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Mock Exams

HL Mock Exam A

Question 1

(a) Consider the following second order difference equation:

\begin{align}T_{n+2}-4T_{n+1} -5 T_n =C(n)\end{align}

where $$T_1=1$$, $$T_2=59$$ and $$C(n)$$ is a function of $$n\in\mathbb{N}$$.

(i) Solve this equation when $$C(n)=0$$.

(ii) Using your result from part (i), solve this equation instead when $$C(n)=8n+18$$.

(i) $$T_n=9(-1)^n+2(5^n)$$

(ii) $$T_n=\dfrac{43}{6}(-1)^n+\dfrac{67}{30}(5^n)-n-2$$

(i)

Characteristic Equation

\begin{align}x^2-4x-5=0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}(x+1)(x-5)=0\end{align}

\begin{align}x=-1\mbox{ or }x=5\end{align}

\begin{align}T_n=p(-1)^n+q(5^n)\end{align}

Inserting that $$T_1=1$$ gives:

\begin{align}p(-1)^1+q(5^1)=1\end{align}

\begin{align}\downarrow\end{align}

\begin{align}-p+5q=1\end{align}

Inserting that $$T_2=59$$ gives:

\begin{align}p(-1)^2+q(5^2)=59\end{align}

\begin{align}\downarrow\end{align}

\begin{align}p+25q=59\end{align}

We therefore have the following two equations with two knowns:

\begin{align}-p+5q=1\end{align}

\begin{align}p+25q=59\end{align}

\begin{align}\downarrow\end{align}

\begin{align}30q=60\end{align}

\begin{align}\downarrow\end{align}

\begin{align}q=2\end{align}

Inserting this into $$-p+5q=1$$ gives:

\begin{align}p&=5q-1\\&=5(2)-1\\&=9\end{align}

\begin{align}\downarrow\end{align}

\begin{align}T_n=9(-1)^n+2(5^n)\end{align}

(ii)

Particular Solution

\begin{align}T_n=rn+s\end{align}

and

\begin{align}T_{n+1} = r(n+1)+s\end{align}

and

\begin{align}T_{n+2} = r(n+2)+s\end{align}

Inserting these into the difference equation gives

\begin{align}r(n+2)+s-4[r(n+1)+s]-5(rn+s)=8n+18\end{align}

\begin{align}\downarrow\end{align}

\begin{align}rn+2r+s-4rn-4r-4s-5rn-5s=8n+18\end{align}

\begin{align}\downarrow\end{align}

\begin{align}(-8r)n+(-2r-8s)=8n+18\end{align}

\begin{align}\downarrow\end{align}

\begin{align}-8r=8\mbox{ and } -2r-8s=18\end{align}

\begin{align}\downarrow\end{align}

\begin{align}r=-1\mbox{ and }s=-2\end{align}

\begin{align}\downarrow\end{align}

\begin{align}T_n=-n-2\end{align}

\begin{align}T_n=p(-1)^n+q(5^n)-n-2\end{align}

Inserting that $$T_1=1$$ gives:

\begin{align}p(-1)^1+q(5^1)-1-2=1\end{align}

\begin{align}\downarrow\end{align}

\begin{align}-p+5q-3=1\end{align}

\begin{align}\downarrow\end{align}

\begin{align}-p+5q=4\end{align}

Inserting that $$T_2=59$$ gives:

\begin{align}p(-1)^2+q(5^2)-2-2=59\end{align}

\begin{align}\downarrow\end{align}

\begin{align}p+25q-4=59\end{align}

\begin{align}\downarrow\end{align}

\begin{align}p+25q=63\end{align}

We therefore have the following two equations with two knowns:

\begin{align}-p+5q=4\end{align}

\begin{align}p+25q=63\end{align}

\begin{align}\downarrow\end{align}

\begin{align}30q=67\end{align}

\begin{align}\downarrow\end{align}

\begin{align}q=\frac{67}{30}\end{align}

Inserting this into $$-p+5q=4$$ gives:

\begin{align}p&=5q-4\\&=5\left(\frac{67}{30}\right)-4\\&=\frac{67}{6}-4\\&=\frac{43}{6}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}T_n=\frac{43}{6}(-1)^n+\frac{67}{30}(5^n)-n-2\end{align}

(b) Newton’s Law of Cooling states the following:

$\,$

The rate of decrease of the temperature of an object of temperature $$T$$ in a colder room of temperature $$T_R$$ is proportional to the difference between both temperatures.

$\,$

(i) Write down a differential equation that correctly models this law.

$\,$

This law can also be modelled with the following difference equation:

\begin{align}T_{n+1}=T_n-k(T_n-T_R)\end{align}

where $$k$$ is a constant and $$T_n$$ is the temperature of the object in the $$n$$th minute ($$n\geq0$$ and $$n\in\mathbb{Z}$$).

(ii) Show that the solution to this equation is:

\begin{align}T_n = (T_0-T_R)(1-k)^n+T_R\end{align}

$\,$

A cup of hot coffee of temperature $$75^{\circ}\mbox{C}$$ is sitting on a table within a room of temperature $$20^{\circ}\mbox{C}$$.

After one minute, the coffee’s temperature has reduced to $$60^{\circ}\mbox{C}$$.

(iii) By modelling this scenario using the above difference equation, find the temperature of the coffee after a further three minutes.

(iv) Would you expect to get the same answer to part (iii) if you had instead modelled the scenario using the differential equation of part (i)? Explain why.

(i)

\begin{align}\frac{dT}{dt}=-k(T-T_R)\end{align}

where $$k>0$$.

($$k$$ and $$k<0$$ rather than $$-k$$ and $$k>0$$ is also acceptable.)

(iii) $$35.39^{\circ}\mbox{C}$$

(iv) No. The model with the differential equation assumes that the temperature is changing gradually over time. The model with the difference equation, on the other hand, assumes that the temperature is changing only at the end of every minute.

(i)

\begin{align}\frac{dT}{dt}=-k(T-T_R)\end{align}

where $$k>0$$.

($$k$$ and $$k<0$$ rather than $$-k$$ and $$k>0$$ is also acceptable.)

