2007 Applied Mathematics Paper: Solution

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Solution to the 2007 Honors Applied Mathematics Paper.

Question 1

(a)
A particle is projected vertically downwards from the top of a tower with speed m/s. It 
takes the particle 4 seconds to reach the bottom of the tower.
During the third second of its motion the particle travels 29.9metres.
Find
 (i) the value of u
 (ii) the height of the tower.

(i)

Where $t$ is the time, consider the second between $t=2$ and $t=3$ .

$u_1=u+2g$

$s=29.9=(u+2g)1+\frac{g}{2}=u+\frac{5g}{2}$

$u=29.9-\frac{5g}{2}=5.4m/s$

(ii)

Let $u=5.4$ and $t=4$ , find $s$ .

$s=ut+\frac{1}{2}at^2$

$s=5.4\times4+8g=100m$

(b)
A train accelerates uniformly from rest to a speed m/s.
It continues at this speed for a period of time and then decelerates uniformly to rest.
In travelling a total distance  mnetres the train accelerates through a 
distance  metres and decelerates through a distance  metres, where  and .
(i) Draw a speed-time graph for the motion of the train.
(ii) If the average speed of the train for the whole journey is , 
find the value of .

(i)

(ii)

WARNING: Messy algebra follows (if anyone can show a neater way, please do)

first, define the middle distance, i.e., the distance travelled at full speed, let's call it $xd$ , with $x<1$ like q and p. This gives us some valuable equations:

$pd+xd+qd=d$	$pd=\frac{v}{2}\times t_1$	$qd=\frac{v}{2}\times t_3$	$xd=v\times t_2$
$p+x+q=1$	$t_1=\frac{2pd}{v}$	$t_3=\frac{2qd}{v}$	$t_2=\frac{xd}{v}$

now we use these values to get an equation for the average speed, and solve:

$\frac{d}{t}=\frac{d}{t_1+t_2+t_3} =\frac{d}{\frac{2pd}{v}+\frac{xd}{v}+\frac{2qd}{v}} =\frac{v}{2p+x+2q}$

$\frac{v}{p+q+b}=\frac{v}{2p+x+2q}$

$\frac{1}{p+q+b}=\frac{1}{2p+x+2q}$

$p+q+b=2p+x+2q$

$b=p+q+x=1$

also, as a quick check you could imagine there was in fact no acceleration or deceleration, which would mean both p and q are zero, and the average speed is $\frac{v}{b}$ , which must be v. This is a handy check, but I doubt it would be accepted as a method for finding b.

Question 2

(a)
Ship B is traveling west at 24 km/h. Ship A is travelling north at 32 km/h.
At a certain instant ship B is 8 km north-east of ship A.
(i)Find the velocity of ship A relative to ship B.
(ii)Calculate the length of time, to the nearest minute, for which the ships are less than or 
equal to 8 km apart.

$V_{AB}=V_A-V_B=32\vec{j}-(-24\vec{i})=24\vec{i}+32\vec{j}$

Draw a diagram, draw the triangle of side x, 8 and 8. Find the angle through use of slopes:

Slope of $24\vec{i}+32\vec{j}=\frac{32}{24}=\frac{4}{3}$

Angle with x-axis $=tan^{-1}\frac{4}{3}=53.13^o$

The angle of the line of length 8 we already know to be $45^o$ so the angle in our triangle is $53.13-45=8.13^o$ .

Sine rule: $\frac{x}{sin163.74}=\frac{8}{sin8.13}$

$x=\frac{8sin163.74}{sin8.13}=15.84m$

... and, at a velocity of $\sqrt{24^2+32^2}=40km/h$ , the time needed to cover this distance is: $\frac{15.84}{40}=0.396h=23.76min \rightarrow 24min$

(b)
A man can swim at 3 m/s in still water.
He swims across a river of width 30 metres.
He sets out at an angle of 30 to the bank.
The river flows with a constant speed of 5 m/s parallel to the straight banks.
In crossing the river he is carried downstream a distance  metres.
Find the value of  correct to two places of decimals.

