm m}^{-1} \times 1.6 \times 10^{-19} \ {\rm C} = 3.2 \times 10^{-15} \ {\rm N}. $$ The acceleration of the charge is $$ a = \frac{3.2 \times 10^{-15} \ {\rm N}}{9.1 \times 10^{-31} \ {\rm kg}} = 3.52 \times 10^{15} \ {\rm m} \ {\rm s}^{-2}. $$ Given this acceleration it takes $$ t=\sqrt{\frac{2d}{a}}=\sqrt{\frac{1 \ {\rm m}}{3.52 \times 10^{15} \ {\rm m} \ {\rm s}^{-2}}}=1.686 \times 10^{-8} \ {\rm s} $$ to travel 50 cm. The answer is (C).

8. If a function $y = f(x)$ has a stationary point at $(x_0, y_0)$, what are the co-ordinates of the corresponding stationary point of the function $y = af(bx + c)$?

$$ \dv{y}{x}= a \dv{f}{u} \dv{(bx+c)}{x}= ab \dv{f}{u} $$ where we have used the chain rule with $u=bx+c$. Since $f$ has a stationary point at $(x_0,y_0)$, $y$ has a stationary point when $bx+c=x_0$ or $x=(x_0-c)/b$. The corresponding value for $y$ is $$ y\qty(\frac{x_0-c}{b})=a f\qty(x_0)=ay_0. $$ The answer is (C).

9. As it appears to move across the sky, the Sun moves through an angle equal to that subtended by its diameter in about two minutes, as in the diagram. In a solar eclipse, the Moon covers the Sun almost exactly in the sky. Using this, what is the approximate ratio of the Moon’s radius to its orbital distance from Earth?

The angular velocity of the sun (from the perspective of an observer on the earth) is $ \omega_{\rm Sun}=2\pi/24 \ {\rm hours} = 0.00436 \ {\rm radians} \ {\rm min}^{-1}. $. The angle inscribed in 1 min is $\theta_{\rm 1 min}= 0.00436 \ {\rm radians}$. Since the moon covers the sun almost exactly during an eclipse we can write $$ \tan\theta_{\rm 1 min} = \frac{R_M}{d_M}=\frac{R_S}{d_S}. $$ $\theta_{\rm 1 min}$ is for all purposes a small angle so we can use the small angle approximation $\tan\theta_{\rm 1 min}\approx\theta_{\rm 1 min}$. Hence $R_M/d_M=0.00436$ and the answer is (D).

10. What is the next number in the sequence $0,\frac{3}{4},\frac{3}{8},\frac{9}{16},\frac{15}{32},\frac{33}{64}$.

$$ \begin{array}{cccccccccc} n & = & 0 & 1 & 2 & 3 & 4 & 5 & {\color{red} 6} & \\ a_n & = & 0 & \frac{3}{4} & \frac{3}{8} & \frac{9}{16} & \frac{15}{32} & \frac{33}{64} & {\color{red} \frac{63}{128}} & \text{(C)}\\ a_n-a_{n-1} & = & - & \frac{3}{4} & -\frac{3}{8} & \frac{3}{16} & -\frac{3}{32} & \frac{3}{64} & {\color{red} -\frac{3}{128}} & \\ \end{array} $$

11. Two moons occupy circular orbits around a planet. The smaller moon has mass $1.5 \times 10^{15} \ {\rm kg}$ and orbital radius $2.3 \times 10^4 \ {\rm km}$. The larger moon has mass $1.1 \times 10^{16} \ {\rm kg}$ and orbital radius $9.4 \times 10^3 \ {\rm km}$. If the gravitational force exerted by the planet on the smaller moon is $10^{14} \ {\rm N}$, what force does the planet exert on the larger moon?

\begin{align*} F_s & = G \frac{Mm_s}{R_s^2} \\ F_l & = G \frac{Mm_l}{R_l^2} \\ \frac{F_l}{F_s} & =\frac{m_l R_s^2}{m_s R_l^2} \\ F_l & = F_s \frac{m_l R_s^2}{m_s R_l^2} = 10^{14} \ {\rm N} \frac{1.1 \times 10^{16} \times \qty(2.3 \times 10^4)^2}{1.5 \times 10^{15} \times \qty(9.4 \times 10^3 )^2} = 4.39 \times 10^{15} \ {\rm N} \quad \text{(E)} \end{align*}