(ii)

\begin{align}T_{n+1}=T_n-k(T_n-T_R)\end{align}

\begin{align}\downarrow\end{align}

\begin{align}T_{n+1}=(1-k)T_n+kT_R\end{align}

A general first order difference equation

\begin{align}T_{n+1}=AT_n+B\end{align}

has the following solution:

\begin{align}T_n = T_0A^n +\frac{B(1-A^n)}{1-A} \end{align}

Comparing to the above, $$A=1-k$$ and $$B=kT_R$$.

Therefore, our equation has the following solution:

\begin{align}T_n &=T_0(1-k)^n + \frac{kT_R[1-(1-k)^n]}{1-(1-k)}\\&=T_0(1-k)^n+T_R[1-(1-k)^n]\\&=T_0(1-k)^n+T_R-T_R(1-k)^n\\&=(T_0-T_R)(1-k)^n+T_R\end{align}

as required.

(iii) We are told that $$T_0=75^{\circ}\mbox{C}$$ and $$T_R=20^{\circ}\mbox{C}$$.

\begin{align}T_n &=(75-20)(1-k)^n+20\\&=55(1-k)^n+20\end{align}

We are also told that $$T_1=60^{\circ}\mbox{C}$$.

\begin{align}55(1-k)^1+20=60\end{align}

\begin{align}\downarrow\end{align}

\begin{align}55(1-k)=40\end{align}

\begin{align}\downarrow\end{align}

\begin{align}55-55k=40\end{align}

\begin{align}\downarrow\end{align}

\begin{align}k&=\frac{55-40}{55}\\&=\frac{15}{55}\\&=\frac{3}{11}\end{align}

Therefore, the equation becomes

\begin{align}T_n &=55\left(1-\frac{3}{11}\right)^n+20\\&=55\left(\frac{8}{11}\right)^n+20\end{align}

After a further three minutes, i.e. after the fourth minute:

\begin{align}T_4 &=55\left(\frac{8}{11}\right)^4+20\\&\approx35.39^{\circ}\mbox{C}\end{align}

(iv) No. The model with the differential equation assumes that the temperature is changing gradually over time. The model with the difference equation, on the other hand, assumes that the temperature is changing only at the end of every minute.

Question 2

The network below shows five towns in addition to the lengths (in kilometres) of various roads that connect them.

(i) What is the total degree of this network?

(ii) Represent the information found within this network in the form of an adjacency matrix.

(iii) Represent the information found within this network in the form of a distance matrix.

(iv) Draw any two spanning trees of this network.

$\,$

Kruskal’s algorithm and Prim’s algorithm are both used to find the minimum spanning tree of a network.
However, only Prim’s algorithm can be used to find the minimum spanning tree if the network is given in the form of a matrix.

(v) State two other differences between Kruskal’s algorithm and Prim’s algorithm.

(vi) Using a suitable matrix, find the minimum spanning tree of the above network. Relevant supporting work must be shown.

(vii) Pipeline is to be laid underneath certain roads shown in this network, at a cost of €$$3{,}100$$ per kilometre, such that each town is connected to the new pipeline. What is the minimum amount of money that needs to be spent in order to achieve this goal?

(i) $$16$$

(ii) $$\left(\begin{array}{ccccc} 0 & 1 & 1 & 1 & 1 \\ 1 & 0 & 1 & 1 & 0\\ 1 & 1 & 0 & 0 & 1 \\ 1 & 1 & 0 & 0 & 1 \\ 1 & 0 & 1 & 1 & 0 \end{array}\right)$$

(iii)

$\begin{array} {|c|c|}\hline & A & B & C & D & E \\ \hline A & – & 11 &17 &23 &6 \\ \hline B & 11 &- & 20 & 5 &- \\ \hline C &17 & 20 &- &- & 18 \\ \hline D &23 & 5 &- &- & 12\\ \hline E &6 &- & 18 & 12 &- \\ \hline \end{array}$

(iv)

(Any two trees that contain all five nodes are considered correct.)

(v) Kruskal’s algorithm starts with an edge whereas Prim’s algorithm starts with a node. Kruskal’s algorithm is faster for sparse networks whereas Prim’s algorithm is faster for dense networks.

(vi)

(vii) €$$120{,}900$$

(i) The number of edges is $$8$$. Therefore, the total degree of the network is $$2\times 8=16$$ (hand-shaking lemma).

(ii)

\begin{align}\left(\begin{array}{ccccc}
0 & 1 & 1 & 1 & 1 \\ 1 & 0 & 1 & 1 & 0\\ 1 & 1 & 0 & 0 & 1 \\ 1 & 1 & 0 & 0 & 1 \\ 1 & 0 & 1 & 1 & 0
\end{array}\right)\end{align}

(iii)

$\begin{array} {|c|c|}\hline & A & B & C & D & E \\ \hline A & – & 11 &17 &23 &6 \\ \hline B & 11 &- & 20 & 5 &- \\ \hline C &17 & 20 &- &- & 18 \\ \hline D &23 & 5 &- &- & 12\\ \hline E &6 &- & 18 & 12 &- \\ \hline \end{array}$

(iv)

(Any two trees that contain all five nodes are considered correct.)

(v) Kruskal’s algorithm starts with an edge whereas Prim’s algorithm starts with a node. Kruskal’s algorithm is faster for sparse networks whereas Prim’s algorithm is faster for dense networks.

(vi)

Distance Matrix

$\begin{array} {|c|c|}\hline & A & B & C & D & E \\ \hline A & – & \xcancel{11} &\xcancel{17} &\xcancel{23} &\xcancel{6} \\ \hline B & \enclose{circle}{11} &- & \xcancel{20} & \xcancel{5} &- \\ \hline C &\enclose{circle}{17} & 20 &- &- & 18 \\ \hline D &\xcancel{23} & \enclose{circle}{5} &- &- & \xcancel{12}\\ \hline E &\enclose{circle}{6} &- & \xcancel{18} & \xcancel{12} &- \\ \hline \end{array}$

Minimum Spanning Tree

(vii) The total weight of the MST is $$5+11+6+17=39\mbox{ km}$$. The total cost is therefore $$39\times3{,}100=120{,}900$$ euro.

Question 3

(a) An object is launched vertically upwards. It travels $$x$$ metres in the first second and $$0.5x$$ metres in the next second.