First, find the time needed to cross the river:

$\frac{d}{v}=\frac{30}{3sin30}=\frac{30}{1.5}=20min$

Now, use this time to find the distance travelled in the other direction:

$vt=(5-3cos30)20=48.04m$

Question 3

(a)
A particle is projected with a speed of  m/s at an angle  to the horizontal.
Find the two values of  that will give a range of 12.5 m.

$u_j=7\sqrt{5}sin\alpha$

$s_j=7\sqrt{5}sin\alpha t - \frac{gt^2}{2}$

but for the range we put $s_j=0$ and $t\ne0$

$7\sqrt{5}sin\alpha - \frac{gt}{2}=0$

$\frac{gt}{2} = 7\sqrt{5}sin\alpha$

$t=\frac{2\times7\sqrt{5}sin\alpha}{g}=\frac{14\sqrt{5}sin\alpha}{g}$

$u_i=7\sqrt{5}cos\alpha$

$s_i=7\sqrt{5}cos\alpha t=7\sqrt{5}cos\alpha \times \frac{14\sqrt{5}sin\alpha}{g} =\frac{2sin\alpha cos\alpha \times 245}{g}=\frac{245sin2\alpha}{g}$

$\frac{245sin2\alpha}{g}=12.5$

$sin2\alpha=\frac{12.5g}{245}=\frac{1}{2}$

remember $0^o<\alpha<90^o$ but $0^o<2\alpha<180^o$

$2\alpha=30^o$ or $150^o$

$\alpha=15^o$ or $75^o$

(b)
A plane is inclined at an angle 45 to the horizontal. A particle is projected up the 
plane with initial speed  at an angle  to the horizontal.
The plane of projection is vertical and contains the line of greatest slope.
The particle is moving horizontally when it strikes the inclined plane.
Show that

$u_j=usin\theta$

$v_j=usin\theta-gt=0$ [because the movement is horizontal, no vertical velocity]

$gt=usin\theta$

$t=\frac{usin\theta}{g}$

$s_j=usin\theta t-\frac{gt^2}{2}=usin\theta\frac{usin\theta}{g}-\frac{g\left(\frac{usin\theta}{g}\right)^2}{2} =\frac{u^2sin^2\theta}{g}-\frac{gu^2sin^2\theta}{2g^2} =\frac{u^2sin^2\theta}{g}-\frac{u^2sin^2\theta}{2g} =\frac{u^2sin^2\theta}{2g}$

$s_i=ucos\theta t=ucos\theta\frac{usin\theta}{g} =\frac{u^2sin\theta cos\theta}{g}$

but, since the plane is inclined at $45^o$ , $s_i=s_j$ at impact.

$s_i=s_j$

$\frac{u^2sin\theta cos\theta}{g}=\frac{u^2sin^2\theta}{2g}$

$cos\theta=\frac{sin\theta}{2}$

$tan\theta=2$

Question 4

(a)
A particle slides down a rough plane inclined at 45 to the horizontal. The coefficient 
of friction between the particle and the plane is .
Find the time of descending a distance 4 metres from rest.

F downaward : $\frac{mg}{\sqrt{2}}-\frac{3mg}{4\sqrt{2}}=\frac{mg}{4\sqrt{2}}$

$F=ma$

$\frac{mg}{4\sqrt{2}}=ma$

$\frac{g}{4\sqrt{2}}=a$

when $s=4$ , $u=0$ and $a=\frac{g}{4\sqrt{2}} \qquad \qquad t=?$

$s=ut+\frac{1}{2}at^2$

$4=0t+\frac{1}{2}(\frac{g}{4\sqrt{2}})t^2$

$t^2=\frac{32\sqrt{2}}{g}$

$t=2.1min$

(b)
A light inextensible string passes over a small fixed pulley A, under a small moveable pulley B, of 
mass  kg, and then over a second small fixed pulley C.
A particle of mass 4 kg is attached to one end of the string and a particle of mass 6 kg is attached 
to the other end.
The system is released from rest.
(i) On separate diagrams show the forces acting on each particle and on the moveable pulley B.
(ii) Find, in terms of , the tension in the string.
(iii) If =9.6 kg prove that pulley B will remain at rest while the other two particles are in 
motion.