12. What is the derivative of $y = x^6 + 6x^5 + 12x^4 + 8x^3$?

$$ \dv{y}{x}=6x^5+30x^4+48x^3+24x^2=6x^2 \qty(x^3+5x^2+8x+4) $$ For the cubic set $5=r-x$ to convert it into a quadratic $$ rx^2+8x+4=0. $$ Its solutions are $x=\frac{-8\pm\sqrt{64-16r}}{2r}$. For $r=3$ one of the solutions is $x=-2$ which satisfies the condition $5=r-x$. Then $$ x^3+5x^2+8x+4=(x+2)(x^2+ax+b) $$ which holds if $b=2$ and $2+a=5$ i.e. $a=3$. The quadratic $x^2+3x+2$ can be factorised as $(x+1)(x+2)$. Thus $$ \dv{y}{x}=6x^2(x+1)(x+2)^2=(6x+6)(x^2+2x)^2. $$ The answer is (E).

13.

1. Draw the functions $y_1(x) = x^2 - 1$, $y_2(x)=4x - 2$ and $y_3(x) = -\frac{x}{2}-2$ on a common set of axes. Label where they cross the axes.
2. Work out the x-values of the intersection points of these three functions.
3. Write down a single integral which describes a finite area bounded by two of the three functions. You do not need to evaluate the integral.

$$ A= \int_{2-\sqrt{3}}^{2+\sqrt{3}} \qty(4x-2 -(x^2-1)) dx = \int_{2-\sqrt{3}}^{2+\sqrt{3}} \qty(-x^2+4x-1) dx. $$

14. The Trojan asteroids share Jupiter’s orbit around the Sun: approximately circular with a mean radius 5.2 AU ($1 \ {\rm AU } = 1.5 \times 10^{11} \ {\rm m}$ is the mean radius of the Earth’s orbit around the Sun). The Trojans are clustered around two points labelled L4 and L5, where the L4 point is $60^\circ$ ahead of Jupiter in its orbit and the L5 point is $60^\circ$ behind Jupiter in its orbit.

1. Determine the mean distance between the asteroids 588 Achilles (at the L4 point) and 617 Patroclus (at the L5 point).
2. A spacecraft travels in a straight line between the two asteroids, accelerating at $10 \ {\rm m} \ {\rm s}^{-2}$ until the half-way point between the asteroids, and decelerating at $10 \ {\rm m} \ {\rm s}^{-2}$ from there to the end-point. Assuming that the asteroids are approximately stationary on the timescale of the journey, and neglecting any gravitational effects of Jupiter or the Sun, find the total travel time.
3. Explain why the assumption that the asteroids are approximately stationary during the journey is well-justified.

$|L4L5|=2 \times 5.2 \cos(30^\circ) \ {\rm AU} = 5.2 \times \sqrt{3} \ {\rm AU} = 9 \ {\rm AU} \; (\text{1 s.f.})$ Using the accelation $a=10 \ {\rm m} \ {\rm s}^{-2}$ the time it takes to reach the half-way point is $t=\sqrt{|L4L5|/a}=4.25 \ {\rm days}$. So the total travel time is $8.5 \ {\rm days}$. From Kepler's third law the square of the orbital period $T$ is proportional to the cube of the orbital radius $R$. Given that the earth has an orbital period of 365 days $T_{\rm Jupiter}=365\sqrt{5.2^3}=4328 \ {\rm days}$. It is therefore a good assumption that in the 8.5 days of travel time the asteroids (which have the same orbital period as Jupiter) have hardly moved.

15. A projectile is launched at speed $v$ and angle $\theta$ (as measured from the horizontal) outwards from the top of a high cliff.

1. Sketch the trajectory of the projectile for launch angles $\theta = 5^\circ, 45^\circ$ and $85^\circ$. Use $x(t)$ for the horizontal displacement from the launch point and $y(t)$ for the vertical displacement from the launch point.
2. Using separate axes, now sketch the absolute distance, $r(t) = \sqrt{x(t)^2 + y(t)^2}$, from its launch point as a function of time for all of the three launch angles above.
3. Obtain an expression for $r(t)$. For which angles does $r(t)$ have a stationary point?
4. For angles below these, what happens to $r(t)$ as time increases?

\begin{align*} \def\half{\frac{1}{2}}\def\quarter{\frac{1}{4}} x(t) &= v t \cos\theta \\ y(t) & = h + v t \sin\theta - \frac{1}{2} g t^2 \\ r(t) & = \qty[x(t)^2+(y(t)-h)^2]^\half = \qty[v^2t^2+\quarter g^2t^4-gvt^3\sin\theta]^\half. \end{align*}

For the stationary points $$ \dv{r(t)^2}{t}= 2 r(t) \dv{r}{t} = 2 r(t) \half r(t)^{-1} \qty(2v^2t+g^2t^3-3gvt^2\sin\theta)=t(2v^2+g^2t^2-3gvt \sin\theta). $$ The discriminant is $D=9g^2v^2\sin^2\theta - 8 g^2 v^2$. This is negative when $\sin\theta < 2\sqrt{2}/3$, $\theta < 70.5^\circ$. For angles below $70.5^\circ$ $r(t)$ increases as $t$ increases.