By modelling this scenario in the absence of wind and air resistance:

(i) What is the value of $$x$$?

(ii) What is the object’s maximum height?

(i) $$19.6\mbox{ m}$$

(ii) $$30.625\mbox{ m}$$

(i)

First Second

\begin{align}u=U && a=-9.8 && s=x && t=1\end{align}

\begin{align}s=ut+\frac{1}{2}at^2\end{align}

\begin{align}\downarrow\end{align}

\begin{align}x=U(1)+\frac{1}{2}(-9.8)(1^2)\end{align}

\begin{align}\downarrow\end{align}

\begin{align}x=U-4.9\end{align}

Total Two Seconds

\begin{align}u=U && a=-9.8 && s=1.5x && t=2\end{align}

\begin{align}s=ut+\frac{1}{2}at^2\end{align}

\begin{align}\downarrow\end{align}

\begin{align}1.5x=U(2)+\frac{1}{2}(-9.8)(2^2)\end{align}

\begin{align}\downarrow\end{align}

\begin{align}1.5x=2U-19.6\end{align}

We therefore have the following two equations with two unknowns:

\begin{align}x=U-4.9\end{align}

\begin{align}1.5x=2U-19.6\end{align}

\begin{align}\downarrow\end{align}

\begin{align}2x=2U-9.8\end{align}

\begin{align}1.5x=2U-19.6\end{align}

\begin{align}\downarrow\end{align}

\begin{align}0.5x=9.8\end{align}

\begin{align}\downarrow\end{align}

\begin{align}x&=\frac{9.8}{0.5}\\&=19.6\mbox{ m}\end{align}

(ii)

\begin{align}x=U-4.9\end{align}

\begin{align}\downarrow\end{align}

\begin{align}U&=x+4.9\\&=19.6+4.9\\&=24.5\mbox{ m/s}\end{align}

Maximum Height

\begin{align}u=24.5 && a=-9.8 && v=0 &&s=?\end{align}

\begin{align}v^2=u^2+2as\end{align}

\begin{align}\downarrow\end{align}

\begin{align}0^2=24.5^2+2(-9.8)s\end{align}

\begin{align}\downarrow\end{align}

\begin{align}s&=-\frac{24.5^2}{2(-9.8)}\\&=30.625\mbox{ m}\end{align}

(b) An object that is projected in the absence of wind and air resistance is said to be undergoing uniform acceleration.

(i) Show that the range $$R$$ of such an object that is projected from horizontal ground with an initial speed $$u$$ and an angle of projection $$\alpha$$ is:

\begin{align}R = \frac{u^2 \sin 2\alpha}{g}\end{align}

(ii) Hence, or otherwise, show that this range is maximised when $$\alpha = 45^{\circ}$$.

Object $$A$$ and object $$B$$ are both projected with an initial speed of $$10\mbox{ m/s}$$.

Object $$A$$ is launched from a rugby field and object $$B$$ is launched from the floor of a warehouse whose ceiling is of height $$h$$.

(iii) Assuming each object’s acceleration is uniform, what is the minimum value of $$h$$ such that both objects still have the same maximum range?

(iii) $$\approx 2.55\mbox{ m}$$

(i)

Time of Flight

\begin{align} u_y = u\sin \alpha && a_y=-g && s_y=0 && t=\mathord{?} \end{align}

\begin{align}\downarrow\end{align}

$s_y=u_yt+\frac{1}{2}a_yt^2$

\begin{align}\downarrow\end{align}

\begin{align}0=u \sin \alpha t +\frac{1}{2}(-g)t^2\end{align}

\begin{align}\downarrow\end{align}

\begin{align}t\left(t-\frac{2 u \sin \alpha}{g}\right)=0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}t=\frac{2 u \sin \alpha}{g}\end{align}

Range

\begin{align} u_x = u\cos \alpha && a_x=0 && t=\frac{2u \sin \alpha}{g} && s_x=R \end{align}

\begin{align}\downarrow\end{align}

$s_x=u_xt+\frac{1}{2}a_xt^2$

\begin{align}\downarrow\end{align}

\begin{align}R &= u \cos A\left(\frac{2u\sin \alpha}{g}\right) + \frac{1}{2}(0)\left(\frac{2u\sin \alpha}{g}\right)^2\\&= \frac{2 u^2 \sin \alpha \cos \alpha}{g}\\&=\frac{u^2(2 \sin \alpha\cos \alpha)}{g}\\&=\frac{u^2 \sin 2 \alpha}{g}\end{align}

as required.

(ii) The range is a maximum when $$\sin2\alpha=1$$.

\begin{align}\sin2\alpha=1\end{align}

\begin{align}\downarrow\end{align}

\begin{align}2\alpha=\sin^{-1}(1)\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\alpha&=\frac{1}{2}\sin^{-1}(1)\\&=\frac{1}{2}(90^{\circ})\\&=45^{\circ}\end{align}

as required.

(iii)

Maximum Height

\begin{align} u_y = u\sin \alpha&& a_y=-g && v_y=0 && s_y=H\end{align}

\begin{align}\downarrow\end{align}

$v_y^2 = u_y^2+2a_ys_y$

\begin{align}\downarrow\end{align}

\begin{align}0^2 = u^2 \sin^2 \alpha+ 2(-g)H\end{align}

\begin{align}\downarrow\end{align}

\begin{align}H = \frac{u^2 \sin^2 \alpha}{2g}\end{align}

We are told that $$u=10$$ and $$\alpha=45^{\circ}$$.

\begin{align}H &= \frac{(10^2) (\sin^2 (45^{\circ}))}{2(9.8)}\\&\approx2.55\mbox{ m}\end{align}

Question 4

A car has a maximum acceleration rate $$a_1$$ and a maximum deceleration rate $$a_2$$. The driver wishes to reach a destination that is a distance $$d$$ away on a straight road which has a speed limit $$V$$.

The driver will begin and end his journey at rest and complete the trip in the fastest possible time.

(i) Represent this motion on a time-velocity graph.

(ii) Show that the fastest time that this trip can be completed in is

\begin{align}t = \frac{V}{2a_1} + \frac{V}{2a_2} + \frac{d}{V}\end{align}

$\,$

On German motorways, collectively referred to as the Autobahn, there is no speed limit $$V$$.