(i)

(ii)

$F=ma$	$F=ma$	$F=ma$
$T-4g=4a$	$mg-2T=m\frac{a-b}{2}$	$6g-T=6b$
$\frac{T-4g}{4}=a$		$\frac{6g-T}{6}=b$

$mg-2T=m\frac{a-b}{2}$

$mg-2T=m\frac{\frac{T-4g}{4}-\frac{6g-T}{6}}{2}$

$mg-2T=m\frac{5T-24g}{24}$

$24mg-48T=5mT-24mg$

$48T+5mT=24mg+24mg$

$T(48+5m)=48mg$

$T=\frac{48mg}{48+5m}$

(iii)

If the force is 0, then there will be no acceleration:

$mg-2T=9.6g-2g(\frac{48\times9.6}{5\times9.6+48}) =9.6g(1-2\frac{48}{96}) =9.6g(1-1) =9.6g(0) =0$

Question 5

(a)
A smooth sphere P, of mass 2 kg, moving with speed 9 m/s collides directly with a smooth sphere Q, 
of mass 3 kg, moving in the same direction with speed 4 m/s.
The coefficient of restitution between the spheres is .
(i) Find, in terms of , the speed of each sphere after the collision.
(ii) Show that the magnitude of the momentum transferred from one sphere to the other is .

(i)

$m_1u_1+m_2u_2=m_1v_1+m_2v_2$

$18+12=2v_1+3v_2$

$30=2v_1+3v_2$

$30-3v_2=2v_1$

$\frac{30-3v_2}{2}=v_1$

$-e=\frac{v_1-v_2}{u_1-u_2}$

$-e=\frac{v_1-v_2}{9-4}=\frac{\frac{30-3v_2}{2}-v_2}{5} =\frac{30-5v_2}{10}$

$-e=\frac{30-5v_2}{10}$

$-10e=30-5v_2$

$10e+30=5v_2$

$2e+6=v_2$

$v_1=\frac{30-3v_2}{2}=\frac{30-3(2e+6)}{2}=\frac{30-6e-18)}{2}=\frac{12-6e)}{2}=6-3e$

Speed of sphere P $=v_1=6-3e$ .

Speed of sphere Q $=v_2=2e+6$ .

(ii)

The magnitude of the momentum transferred, is the difference between the momentum before and the momentum after (for either sphere).

$=|2(9)-2(6-3e)|=|6+6e|=6(1+e)$

or $=|3(4)-3(2e+6)|=|-6-6e|=6(1+e)$

(b)
A smooth sphere A, of mass 4 kg, moving with speed m collides with a stationary smooth 
sphere B of mass 8 kg. The direction of motion of A, before impact makes an angle  
with the line of centres at impact.
The coefficient of restitution between the spheres is .
Find in terms of  and 
(i) the speed of each sphere after the collision
(ii) the angle through which the 4 kg sphere is deflected as a result of the collision
(iii) the loss in kinetic energy due to the collision.

(i)

$m_1u_1+m_2u_2=m_1v_1+m_2v_2$

$4ucos\alpha+8(0)=4v_1+8v_2$

$ucos\alpha=v_1+2v_2$

$ucos\alpha-2v_2=v_1$

$-e=\frac{v_1-v_2}{u_1-u_2}$

$-\frac{1}{2}=\frac{v_1-v_2}{ucos\alpha-0}$

$-\frac{1}{2}=\frac{ucos\alpha-2v_2-v_2}{ucos\alpha}$

$-\frac{1}{2}=\frac{ucos\alpha-3v_2}{ucos\alpha}$

$-ucos\alpha=2ucos\alpha-6v_2$

$3ucos\alpha=6v_2$

$\frac{ucos\alpha}{2}=v_2$

$ucos\alpha-2v_2=v_1$

$ucos\alpha-2\frac{ucos\alpha}{2}=v_1$

$v_1=ucos\alpha-ucos\alpha=0$

(ii)

the sphere A has no horizontal velocity, therefore it is travelling purely upward, therefore the angle of deflection is $90-\alpha$ .