16. Suppose $f(t)=4t$ and $\def\threeq{\frac{3}{2}}g(x)=\threeq(3x-x^2)$. Consider the inequality $$ \dv{g(x)}{x}> \int_{\threeq}^x f(t)dt.$$ For which values of $x$ is this inequality satisfied?

\begin{align*} \dv{g(x)}{x} & =\threeq \qty(3-2x) \\ \int_{\threeq}^x f(t)dt & = 4\int_{\threeq}^x t dt = 2 \qty[t^2]_{\threeq}^x = 2 \qty(x-\threeq)\qty(x + \threeq) \\ \threeq \qty(3-2x) & > 2 \qty(x-\threeq)\qty(x + \threeq) \\ \threeq \qty(\threeq - x) & > \qty(x-\threeq)\qty(x + \threeq). \end{align*} If $x < \threeq$, $x > - 3$ the inequality is satisfied. If $x >\threeq $ we must also have $ x < -3$ which is not possible. Therefore the inequality is satisfied for $-3 < x < \threeq$.

17. Four circles of radius $r_1$ are inscribed inside a square of side $4r_1$ as shown in the diagram below.

1. What is the radius $r_2$ of the largest circle that can fit in the space at the centre of the square, bounded by the outer circles?
2. If 8 spheres of radius $r_1$ are now similarly arranged inside a cube of edge length $4r_1$, what is the radius $r_3$ of the largest sphere that can fit in the space at the centre of the cube?

There is an error in (a); the question should be “ [w]hat is the radius of the smallest circle that can fit in the space at the centre of the square... ” . For the smallest circle to fit we must have $(2r_1)^2=2(r_1+r_2)^2$, $r_2=(\sqrt{2}-1)r_1$.

To pack eight spheres in a cube with edges of size $4r_1$, the first sphere will have its centre at coordinate $\qty(r_1,r_1,r_1)$ if we treat corner $K_1$ as the origin. Therefore $|K_1C_1|=\sqrt{3}r_1$; by symmetry $|K_8C_8|=\sqrt{3}r_1$. Since $|K_1K_8|=4\sqrt{3}r_1$ and points $K_1'$, $K_8'$ are a distance $r_1$ from, respectively, $C_1$, $C_8$ we have $$|K_1'K_8'|=|K_1K_8|-|K_1C_1|-|C_1K_1'|-|K_8C_8|-|C_8K_8'|=4\sqrt{3}r_1-2\sqrt{3}r_1-2r_1=2\qty(\sqrt{3}-1)r_1.$$ Therefore the radius of the red sphere is $r_3=\qty(\sqrt{3}-1)r_1$.

18. Consider the function $$ f(x)=-\frac{P}{x^3}+\frac{Q}{x^2}-\frac{R}{x}$$ in the region $x>0$ wher $P$, $Q$ and $R$ are all positive constants.

1. Find an inequality satisfied by $P$, $Q$ and $R$ in order for $f(x)$ to have at least one real root.
2. Find a relationship between $P$, $Q$ and $R$ in order for $f(x)$ to have exactly one stationary point.
3. If the relationship of the previous part holds, so that exactly one stationary point exists, what is the nature of that stationary point and at what value of $x$ (expressed in terms of $P$, $Q$ and $R$) is it? It is not necessary to work out a second derivative to answer this.

\begin{align*} f(x) & =-\frac{P}{x^3}+\frac{Q}{x^2}-\frac{R}{x} \\ -x^3f(x) & =Rx^2-Qx+P \\ Q & \geq 2\sqrt{RP} &\text{for at least one real root} \\ -3x^2f(x)-x^3\dv{f}{x} & =2Rx-Q \\ -3x^2f(x) & =2Rx-Q &\text{since $\dv{f}{x}=0$} \\ 3\qty(Rx^2-Qx+P) & = 2Rx^2-Qx \\ Q & = \sqrt{3RP} & \text{for exactly one stationary point} \\ \end{align*} A repeating stationary point is an inflexion point since in this case the second derivative is always zero. To see this consider a function $f(x)$ with $\dv*{f}{x}=(x-a)^2g(x)$. $x=a$ is a repeating stationary point and $g(x)$ is some other function. Then $$ \dv[2]{f}{x}=2(x-a)g(x)+(x-a)^2 \dv{g}{x} $$ and so $[\dv*[2]{f}{x}]_{x=a}=0$, i.e. $x=a$ is an inflexion point.