The same car is to complete a trip on such a motorway. The driver will again begin and end his journey at rest and complete the trip in the fastest possible time.

(iii) Represent this motion on a time-velocity graph.

(iv) Show that the fastest time that this trip can be completed in is instead

$t=\sqrt{\frac{2a_2d}{a_1(a_1+a_2)}}+\sqrt{\frac{2a_1d}{a_2(a_1+a_2)}}$

(v) Conduct a dimensional analysis of the equations derived in parts (ii) and (iv).

(i)

(iii)

(v) All of the terms in both equations have units of time and are therefore both dimensionally correct.

(i)

(ii)

First Stage

\begin{align} u = 0 && v=V && a = a_1 && s=s_1 && t=t_1 \end{align}

\begin{align}\downarrow\end{align}

$v=u+at$

\begin{align}\downarrow\end{align}

$V=0+a_1t_1$

\begin{align}\downarrow\end{align}

\begin{align}t_1 = \frac{V}{a_1}\end{align}

and

$s=\left(\frac{u+v}{2}\right)t$

\begin{align}\downarrow\end{align}

\begin{align}s_1 &= \left(\frac{0+V}{2}\right)(t_1)\\&=\frac{Vt_1}{2}\end{align}

Second Stage

\begin{align} u = v= V && a=0 && s=s_2 && t=t_2\end{align}

\begin{align}\downarrow\end{align}

$v=u+at$

\begin{align}\downarrow\end{align}

$V = V+(0)t_2$

and

$s=\left(\frac{u+v}{2}t\right)$

\begin{align}\downarrow\end{align}

\begin{align}s_2 &= \left(\frac{V+V}{2}\right)t_2\end{align}

\begin{align}\downarrow\end{align}

\begin{align}s_2 = Vt_2\end{align}

Third Stage

\begin{align} u = V && v=0 && a = -a_2 && s=s_3 && t=t_3 \end{align}

\begin{align}\downarrow\end{align}

$v=u+at$

\begin{align}\downarrow\end{align}

$0=V+(-a_2)t_3$

\begin{align}\downarrow\end{align}

\begin{align}t_3 = \frac{V}{a_2}\end{align}

and

$s=\left(\frac{u+v}{2}\right)t$

\begin{align}\downarrow\end{align}

\begin{align}s_3 &= \left(\frac{V+0}{2}\right)(t_3)\\&=\frac{Vt_3}{2}\end{align}

We also know that $$s_1+s_2+s_3=d$$. Therefore

$\frac{Vt_1}{2}+s_2+\frac{Vt_3}{2} = d$

\begin{align}\downarrow\end{align}

\begin{align}s_2 = d-\frac{Vt_1}{2}-\frac{Vt_3}{2}\end{align}

Therefore

\begin{align}t_2 &= \frac{s_2}{V}\\&= \frac{d}{V}-\frac{t_1}{2}-\frac{t_3}{2}\\&= \frac{d}{V}-\frac{1}{2}\left(\frac{V}{a_1}\right)-\frac{1}{2}\left(\frac{V}{a_2}\right)\\&= \frac{d}{V}-\frac{V}{2a_1}-\frac{V}{2a_2}\end{align}

Hence, the quickest possible duration of this trip is

\begin{align}t_1+t_2+t_3&=\frac{V}{a_1} +\left(\frac{d}{V}-\frac{V}{2a_1}-\frac{V}{2a_2}\right)+\frac{V}{a_2}\\&=\frac{V}{2a_1}+\frac{V}{2a_2}+\frac{d}{V}\end{align}

as required.

(iii)

(iv)

First Stage

\begin{align} u = 0 && v=V && a = a_1 && s=s_1 && t=t_1 \end{align}

\begin{align}\downarrow\end{align}

$v=u+at$

\begin{align}\downarrow\end{align}

$V=0+a_1t_1$

\begin{align}\downarrow\end{align}

\begin{align}V=a_1t_1\end{align}

and

$s=\left(\frac{u+v}{2}\right)t$

\begin{align}\downarrow\end{align}

\begin{align}s_1 &= \left(\frac{0+V}{2}\right)(t_1)\\&=\frac{Vt_1}{2}\end{align}

Second Stage

\begin{align} u = V && v=0 && a = -a_2 && s=s_2 && t=t_2 \end{align}

\begin{align}\downarrow\end{align}

$v=u+at$

\begin{align}\downarrow\end{align}

$0=V+(-a_2)t_2$

\begin{align}\downarrow\end{align}

\begin{align}V=a_2t_2\end{align}

and

$s=\left(\frac{u+v}{2}\right)t$

\begin{align}\downarrow\end{align}

\begin{align}s_2 &= \left(\frac{V+0}{2}\right)(t_2)\\&=\frac{Vt_2}{2}\end{align}

We therefore have the following five equations:

\begin{align}V=a_1t_1 && V=a_2t_2\end{align}

\begin{align}s_1=\frac{Vt_1}{2} && s_2 = \frac{Vt_2}{2}\end{align}

\begin{align}s_1+s_2 = d\end{align}

\begin{align}\downarrow\end{align}

\begin{align}a_1t_1=a_2t_2\end{align}

\begin{align}\frac{Vt_1}{2} + \frac{Vt_2}{2} = d\end{align}

\begin{align}\downarrow\end{align}

\begin{align}a_1t_1=a_2t_2\end{align}

\begin{align}\frac{a_1t_1^2}{2} + \frac{a_1t_1t_2}{2} = d\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\frac{a_1t_1^2}{2} + \frac{a_1t_1\left(\frac{a_1}{a_2}t_1\right)}{2} = d\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\frac{a_1t_1^2}{2} + \frac{a_1^2t_1^2}{2a_2} = d\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\frac{a_1a_2t_1^2}{2a_2} + \frac{a_1^2t_1^2}{2a_2} = d\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\frac{t_1^2}{2}\left(\frac{a_1a_2+a_1^2}{a_2}\right)=d\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\frac{t_1^2}{2}\left(\frac{a_1(a_1+a_2)}{a_2}\right)=d\end{align}

\begin{align}\downarrow\end{align}

\begin{align}t_1 = \sqrt{\frac{2a_2d}{a_1(a_1+a_2)}}\end{align}

and

\begin{align}t_2 &= \frac{a_1}{a_2}t_1\\&=\frac{a_1}{a_2}\sqrt{\frac{2a_2d}{a_1(a_1+a_2)}}\\&=\sqrt{\frac{a_1^2}{a_2^2}\frac{2a_2d}{a_1(a_1+a_2)}}\\&=\sqrt{\frac{2a_1d}{a_2(a_1+a_2)}}\end{align}

Therefore, the fastest time in which the trip could be completed is

\begin{align}t&=t_1+t_2 \\&= \sqrt{\frac{2a_2d}{a_1(a_1+a_2)}} + \sqrt{\frac{2a_1d}{a_2(a_1+a_2)}}\end{align}

as required.