(iii)

The loss in kinetic energy is $E_k=E_k before - E_k after$ .

$=\frac{1}{2}m_1u_1^2+0-0-\frac{1}{2}m_2v_2^2$

$=2u^2cos^2\alpha+0-0-4\frac{u^2cos^2\alpha}{4}$

$=2u^2cos^2\alpha-u^2cos^2\alpha$

$=u^2cos^2\alpha$

Question 6

(a)
A particle of mass  kg is suspended from a fixed point  by a light elastic string.
The extension of the string is  when the particle is in equilibrium.
The particle is then displaced vertically from the equilibrium position a distance not greater than  and is then released from rest.
(i)Show that the motion of the particle is simple harmonic.
(ii)Find, in terms of , the period of the motion.

(i)

Let the displacement be $x$ . By Hookes Law:

$F\ =\ -kx$

$ma\ =\ -kx$

$a\ =\ \frac{-kx}{m}$

Therefore the acceleration of the particle is proportional to its displacement and directed towards its' point of equilibrium. This is consistent with Simple Harmonic Motion.

(ii)

$F\ =\ -kd$

$mg\ =\ -kd$

$k\ =\ \frac{-mg}{d}$

From part (i):

$a\ =\ \frac{-kx}{m}$

$...\ a\ =\ \frac{xg}{d}$

$a =\ -\omega^{2}x$

$\frac{xg}{d}\ =\ -\omega^{2}x$

$\omega =\sqrt{\frac{g}{d}}$

$T =\ \frac{2\pi}{\omega}\ =\ \2\pi\sqrt{\frac{d}{g}}$

(b)
A bead slides on a smooth fixed circular hoop, of radius , in a vertical plane.
The bead is projected with speed  from the highest point .
It impinges upon and coalesces with another bead of equal mass at .
 is the vertical diameter of the hoop.
Show that the combined mass will not reach the point  in subsequent motion.

$\frac{1}{2}mv^{2}\ +\ mgh =\ Constant$

Compare position $c$ with position $d$ .

$\frac{1}{2}m(\sqrt{10gr})^{2}\ +\ mg(2r) =\ \frac{1}{2}mv^{2}\ +\ mg(0)$

$5grm\ +\ 2grm\ = \frac{mv^{2}}{2}$

$v\ =\ \sqrt{14gr}$

$...\ m_1u_1\ +\ m_2u_2\ =\ m_1v_1\ +\ m_2v_2$

Consider before and after the particles coalesce.

$m(\sqrt{14gr})\ +\ m(0)\ =\ 2mv_1$

$v_1\ =\ \frac{1}{2}\sqrt{14gr}$

$...\ \frac{1}{2}mv^{2}\ +\ mgh =\ Constant$

Find the subsequent height to which the combined mass rises.

$\frac{1}{2}(2m)(\frac{1}{2}\sqrt{14gr}) +\ (2m)g(0) =\ \frac{1}{2}(2m)v_2^{2}\ +\ (2m)gh$

$\frac{7gr}{2}\ =\ 2gh\ +\ v_2^{2}$

Maximum height is when the particle has a velocity of zero.

$<tex>\frac{7gr}{2}\ =\ 2gh$

$h\ =\ \frac{7r}{4}$

$\frac{7r}{4}\ <\ 2r\ ...\ Height\ of\ c$

The combined mass subsequently rises to a maximum height which is less than the height of c.

Therefore, the combined mass will not reach the point $c$ in subsequent motion (QED).

Question 7

(a)
 and  are two uniform rods, each of weight , freely hinged at .
 and 
The rods are in equilibrium in a vertical plane.
The ends  and  rest on a smooth horizontal plane and are connected by a light inextensible string of length .
Find the tension in the string.

(b)
A uniform disc of radius 25cm and mass 100 kg rests in a vertical plane perpendicular to a kerb stone 10 cm high.
A force F is applied to the disc at an angle  to the horizontal, where:

(i) Draw a diagram showing all the forces acting on the disc.
(ii) Find the least value of F required to raise the disc over the kerb stone.

(i)

(ii)

Let the system be in equilibrium.

