Measurement, Vectors & Kinematics
Mastery Examination (Level 5)
Time limit: 90 minutes
Total marks: 60
Instructions: Answer all three questions. Show full derivations, justify sign conventions, and state where you use dimensional or vector reasoning. Calculator permitted; take .
Question 1 — Dimensional Modelling & Error Propagation (20 marks)
A student models the terminal speed of a small sphere of radius falling through a viscous fluid. She hypothesises that the drag force has the form where is the fluid's dynamic viscosity (SI units ), is the speed, is a dimensionless constant.
(a) Using dimensional analysis, determine the exponents , , . State the resulting form of . (6)
(b) At terminal velocity the drag balances the net downward force (sphere density , fluid density ). Assuming , derive an explicit expression for the terminal speed . (5)
(c) In an experiment the following are measured with uncertainties: , , and the density difference . Compute and its percentage uncertainty (propagate errors). Give the final answer to the correct number of significant figures. (6)
(d) State one systematic and one random error that could arise in this experiment, and explain how each would affect the measured . (3)
Question 2 — Vector Algebra, Torque & Work (18 marks)
Three points are given: , , (metres).
(a) Find the unit vector along and the angle using the dot product. (6)
(b) A force acts at . Taking as the pivot, compute the torque (with ) and its magnitude. Confirm your torque direction is consistent with the right-hand rule. (6)
(c) The same force moves a particle along displacement . Compute the work done, and hence the component of along . (3)
(d) Compute the area of triangle using a cross product, and prove in general that equals the triangle's area. (3)
Question 3 — Kinematics: Calculus, Projectiles & Relative Motion (22 marks)
(a) Starting from the definitions and constant , derive by integration the two SUVAT relations and . Then eliminate to obtain . (6)
(b) A projectile is launched from ground level with speed at angle above the horizontal. From the independent horizontal/vertical equations, derive expressions for the time of flight , maximum height , and range . Prove that the range is maximised at . (6)
(c) A ball is thrown at , . Compute , , , and the speed and direction of the velocity vector at . (6)
(d) A river flows east at . A boat can move at relative to the water. The pilot wants to cross a wide river (banks run east–west… i.e. river flows east, crossing is north) and land directly opposite the start. Determine the heading the pilot must steer (angle west of north), the boat's ground speed, and the crossing time. Use a Galilean velocity-addition argument. (4)
Answer keyMark scheme & solutions
Question 1
(a) Dimensions: , , , . Match powers:
- M: (1)
- T: (2)
- L: (2)
So , giving (Stokes' law form). (1) — [6]
(b) At terminal velocity: . (2) — [5]
(c) Substitute m, , , : Numerator . Divide by 9: . (2)
Error propagation ():
= \frac{30}{1200} + 2\cdot\frac{0.05}{2.00} + \frac{0.02}{1.00}.$$ $= 0.0250 + 0.0500 + 0.0200 = 0.0950 = 9.5\%.$ **(3)** So $v_T = 0.01045\ \text{m/s}$, uncertainty $\approx 9.5\%$ → $\Delta v_T \approx 0.00099 \approx 0.0010$ m/s. Final: $v_T = (1.05 \pm 0.10)\times10^{-2}\ \text{m/s}$ (3 s.f. from data). **(1)** — **[6]** **(d)** *Systematic:* fluid temperature drift lowers $\eta$ over time → $v_T$ systematically higher (consistent bias). *Random:* human reaction-time error in timing the fall → scatter in measured $v_T$, reduced by averaging repeats. **(3)** — **[20]** --- ## Question 2 **(a)** $\vec{AB} = B-A = (2,2,-3)$. $|\vec{AB}|=\sqrt{4+4+9}=\sqrt{17}\approx4.123$. *(2)* Unit vector $\hat{AB} = \tfrac{1}{\sqrt{17}}(2,2,-3)$. *(1)* $\vec{AC}=C-A=(-1,4,-1)$, $|\vec{AC}|=\sqrt{1+16+1}=\sqrt{18}=3\sqrt2$. $\vec{AB}\cdot\vec{AC} = (2)(-1)+(2)(4)+(-3)(-1) = -2+8+3 = 9$. *(2)* $\cos\angle BAC = \dfrac{9}{\sqrt{17}\cdot 3\sqrt2}=\dfrac{9}{3\sqrt{34}}=\dfrac{3}{\sqrt{34}}=0.5145$; $\angle BAC = 59.0^\circ$. **(1)** — **[6]** **(b)** $\vec\tau = \vec{AB}\times\vec F$, $\vec{AB}=(2,2,-3)$, $\vec F=(4,-3,2)$: $$\vec\tau = \begin{vmatrix}\hat i&\hat j&\hat k\\2&2&-3\\4&-3&2\end{vmatrix} = \hat i(2\cdot2 -(-3)(-3)) - \hat j(2\cdot2-(-3)(4)) + \hat k(2(-3)-2\cdot4).