(v)

$t=\frac{V}{2a_1}+\frac{V}{2a_2}+\frac{d}{V}$

\begin{align}
\begin{gathered}[t]
\frac{V}{2a_1}\\ \, \\ \Downarrow\\ \, \\\frac{\mbox{m/s}}{\mbox{m/s}^2}\\=\mbox{s}
\end{gathered}&&\begin{gathered}[t]
\frac{V}{2a_2}\\ \, \\ \Downarrow\\ \, \\\frac{\mbox{m/s}}{\mbox{m/s}^2}\\=\mbox{s}
\end{gathered}&& \begin{gathered}[t]
\frac{d}{V}\\ \, \\ \Downarrow\\ \, \\\frac{\mbox{m}}{\mbox{m/s}}\\=\mbox{s}\end{gathered}\end{align}

All terms have the same units as the left hand side, i.e. time. Therefore, this equation is dimensionally correct.

$\,$

\begin{align}t= \sqrt{\frac{2a_2d}{a_1(a_1+a_2)}} + \sqrt{\frac{2a_1d}{a_2(a_1+a_2)}}\end{align}

\begin{align}
\begin{gathered}[t]
\sqrt{\frac{2a_2d}{a_1(a_1+a_2)}}\\ \, \\ \Downarrow\\ \, \\\sqrt{\frac{(\mbox{m/s}^2)(\mbox{m})}{(\mbox{m/s}^2)(\mbox{m/s}^2+\mbox{m/s}^2)}}\\=\sqrt{\frac{(\mbox{m/s})^2}{(\mbox{m/s}^2)^2}}\\=\frac{\mbox{m/s}}{\mbox{m/s}^2}\\=\mbox{s}
\end{gathered}&& \begin{gathered}[t]
\sqrt{\frac{2a_1d}{a_2(a_1+a_2)}}\\ \, \\ \Downarrow\\ \, \\\sqrt{\frac{(\mbox{m/s}^2)(\mbox{m})}{(\mbox{m/s}^2)(\mbox{m/s}^2+\mbox{m/s}^2)}}\\=\sqrt{\frac{(\mbox{m/s})^2}{(\mbox{m/s}^2)^2}}\\=\frac{\mbox{m/s}}{\mbox{m/s}^2}\\=\mbox{s}
\end{gathered}\end{align}

All terms have the same units as the left hand side, i.e. time. Therefore, this equation is dimensionally correct.

Question 5

(a) An object is undergoing uniform circular motion, as shown in the image below.

(i) An object that is undergoing uniform linear motion is said to be moving with a constant acceleration. Is this also true in regards to uniform circular motion? Explain your reasoning.

(ii) Show that acceleration of an object undergoing uniform circular motion has a magnitude of $$\dfrac{v^2}{r}$$, where $$v$$ is the object’s linear speed and $$r$$ is the radius of the circular trajectory.

(i) No.

(i) No. The centripetal acceleration vector is constantly changing direction with time. Uniform circular motion instead implies that 1) the magnitude of the centripetal acceleration is constant and that 2) the tangential acceleration is zero.

(ii)

Initial Location $$\mathbf{t}$$ seconds later $\vec{s}(\theta) = -r \cos \theta \vec{i} +r \sin \theta \vec{j}$

Inserting $$\theta=\omega t$$ gives

$\vec{s}(t) = -r \cos (\omega t) \vec{i} +r \sin (\omega t) \vec{j}\tag{1}$

\begin{align}\downarrow\end{align}

\begin{align}\vec{v} &= \frac{d\vec{s}}{dt}\\&= -r[-\omega \sin (\omega t) \vec{i}] +r [\omega\cos (\omega t) \vec{j}]\\&=r\omega \sin (\omega t) \vec{i} +r\omega \cos (\omega t) \vec{j}\end{align}

and

\begin{align}\vec{a}_c &= \frac{d\vec{v}}{dt}\\&=r \omega[\omega \cos (\omega t) \vec{i}] +r \omega[-\omega \sin (\omega t) \vec{j}]\\&=-\omega^2 [-r\cos (\omega t) \vec{i} +r\sin (\omega t) \vec{j}]\\&=-\omega^2 \vec{s}\end{align}

where we have reinserted equation $$(1)$$. Therefore

\begin{align}|\vec{a}_c| &= |-\omega^2 \vec{s}|\\&=r\omega^2\\&r\left(\frac{v}{r}\right)^2\\&=\frac{v^2}{r} \end{align}

as required, where we have inserted that $$v=r\omega$$.

(b) An object of mass $$m$$ hangs freely from an inextensible, light string of length $$l$$.

The object is then pushed to the right, giving it an initial speed $$u$$ in the process.

The object reaches a point $$P$$ directly east of the centre of the circular trajectory.

The motion of this object can be suitably modelled by using a combination of Newton’s 2nd Law in addition to the Principle of Conservation of Energy.

(i) Using this model, show that the tension coefficient of the string at $$P$$ is

\begin{align}T = 2mg-\frac{mu^2}{l}\end{align}

(ii) If $$P$$ is the maximum displacement of the object from its initial position, show that $$u= \sqrt{2gl}$$.