19. Following Bohr, we assume that a hydrogen-like atom may be modelled as a single electron (mass $m$ and charge $e$) in a circular orbit around a much more massive nucleus (charge $+Ze$).

1. By balancing forces, find the speed v of the electron in terms of its orbital radius.
2. Show that the total energy of the electron is equal to the negative of its kinetic energy. You may assume that its potential energy $U$ is given by (where $r$ is the radius of the orbit) $$ U = -\frac{Ze^2}{4\pi \epsilon_0 r}$$
3. Assuming that for the electron the product $mvr = n \hslash$, where $n$ is an integer and $\hbar$ (pronounced h-bar) is a constant, find an expression for the electron energy in terms of $n$ (and which does not depend on either $v$ or $r$).
4. If $E(n = 1) = -13.6 \ {\rm eV}$ for hydrogen, what is $E(n = 3)$ for once-ionised helium (${\rm He}^+$)?

The electrostatic force of attraction $$ F_e = \frac{Ze^2}{4 \pi \epsilon_0 r^2} $$ is balanced by the centripetal force $$ F_c = \frac{m v^2}{r}. $$ Therefore $$ v = {\frac{Ze^2}{4 \pi \epsilon_0 m r}}, \; mv^2 = \frac{Ze^2}{4\pi\epsilon_0 r}. $$ The kinetic energy of the electon is $$ KE = \half m v^2= \frac{Ze^2}{8\pi\epsilon_0 r} ; \tag{I} $$ since the potential energy is $$ U = -\frac{Ze^2}{4\pi \epsilon_0 r} $$ the total energy is $$ E = KE + U =\frac{Ze^2}{8\pi\epsilon_0 r} -\frac{Ze^2}{4\pi \epsilon_0 r} =-\frac{Ze^2}{8 \pi \epsilon_0 r}= -KE. $$ Use $mvr=n \hslash$ and $$ mv^2r=\frac{Ze^2}{4 \pi \epsilon_0 } $$ to write $$ n \hslash v = \frac{Ze^2}{4 \pi \epsilon_0 }, \; v = \frac{Ze^2}{4 \pi n \hslash \epsilon_0 }. $$ Using this expression for $v$ $$ E = -\half m v^2 = -\half m \qty({\frac{Ze^2}{4 \pi n \epsilon_0 }})^2 = -\frac{mZ^2e^4}{32\pi^2n^2\hslash^2 \epsilon_0^2n^2}. $$ Since $$ E(n=1)=-13.6 \ {\rm eV} , \; \frac{mZ^2e^4}{32\pi^2\hslash^2 \epsilon_0^2}=13.6 \ {\rm eV} $$ and ${\rm He}^+$ has one electron and two protons $$ E_{{\rm He}^+}(n=3) = -\half m \qty({\frac{2Ze^2}{12 \pi \epsilon_0 m r}})^2 = - \frac{4}{9} \frac{mZ^2e^4}{32\pi^2\hslash^2 \epsilon_0^2n^2}=-6.04 \ {\rm eV}. $$

20. Two unbiased dice are rolled. The numbers obtained are multiplied.

1. What is the probability that the product is even?
2. Which product has a probability of $\frac{1}{12}$ to occur?
3. What is the probability that the product is greater than 28?
4. Which product(s) has(ve) the highest probability to occur?
5. If the product is known to be even, what is the probability that it is also divisible by 4?

Since $\Pr({\rm even})+\Pr({\rm odd})=1$ and $$\def\oneq{\frac{1}{4}} \Pr({\rm odd})=\Pr(\text{odd}\land{\color{red}\rm odd})=\half \times \half = \oneq . $$ we have $\Pr({\rm even})=\frac{3}{4}$ (we distinguish the two dice by using a black and a ${\color{red}\rm red}$ color).