$$ $= \hat i(4-9) - \hat j(4+12) + \hat k(-6-8) = (-5,\,-16,\,-14)\ \text{N·m}.$ **(4)** $|\vec\tau| = \sqrt{25+256+196}=\sqrt{477}\approx 21.8\ \text{N·m}$. **(1)** Right-hand rule: fingers from $\vec{AB}$ curling to $\vec F$; the resulting vector points into the octant with all negative components — consistent with $\vec\tau$ found. **(1)** — **[6]** **(c)** $W = \vec F\cdot\vec{AC} = (4)(-1)+(-3)(4)+(2)(-1) = -4-12-2 = -18\ \text{J}$. **(2)** Component of $\vec F$ along $\vec{AC}$: $F_\parallel = W/|\vec{AC}| = -18/(3\sqrt2) = -3\sqrt2 = -4.24\ \text{N}$. **(1)** — **[3]** **(d)** $\vec{AB}\times\vec{AC}$: $\vec{AB}=(2,2,-3)$, $\vec{AC}=(-1,4,-1)$: $$=\hat i(2\cdot(-1)-(-3)(4)) - \hat j(2\cdot(-1)-(-3)(-1)) + \hat k(2\cdot4-2\cdot(-1)) =(10,\,5,\,10).$$ $|\vec{AB}\times\vec{AC}|=\sqrt{100+25+100}=\sqrt{225}=15$. Area $=\tfrac12(15)=7.5\ \text{m}^2$. **(2)** Proof: $|\vec{AB}\times\vec{AC}| = |\vec{AB}||\vec{AC}|\sin\theta$ = base×height of the parallelogram spanned by the two sides; the triangle is half that parallelogram, hence Area $=\tfrac12|\vec{AB}\times\vec{AC}|$. **(1)** — **[18]** --- ## Question 3 **(a)** From $a=dv/dt$ constant: $\int_u^v dv = \int_0^t a\,dt \Rightarrow v-u=at \Rightarrow v=u+at$. *(2)* Then $v=dx/dt=u+at$: $\int_0^x dx=\int_0^t(u+at)dt \Rightarrow x=ut+\tfrac12at^2$. *(2)* Eliminate $t=(v-u)/a$: $v^2 = u^2+2ax$ via $x=ut+\tfrac12at^2$ substitution → $2ax=2u(v-u)+(v-u)^2=v^2-u^2$. *(2)* — **[6]** **(b)** $u_x=u\cos\theta$, $u_y=u\sin\theta$, $a_y=-g$. Time of flight (returns to $y=0$): $0=u\sin\theta\,T-\tfrac12gT^2\Rightarrow T=\dfrac{2u\sin\theta}{g}$. *(2)* Max height (at $v_y=0$): $H=\dfrac{u^2\sin^2\theta}{2g}$. *(2)* Range: $R=u\cos\theta\cdot T=\dfrac{u^2\sin2\theta}{g}$. Max when $\sin2\theta=1\Rightarrow2\theta=90^\circ\Rightarrow\theta=45^\circ$. *(2)* — **[6]** **(c)** $u=25$, $\theta=40^\circ$: $u_x=25\cos40=19.15$, $u_y=25\sin40=16.07$ m/s. $T=\dfrac{2(16.07)}{9.8}=3.28\ \text{s}$. **(1)** $H=\dfrac{16.07^2}{2(9.8)}=13.18\ \text{m}$. **(1)** $R=\dfrac{25^2\sin80^\circ}{9.8}=\dfrac{625(0.9848)}{9.8}=62.8\ \text{m}$. **(2)** At $t=1.5$ s: $v_x=19.15$, $v_y=16.07-9.8(1.5)=16.07-14.7=1.37$ m/s. Speed $=\sqrt{19.15^2+1.37^2}=19.20\ \text{m/s}$; angle $=\arctan(1.37/19.15)=4.1^\circ$ above horizontal. **(2)** — **[6]** **(d)** To land directly opposite (net eastward velocity = 0), the boat's westward component must cancel the current: $5\sin\phi = 3 \Rightarrow \phi=\arcsin(0.6)=36.87^\circ$ west of north. **(2)** Ground speed (north component): $5\cos\phi=5(0.8)=4\ \text{m/s}$. **(1)** Crossing time $=120/4=30\ \text{s}$. Galilean addition: $\vec v_{boat,ground}=\vec v_{boat,water}+\vec v_{water,ground}$; choosing $\vec v_{boat,water}=(-3,4)$ gives ground velocity $(0,4)$ — purely northward. **(1)** — **[22]** --- ```verify [ {"claim":"Terminal speed v_T = 0.010453 m/s for given data","code":"r=2.00e-3; eta=1.00; drho=1200; g=Rational(98,10); vT=2*drho*g*r**2/(9*eta); result=abs(float(vT)-0.010453)<1e-5"}, {"claim":"Percentage uncertainty in v_T is 9.5%","code":"pct=Rational(30,1200)+2*Rational(5,200)+Rational(2,100); result=pct==Rational(95,1000)"}, {"claim":"Angle BAC has cos = 3/sqrt(34) approx 59 deg","code":"import sympy as sp;AB=Matrix([2,2,-3]);AC=Matrix([-1,4,-1]);c=(AB.dot(AC))/(sp.sqrt(AB.dot(AB))*sp.sqrt(AC.dot(AC)));result=sp.simplify(c-3/sqrt(34))==0"}, {"claim":"Torque r x F = (-5,-16,-14), magnitude sqrt(477)","code":"AB=Matrix([2,2,-3]);F=Matrix([4,-3,2]);t=AB.cross(F);result=(t==Matrix([-5,-16,-14])) and (t.dot(t)==477)"}, {"claim":"Triangle ABC area = 7.5","code":"AB=Matrix([2,2,-3]);AC=Matrix([-1,4,-1]);cr=AB.cross(AC);area=Rational(1,2)*sqrt(cr.dot(cr));result=area==Rational(15,2)"}, {"claim":"Projectile T=3.28s, R=62.8m for u=25,theta=40deg","code":"u=25;th=rad(40);g=9.8;T=2*u*sin(th)/g;R=u**2*sin(2*th)/g;result=abs(float(T)-3.279)<0.01 and abs(float(R)-62.8)<0.2"}, {"claim":"River crossing he