(i) \begin{align} \vec{T} = T\vec{i} && \vec{W}=W\vec{j}\end{align}

and

\begin{align} \vec{a}= a_c\vec{i} + a_t\vec{j} \end{align}

\begin{align}\downarrow\end{align}

$T\vec{i}+W\vec{j}= m(a_c\vec{i}+a_t\vec{j})$

We know that $$r=l$$, $$W=-mg$$ and

\begin{align}|\vec{a_c}| &= \frac{|\vec{v}|^2}{r}\\&=\frac{|\vec{v}|^2}{l}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}a_c = -\frac{v^2}{l}\end{align}

Therefore

$T\vec{i}-mg\vec{j}= m\left(-\frac{v^2}{l}\vec{i}+a_t\vec{j}\right)$

\begin{align}\downarrow\end{align}

\begin{align}T = -\frac{mv^2}{l}\end{align}

Initial Energy

\begin{align}U_{i}&=mgh\\&=mg(0)\\&=0 \mbox{ J}\end{align}

and

\begin{align}K_i &= \frac{1}{2}mu^2\end{align}

Final Energy

\begin{align}U_f&=mgl\end{align}

and

\begin{align}K_f &= \frac{1}{2}mv^2\end{align}

\begin{align}U_i+K_i &=U_f+K_f \end{align}

\begin{align}\downarrow\end{align}

\begin{align}0+\frac{1}{2}mu^2 &=mgl+\frac{1}{2}mv^2\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\frac{1}{2}u^2 &=gl+\frac{1}{2}v^2\end{align}

\begin{align}\downarrow\end{align}

\begin{align}v^2=u^2-2gl\end{align}

Inserting this into our equation for the tension gives

\begin{align}T = -\frac{m(u^2-2gl)}{l}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}T = 2mg-\frac{mu^2}{l}\end{align}

as required.

(ii) Setting $$v=0$$ in the following equation

\begin{align}v^2=u^2-2gl\end{align}

gives

\begin{align}u=\pm \sqrt{2gl}\end{align}

As $$u$$ is pointing in the positive $$x$$ direction for standard $$x$$-$$y$$ axes, we take the positive root. Therefore, $$u=\sqrt{2gl}$$, as required.

Question 6

The activities of a particular project, and the order in which they can be completed, are represented by the following precedence table.

Activity Depends directly on...

$$A$$

-

$$B$$

-

$$C$$

$$A$$

$$D$$

$$A$$

$$E$$

$$B,C,D$$

$$F$$

$$B,D$$

(i) Construct the corresponding activity network of this project. In your diagram, highlight the location of both the sink and source node. All other nodes may remain unlabelled. All activities, on the other hand, must be labelled.

$\,$

The time, in hours, that it takes to complete each of these activities is shown below.

Activity Time (hours)

$$A$$

$$8$$

$$B$$

$$11$$

$$C$$

$$6$$

$$D$$

$$15$$

$$E$$

$$3$$

$$F$$

$$5$$

These values allow us to calculate both the early time and late time of each event in our activity network.

(ii) Define what is meant by either the early time or the late time of an event.

(iii) Using the critical path method, together with the table above, reconstruct the activity network of part (i). This time, in addition, label 1) the weight of each activity and 2) the early and late times of each event.

(iv) Is activity $$B$$ a critical activity? Explain your reasoning.

$\,$

The fastest time in which a project can possibly be completed is known as the critical time.

(v) If only one person works on this project, can the project be completed in the critical time? Explain your reasoning.

(i)

(Any network that is isomorphic to this network is considered correct.)

(ii) The early time of an event is the earliest time in which all of the preceding activities could be finished.

(Defining late time instead is also considered correct.)

(iii)

(iv) No.

(v) No.

(i)

(Any network that is isomorphic to this network is considered correct.)

(ii) The early time of an event is the earliest time in which all of the preceding activities could be finished.

(Defining late time instead is also considered correct.)

(iii)

(iv) No, as it has a non-zero float of $$23-(0+11)=12$$.

(v) The lower bound is:

\begin{align}\frac{8+11+6+15+3+5}{28}\approx1.7\end{align}

Therefore, we need at least two people (and possibly more) if we wish to complete this project in the critical time.

Question 7

(a) A particular system is composed of both an object and horizontal ground.

(i) Explain what is meant by both an open and a closed system.

$\,$

The object impacts the ground with a velocity of $$-5\vec{j}\mbox{ m/s}$$. The object loses $$40\%$$ of its energy each time it bounces.

(ii) A student states that we cannot consider this system as being closed as the object loses a large fraction of its energy on impact with the ground. Explain why this student’s reasoning is incorrect.

$\,$

Assuming that this system is indeed closed, find the maximum height of the object after:

(iii) the first bounce.

(iv) the second bounce.

(i) Energy is conserved within a closed system. Energy can instead be transferred from the system to the surroundings in an open system.

(ii) The system is composed of both the object and the ground. Therefore, if energy is transferred from the object to the ground, energy does not leave the system and the system can therefore still be considered closed.

(iii) $$0.76\mbox{ m}$$

(iv) $$0.46\mbox{ m}$$

(i) Energy is conserved within a closed system. Energy can instead be transferred from the system to the surroundings in an open system.

(ii) The system is composed of both the object and the ground. Therefore, if energy is transferred from the object to the ground, energy does not leave the system and the system can therefore still be considered closed.

(iii)

First Impact

\begin{align}E_i&=\frac{1}{2}mu^2\\&=\frac{1}{2}m(-5)^2\\&=12.5m\end{align}

\begin{align}\downarrow\end{align}

\begin{align}E_f&=0.6E_i\\&=(0.6)(12.5m)\\&=7.5m\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\frac{1}{2}mv^2=7.5m\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\frac{1}{2}v^2=7.5\end{align}

\begin{align}\downarrow\end{align}

\begin{align}v&=\pm \sqrt{2(7.5)}\\&\approx \pm 3.87 \mbox{ m/s}\end{align}

As the object will be moving in the positive $$y$$ direction after impact, we take the positive root, i.e. $$v=3.87$$.

After First Impact

$$\vec{u}=3.87\vec{j}\mbox{ m/s}$$

\begin{align}\downarrow\end{align}

\begin{align} u = 3.87 && a = -9.8 && v=0 && s=\mathord{?}\end{align}

\begin{align}\downarrow\end{align}

$v^2=u^2+2as$

\begin{align}0^2 = 3.87^2 + 2(-9.8)s\end{align}

\begin{align}\downarrow\end{align}

\begin{align}s&=-\frac{3.87^2}{2(-9.8)}\\&\approx 0.76 \mbox{ m}\end{align}

(iv) By symmetry, the object will land just before impact with a velocity of $$\vec{v}=-3.87\vec{i}\mbox{ m/s}$$.