Each dice combination has probability $1/36$ (remember that the order is important; $\def\r#1{{\color{red} #1}}(2,\r{5}) \neq (5,\r{2})$). To get a probability of $1/12$ we need a product that is the result of 3 different dice combinations. From the commutative property of integer multiplication, $\cdot$, any number that is the product of two dice throws $n$ and $\r{m}$ is the result of a (minimum) of 2 dice combinations since $$ (n,\r{m}) \neq (m,\r{n}) , \; n \cdot \r{m} = m \cdot\r{n}. $$ To get an odd number like 3 we must use the property $$ (n,\r{n}) = (n,\r{n}) , \; n \cdot\r{n} = n \cdot\r{n}, $$ i.e. the product must be also the square of an integer. With two dice we could have $$ 4 = 1 \cdot \r{4} = 4 \cdot \r{1} = 2 \cdot \r{2} $$ which is 3 combinations. With 9 we cannot write a similar expression while 16 is out of range with six sided dice.

To get a product greater than 28 note that conditional on the black dice roll $n$ we must have $\r{m}>28/n$. Since $m \leq 6$ we must have $n \geq 28/6 = 4.\dot{6}$. This is possible for $n=5,6$. We can use $(5,\r{6}),(6,\r{5}),(6,\r{6})$ and so $\Pr(nm>28)=1/12$.

For (d): $$ \begin{array}{|c|c|c|c|c|c|c|} \hline & \textbf{1} & \textbf{2} & \textbf{3} & \textbf{4} & \textbf{5} & \textbf{6} \\ \hline \textbf{1} & 1 & 2 & 3 & 4 & 5 & \cellcolor{gray}6 \\ \hline \textbf{2} & 2 & 4 & \cellcolor{gray}6 & 8 & 10 & \cellcolor{lightgray}12 \\ \hline \textbf{3} & 3 & \cellcolor{gray}6 & 9 & \cellcolor{lightgray}12 & 15 & 18 \\ \hline \textbf{4} & 4 & 8 & \cellcolor{lightgray}12 & 16 & 20 & 24\\ \hline \textbf{5} & 5 & 10 & 15 & 20 & 25 & 30 \\ \hline \textbf{6} & \cellcolor{gray}6 & \cellcolor{lightgray}12 & 18 & 24 & 30 & 36 \\ \hline \end{array} $$

For (e) note that the probability that a product is even is $$ \begin{array}{|c|c|c|c|c|c|c|} \hline & \textbf{1} & \textbf{2} & \textbf{3} & \textbf{4} & \textbf{5} & \textbf{6} \\ \hline \textbf{1} & 1 & \cellcolor{lightgray}2 & 3 & \cellcolor{lightgray}4 & 5 & \cellcolor{lightgray}6 \\ \hline \textbf{2} & \cellcolor{lightgray}2 & \cellcolor{lightgray}4 & \cellcolor{lightgray}6 & \cellcolor{lightgray}8 & \cellcolor{lightgray}10 & \cellcolor{lightgray}12 \\ \hline \textbf{3} & 3 & \cellcolor{lightgray}6 & 9 & \cellcolor{lightgray}12 & 15 & \cellcolor{lightgray}18 \\ \hline \textbf{4} & \cellcolor{lightgray}4 & \cellcolor{lightgray}8 & \cellcolor{lightgray}12 & \cellcolor{lightgray}16 & \cellcolor{lightgray}20 & \cellcolor{lightgray}24\\ \hline \textbf{5} & 5 & \cellcolor{lightgray}10 & 15 & \cellcolor{lightgray}20 & 25 & \cellcolor{lightgray}30 \\ \hline \textbf{6} & \cellcolor{lightgray}6 & \cellcolor{lightgray}12 & \cellcolor{lightgray}18 & \cellcolor{lightgray}24 & \cellcolor{lightgray}30 & \cellcolor{lightgray}36 \\ \hline \end{array} $$ ($27/36=3/4$ as shown above) and the probability that it is divisible by 4 is $$ \begin{array}{|c|c|c|c|c|c|c|} \hline & \textbf{1} & \textbf{2} & \textbf{3} & \textbf{4} & \textbf{5} & \textbf{6} \\ \hline \textbf{1} & 1 & 2 & 3 & \cellcolor{lightgray}4 & 5 & 6 \\ \hline \textbf{2} & 2 & \cellcolor{lightgray}4 & 6 & \cellcolor{lightgray}8 & 10 & \cellcolor{lightgray}12 \\ \hline \textbf{3} & 3 & 6 & 9 & \cellcolor{lightgray}12 & 15 & 18 \\ \hline \textbf{4} & \cellcolor{lightgray}4 & \cellcolor{lightgray}8 & \cellcolor{lightgray}12 & \cellcolor{lightgray}16 & \cellcolor{lightgray}20 & \cellcolor{lightgray}24\\ \hline \textbf{5} & 5 & 10 & 15 & \cellcolor{lightgray}20 & 25 & 30 \\ \hline \textbf{6} & 6 & \cellcolor{lightgray}12 & 18 & \cellcolor{lightgray}24 & 30 & \cellcolor{lightgray}36 \\ \hline \end{array} $$ ($15/36$). Therefore the probability of this event is $15/27=5/9$.