Take moments about the point of contact between the two objects.

$100g(0.2)\ =\ S(0.2)\ +\ F(0.25)$

$4S\ =\ 400g\ -\ 5F$

The object leaves the ground when its' normal reaction with the ground is zero.

$...\ S\ =\ 0$

$...\ 400g\ -\ 5F\ =\ 0$

$400g\ =\ 5F$

$F = 784 N$

Question 9

(a)
A U-tube whose limbs are vertical and of equal length has mercury poured in until the level is 26.2 cm from the top in each limb.
Water is then poured into one limb until that limb is full.
The relative density of mercury is 13.6.
Find the length of the column of water added to the limb.

(b)
A uniform solid sphere is held completely immersed in 500 cm of water by means of a string tied to a point on the base of the container.
The tension in the string is 0.0784 N.
When 300  of another liquid of relative density 1.2 is added and thoroughly mixed with the water, the volume of the mixture is 800 cm and the tension in the string is 0.1078 N.
Find
(i) the relative density of the mixture
(ii) the mass of the sphere.

Question 10

(a)
Solve the differential equation

given that  when .

$\frac{dy}{dx}\ =\ y^2Sinx$

$\frac{dy}{y^2}\ =\ Sinx\ dx$

$\int \frac{dy}{y^2}\ =\int Sinx\ dx$

$-\frac{1}{y}\ =\ -Cosx\ +\ C$

Using our initial values we find that the constant of integration $C$ is $-1$ .

$-\frac{1}{y}\ =\ -Cosx\ -\ 1$

$\frac{1}{y}\ =\ Cosx\ +\ 1$

$y\ =\ \frac{1}{Cosx + 1}$

(b)
The acceleration of a racing car at a speed of  m/s is

The car starts from rest.
Calculate correct to two decimal places
(i) the speed of the car when it has travelled 1500m from rest
(ii) the maximum speed of the car.

(i)

We require an equation with speed and distance.

Let the distance traveled be $x$ and the speed be $v$

Letting $a$ be the acceleration, let:

$a\ =\ v\frac{dv}{dx}$

$v\frac{dv}{dx}\ =\ 1-\frac{v^2}{3200}$

$v\frac{dv}{dx}\ =\ \frac{3200-v^2}{3200}$

$\frac{v}{3200-v^2}\ dv\ =\ \frac{dx}{3200}$

$\int \frac{v}{3200-v^2}\ dv\ =\int \frac{dx}{3200}$

To integrate the left-hand side use a basic substitution function.

$-\frac{1}{2}ln\left(3200-v^2\right)\ =\ \frac{x}{3200}\ +\ C$

Using our initial values - that $v=0$ when $x=0$ - we find that:

$C\ =\ -\frac{1}{2}ln\left(3200\right)$ .

$-\frac{1}{2}ln\left(3200-v^2\right)\ =\ \frac{x}{3200}\ -\ \frac{1}{2}ln\left(3200\right)$

Find the speed when the distance traveled is 1500m.

$-\frac{1}{2}ln\left(3200-v^2\right)\ =\ \frac{1500}{3200}\ -\ \frac{1}{2}ln\left(3200\right)$

$ln\left(3200-v^2\right)\ =\ ln\left(3200\right)\ -\ \frac{15}{16}$

$ln\left(3200-v^2\right)\ =\ ln\left(\frac{3200}{e^{\frac{15}{16}}}\right)$

$3200\ -\ v^2\ =\ 1253.138$

$v^2\ =\ 1946.861$

$v\ =\ 44.123\ m/s$

(ii)

To determine the maximum speed of the car (also known as the limiting speed), we determine the speed when the time tends to infinity. However, since our equation from part (i) deals with distance traveled and speed, we will let the distance tend to infinity, since in the case of this question, this will yield the same results as letting the time tend to infinity.

$-\frac{1}{2}ln\left(3200-v^2\right)\ =\ \frac{x}{3200}\ -\ \frac{1}{2}ln\left(3200\right)$

$ln\left(3200-v^2\right)\ =\ ln\left(3200\right)\ -\ \frac{x}{1600}$