##### Second Impact

$$\vec{u}=-3.87\vec{i}\mbox{ m/s}$$

\begin{align}E_i&=\frac{1}{2}mu^2\\&=\frac{1}{2}m(-3.87)^2\\&\approx 7.49m\end{align}

\begin{align}\downarrow\end{align}

\begin{align}E_f&=0.6E_i\\&=(0.6)(7.49m)\\&\approx 4.49m \mbox{ J}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\frac{1}{2}mv^2=4.49m\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\frac{1}{2}v^2=4.49\end{align}

\begin{align}\downarrow\end{align}

\begin{align}v&=\pm \sqrt{2(4.49)}\\&\approx \pm 3.00 \mbox{ m/s}\end{align}

As the object will be moving in the positive $$y$$ direction after impact, we take the positive root, i.e. $$v=3$$.

After Second Impact

$$\vec{u}=3\vec{i}\mbox{ m/s}$$

\begin{align}\downarrow\end{align}

\begin{align} u = 3 && a = -9.8 && v=0 && s=\mathord{?}\end{align}

\begin{align}\downarrow\end{align}

$v^2=u^2+2as$

\begin{align}\downarrow\end{align}

\begin{align}0^2 = 3^2 + 2(-9.8)s\end{align}

\begin{align}\downarrow\end{align}

\begin{align}s&=-\frac{3^2}{2(-9.8)}\\&\approx 0.46 \mbox{ m}\end{align}

(b) Sphere $$A$$, of mass $$m$$ and moving with a velocity of $$u\vec{i}$$ along a smooth surface, collides with sphere $$B$$ of mass $$2m$$ also moving with a speed $$u$$ in the direction shown.

Sphere $$A$$ comes to rest after the impact.

At the moment of impact, the line going through the centres of both spheres lies along the $$x$$ axis.

The coefficient of restitution between both spheres is $$\displaystyle\frac{4}{5}$$. (i) What is the angle $$\alpha$$?

(ii) What is the velocity of sphere $$B$$ after the collision in terms of $$u$$?

(i) $$9.59^{\circ}$$

(ii) $$\vec{v}_B = 0.33u\vec{i}+0.99u\vec{j}\mbox{ m/s}$$

##### Sphere A

$$m_A = m$$

Initial Velocity

$$\vec{u}_{Ax}=u \vec{i}$$

$$\vec{u}_{Ay}=0 \vec{j}\mbox{ m/s}$$

Final Velocity

$$\vec{v}_{Ax}=0\vec{i}\mbox{ m/s}$$

$$\vec{v}_{Ay}=0 \vec{j}\mbox{ m/s}$$

##### Sphere B

$$m_B = 2m$$

Initial Velocity

$$\vec{u}_{Bx}=-u\sin \alpha \vec{i}\mbox{ m/s}$$

$$\vec{u}_{By}=u \cos \alpha \vec{j}\mbox{ m/s}$$

Final Velocity

$$\vec{v}_{Bx}=v_{Bx} \vec{i}\mbox{ m/s}$$

$$\vec{v}_{By}=v_{By} \vec{j}$$

(i)

Conservation of Momentum

\begin{align}u \cos\alpha =v_{By}\end{align}

and

\begin{align}m_Au_{Ax}+m_Bu_{Bx} = m_Av_Ax+m_Bv_{Bx}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}mu+(2m)(-u\sin\alpha) = m(0)+(2m)(v_{Bx})\end{align}

\begin{align}\downarrow\end{align}

\begin{align}mu-2mu\sin\alpha = 2mv_{Bx}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}u-2u\sin\alpha = 2v_{Bx}\end{align}

Newton’s Law of Resitution

\begin{align}e=-\frac{v_{Ax}-v_{Bx}}{u_{Ax}-u_{Bx}}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\frac{4}{5}=-\frac{0-v_{Bx}}{u-(-u\sin \alpha)}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}4u+4u\sin\alpha = 5v_{Bx}\end{align}

We therefore have the following equations.

\begin{align}u \cos\alpha =v_{By}\end{align}

\begin{align}u-2u\sin\alpha = 2v_{Bx}\end{align}

\begin{align}4u+4u\sin\alpha = 5v_{Bx}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}u \cos\alpha =v_{By}\end{align}

\begin{align}5u-10u\sin\alpha = 10v_{Bx}\end{align}

\begin{align}8u+8u\sin\alpha = 10v_{Bx}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}u \cos\alpha =v_{By}\end{align}

\begin{align}5u-10u\sin \alpha = 8u+8u\sin\alpha\end{align}

\begin{align}\downarrow\end{align}

\begin{align}u \cos\alpha =v_{By}\end{align}

\begin{align}5-10\sin \alpha = 8+8\sin\alpha\end{align}

\begin{align}\downarrow\end{align}

\begin{align}u \cos\alpha =v_{By}\end{align}

\begin{align}18\sin\alpha =-3\end{align}

\begin{align}\downarrow\end{align}

\begin{align}u \cos\alpha =v_{By}\end{align}

\begin{align}\alpha =\sin^{-1}\left(-\frac{1}{6}\right)\end{align}

\begin{align}\downarrow\end{align}

\begin{align}u \cos\alpha =v_{By}\end{align}

\begin{align}\alpha \approx 9.59^{\circ}\end{align}

(ii) If we now insert this angle into our other equation, this gives

\begin{align}u \cos 9.59^{\circ} =v_{By}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}v_{By} \approx 0.99u\end{align}

and

\begin{align}u-2u\sin\alpha = 2v_{Bx}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}u-2u\sin 9.59^{\circ}= 2v_{Bx}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}v_{Bx}&=\left(\frac{1-2\sin 9.59^{\circ}}{2}\right)u\\&\approx 0.33u\end{align}

Hence, the velocity of sphere $$B$$ after the collision is

\begin{align}\vec{v}_B = 0.33u\vec{i}+0.99u\vec{j}\end{align}

Question 8

(a) The rate of increase of a small town’s population $$P$$ is inversely proportional to the population size itself.

(i) Write a differential equation which suitably models how the population $$P(t)$$ of that town changes with time $$t$$.