21. A ball of mass $2m$ slides along a frictionless track with speed $u$. Starting from a long distance away, it collides elastically with a stationary ball of mass $m$.

1. Calculate the final speeds of both balls (you may neglect any rotation of the balls).
2. If both balls were now positively electrically charged, describe qualitatively either how the results would change or why you would leave the results unaltered.

From the conservation of momentum $$2mu=2mu'+mv ;$$ from the conservation of energy $$ \half 2m u^2 = \half 2m u'^2 + \half m v^2. \tag{I} $$ Since $u'=u-v/2$ \begin{align*} \half 2m u^2 & = \half 2m \qty(u- \frac{v}{2})^2 + \half m v^2 \\ v & =\frac{4}{3} u \\ u' & =\frac{u}{3}. \end{align*}

In the presence of an electric field we have to consider the repulsive force acting on the two, positively charged, balls. The force acting on the larger ball $\va{F}_{2m}=-\va{F}_m$, the force on the smaller ball; since $$\dv{\va{p}_{2m}}{t}=\va{F}_{2m}=-\va{F}_m=-\dv{\va{p}_m}{t}$$ we have $$ \dv{\qty(\va{p}_{2m}+\va{p}_m)}{t}=0$$ i.e. the total momentum of the system is conserved. The only difference is that now the balls will (eventually) move in opposite directions and so $u'$ has a minus sign.

22. Consider the following set of equations: \begin{align*} 2x+y & =z, \tag{I}\\ x^2 & = y, \tag{II}\\ z+2y & = 2x^3. \tag{III} \end{align*} Find the possible values of $x$ which satisfy these equations.

Substitute $z$ using $\text{(I)}$ in $\text{(III)}$: $$ 2x+3y=2x^3.$$ Next combine $\text{(II)}$ and $\text{(III)}$ to obtain $$ 2x+3x^2=2x^3 $$ which can be simplified to $$ x(2x^2-3x-2)=0, \; x(2x+1)(x-2)=0. $$ The solutions are $x=0,-1/2,2$.

23. The number of atoms $N_x$ in a sample of a radioactive substance $x$ decays with time according to the equation, $$ N_x(t)=N_x(0) e^{-\lambda_x t},$$ where $N_x(0)$ is the number of atoms at time $t=0$ and $\lambda_x$ is a constant for substance $x$.

The half-life of a substance is defined as the time taken for $N_x$ to reach half of its initial value. Substance $a$ has a half-life of 1 hour. 36% of its decays emit an alpha particle and 64% of its decays emit a beta particle. Substance $b$ has a half-life of 15 minutes. 56% of its decays emit an alpha particle and 44% of its decays emit a beta particle

If the total particle emission rate of substance $x$ (where $x = a, b$) is $\lambda_xN_x(t)$ and $N_a(0) = N_b(0)$, what time in minutes passes before the beta particle emission rates from the two samples are equal?

The two beta particle emission rates are equal when $$ r_a \lambda_a N_a(t) = r_b \lambda_b N_b(t) $$ where $[r_a]=[r_b]=1$ and $r_a=0.64$, $r_b=0.44$. Rewrite this as $$ r_a \lambda_a e^{-\lambda_a t} = r_b \lambda_b e^{-\lambda_b t}, \; t=\frac{\ln\qty(\frac{r_b\lambda_b}{r_a\lambda_a})}{\lambda_b-\lambda_a}. $$ From the equation for the half-life $t_x$, $$ \lambda_x = \frac{\ln 2}{t_x}; $$ therefore the last equation can be rewritten as $$ t = \frac{t_at_b}{\ln 2} \frac{\ln\qty(\frac{r_bt_a}{r_at_b})}{t_a-t_b} = \frac{60 \times 15}{\ln 2}\frac{\ln\qty(\frac{0.44\times 60}{0.64 \times 15})}{60-15} = \frac{20}{\ln 2} \ln\qty(\frac{11}{4})=29.19 \ {\rm min}. $$