(ii) Show that the solution to the equation of part (i) is given by:

\begin{align}P(t) = \sqrt{2kt+P_0^2}\end{align}

where $$k\in\mathbb{R}$$ and $$P_0=P(0)$$.

The population of the town on March 1st and July 1st was $$100$$ and $$120$$ respectively.

(iii) What was the population of the town at the start of the year?

(i) $$\dfrac{dP}{dt} = \dfrac{k}{P}$$

(iii) $$88$$

(i) $\frac{dP}{dt} = \frac{k}{P}$

(ii) $\frac{dP}{dt} = \frac{k}{P}$

\begin{align}\downarrow\end{align}

\begin{align}P\, dP = k\, dt\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\int P\, dP = k\int  dt\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\frac{P^2}{2}+C = k(t+D)\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\frac{P^2}{2} = kt+E\end{align}

\begin{align}\downarrow\end{align}

\begin{align}P^2 = 2kt+F\end{align}

\begin{align}\downarrow\end{align}

\begin{align}P(t) = \sqrt{2kt+F}\end{align}

Since

\begin{align}P(0) &= \sqrt{2k(0)+F}\\&=\sqrt{F}\end{align}

is equal to $$P_0$$, we must therefore have that

\begin{align}\sqrt{F}= P_0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}F= P_0^2\end{align}

and therefore

\begin{align}P(t) = \sqrt{2kt+P_0^2}\end{align}

as required.

(iii)

Since

\begin{align}P(2) &= \sqrt{2k(2)+P_0^2}\\&=\sqrt{4k+P_0^2}\end{align}

is equal to $$100$$, we must therefore have that

\begin{align}\sqrt{4k+P_0^2}= 100\end{align}

Also, since

\begin{align}P(6) &= \sqrt{2k(6)+F}\\&=\sqrt{12k+F}\end{align}

is equal to $$120$$, we must therefore have that

\begin{align}\sqrt{12k+P_0^2} = 120\end{align}

We therefore have the following two equations for two unknowns.

\begin{align}\sqrt{4k+P_0^2}= 100\end{align}

\begin{align}\sqrt{12k+P_0^2} = 120\end{align}

\begin{align}\downarrow\end{align}

\begin{align}4k+P_0^2= 10{,}000\end{align}

\begin{align}12k+P_0^2 = 14{,}400\end{align}

\begin{align}\downarrow\end{align}

\begin{align}8k =4{,}400\end{align}

\begin{align}\downarrow\end{align}

\begin{align}k &= \frac{4{,}400}{8}\\&=550\end{align}

To find $$P_0$$, we can insert this value into e.g.

\begin{align}4k+P_0^2= 10{,}000\end{align}

\begin{align}\downarrow\end{align}

\begin{align}4(550)+P_0^2= 10{,}000\end{align}

\begin{align}\downarrow\end{align}

\begin{align}P_0&= \sqrt{10{,}000-4(550)}\\&=\sqrt{7{,}800}\\&\approx 88\end{align}

(b) An object of mass $$m_1$$ is placed on a smooth incline that is at an angle $$\theta$$ relative to the horizontal ground.

The object is attached to a hanging object of mass $$m_2$$, as shown below. (i) Show that the common acceleration rate of the system is

\begin{align}a=\left(\frac{m_2-m_1\sin \theta}{m_1+m_2}\right)g\end{align}

(ii) Show that the hanging object will only accelerate downwards when

\begin{align}\frac{m_2}{m_1}> \sin \theta \end{align}

(i)

Hanging Block \begin{align} \vec{T}_2 = T_2 \vec{j} && \vec{W}_2=W_2\vec{j}\end{align}

and

\begin{align} \vec{a}_2 = a_2\vec{j} \end{align}

\begin{align}\downarrow\end{align}

$T_2\vec{j}+W_2\vec{j} = m_2(a_2\vec{j})$

We know that $$W_2=-m_2g$$. Therefore

$T_2\vec{j}-m_2g\vec{j} = m_2(a_2\vec{i})$

\begin{align}\downarrow\end{align}

\begin{align}T_2-m_2g=m_2a_2\end{align}

Inclined Block \begin{align} \vec{W}_1 = W_{1x}\vec{i}+ W_{1y}\vec{j} && \vec{T}_1=T_1\vec{i} && \vec{R}_1 = R_1 \vec{j}\end{align}

and

\begin{align} \vec{a}_1 = a_1\vec{i} \end{align}

\begin{align}\downarrow\end{align}

$W_{1x}\vec{i} + W_{1y}\vec{j}+ T_1\vec{i}+ R_1\vec{j} = m_1(a_1\vec{i})$

We know that $$W_{1x}= -m_1g\sin \theta$$, $$W_{1y}=-m_1 g \cos \theta$$, $$T_1=T_2$$ and $$a_1=-a_2$$. Therefore

$-m_1 g \sin \theta\vec{i} + -m_1g \cos \theta\vec{j}+ T_2\vec{i}+ R_1\vec{j} = m_1(-a_2\vec{i})$

\begin{align}\downarrow\end{align}

$-m_1 g \sin \theta+ T_2 = -m_1 a_2$

We therefore have the following two equations with two unknowns

\begin{align}T_2-m_2g=m_2a_2\end{align}

$-m_1 g \sin \theta+ T_2 = -m_1 a_2$

\begin{align}\downarrow\end{align}

$-m_1 g \sin \theta+ (m_2a_2+m_2g) = -m_1 a_2$

\begin{align}\downarrow\end{align}

\begin{align}a_2 = -\left(\frac{m_2-m_1\sin \theta}{m_1+m_2}\right)g\end{align}

Therefore, the common acceleration rate of the system is

\begin{align}a = \left(\frac{m_2-m_1\sin \theta}{m_1+m_2}\right)g\end{align}

as required.

(ii)

\begin{align}a_2 = -\left(\frac{m_2-m_1\sin \theta}{m_1+m_2}\right)g\end{align}

This will only have a negative value when

\begin{align}m_2-m_1\sin \theta > 0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}m_2>m_1\sin \theta \end{align}

\begin{align}\downarrow\end{align}

\begin{align}\frac{m_2}{m_1}>\sin \theta \end{align}

as required.