Level 5 — MasteryMeasurement, Vectors & Kinematics

Measurement, Vectors & Kinematics

90 minutes60 marksprintable — key stays hidden on paper

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 g=9.8 m/s2g = 9.8\ \text{m/s}^2.


Question 1 — Dimensional Modelling & Error Propagation (20 marks)

A student models the terminal speed of a small sphere of radius rr falling through a viscous fluid. She hypothesises that the drag force has the form Fd=kηarbvc,F_d = k\,\eta^{a}\, r^{b}\, v^{c}, where η\eta is the fluid's dynamic viscosity (SI units Pa⋅s=kg⋅m1s1\text{Pa·s} = \text{kg·m}^{-1}\text{s}^{-1}), vv is the speed, kk is a dimensionless constant.

(a) Using dimensional analysis, determine the exponents aa, bb, cc. State the resulting form of FdF_d. (6)

(b) At terminal velocity the drag balances the net downward force (ρsρf)43πr3g(\rho_s - \rho_f)\tfrac{4}{3}\pi r^3 g (sphere density ρs\rho_s, fluid density ρf\rho_f). Assuming k=6πk=6\pi, derive an explicit expression for the terminal speed vTv_T. (5)

(c) In an experiment the following are measured with uncertainties: r=(2.00±0.05) mmr = (2.00 \pm 0.05)\ \text{mm}, η=(1.00±0.02) Pa⋅s\eta = (1.00 \pm 0.02)\ \text{Pa·s}, and the density difference (ρsρf)=(1200±30) kg⋅m3(\rho_s-\rho_f) = (1200 \pm 30)\ \text{kg·m}^{-3}. Compute vTv_T 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 vTv_T. (3)


Question 2 — Vector Algebra, Torque & Work (18 marks)

Three points are given: A(1,0,2)A(1,\,0,\,2), B(3,2,1)B(3,\,2,\,-1), C(0,4,1)C(0,\,4,\,1) (metres).

(a) Find the unit vector along AB\vec{AB} and the angle BAC\angle BAC using the dot product. (6)

(b) A force F=(4i^3j^+2k^) N\vec{F} = (4\hat{i} - 3\hat{j} + 2\hat{k})\ \text{N} acts at BB. Taking AA as the pivot, compute the torque τ=r×F\vec{\tau} = \vec{r}\times\vec{F} (with r=AB\vec{r}=\vec{AB}) and its magnitude. Confirm your torque direction is consistent with the right-hand rule. (6)

(c) The same force F\vec F moves a particle along displacement AC\vec{AC}. Compute the work done, and hence the component of F\vec F along AC\vec{AC}. (3)

(d) Compute the area of triangle ABCABC using a cross product, and prove in general that 12AB×AC\tfrac12|\vec{AB}\times\vec{AC}| equals the triangle's area. (3)


Question 3 — Kinematics: Calculus, Projectiles & Relative Motion (22 marks)

(a) Starting from the definitions v=dxdtv=\dfrac{dx}{dt} and constant a=dvdta=\dfrac{dv}{dt}, derive by integration the two SUVAT relations v=u+atv = u + at and x=ut+12at2x = ut + \tfrac12 at^2. Then eliminate tt to obtain v2=u2+2axv^2 = u^2 + 2ax. (6)

(b) A projectile is launched from ground level with speed uu at angle θ\theta above the horizontal. From the independent horizontal/vertical equations, derive expressions for the time of flight TT, maximum height HH, and range RR. Prove that the range is maximised at θ=45\theta = 45^\circ. (6)

(c) A ball is thrown at u=25 m/su = 25\ \text{m/s}, θ=40\theta = 40^\circ. Compute TT, HH, RR, and the speed and direction of the velocity vector at t=1.5 st = 1.5\ \text{s}. (6)

(d) A river flows east at 3 m/s3\ \text{m/s}. A boat can move at 5 m/s5\ \text{m/s} relative to the water. The pilot wants to cross a 120 m120\ \text{m} 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: [Fd]=MLT2[F_d]=\text{MLT}^{-2}, [η]=ML1T1[\eta]=\text{ML}^{-1}\text{T}^{-1}, [r]=L[r]=\text{L}, [v]=LT1[v]=\text{LT}^{-1}. MLT2=(ML1T1)a(L)b(LT1)c.\text{MLT}^{-2} = (\text{ML}^{-1}\text{T}^{-1})^a (\text{L})^b (\text{LT}^{-1})^c. Match powers:

  • M: a=1a = 1 (1)
  • T: ac=2c=2a=1-a - c = -2 \Rightarrow c = 2 - a = 1 (2)
  • L: a+b+c=1b=1+ac=1+11=1-a + b + c = 1 \Rightarrow b = 1 + a - c = 1 + 1 - 1 = 1 (2)

So a=b=c=1a=b=c=1, giving Fd=kηrvF_d = k\,\eta r v (Stokes' law form). (1)[6]

(b) At terminal velocity: 6πηrvT=(ρsρf)43πr3g6\pi\eta r v_T = (\rho_s-\rho_f)\tfrac43\pi r^3 g. (2) vT=(ρsρf)43πr3g6πηr=2(ρsρf)gr29η.**(3)**v_T = \frac{(\rho_s-\rho_f)\tfrac43\pi r^3 g}{6\pi\eta r} = \frac{2(\rho_s-\rho_f) g\, r^2}{9\eta}. \quad \text{**(3)**}[5]

(c) Substitute r=2.00×103r=2.00\times10^{-3} m, η=1.00\eta=1.00, Δρ=1200\Delta\rho=1200, g=9.8g=9.8: vT=2(1200)(9.8)(2.00×103)29(1.00)=2(1200)(9.8)(4.0×106)9.v_T = \frac{2(1200)(9.8)(2.00\times10^{-3})^2}{9(1.00)} = \frac{2(1200)(9.8)(4.0\times10^{-6})}{9}. Numerator =2×1200×9.8×4.0×106=0.094080= 2\times1200\times9.8\times4.0\times10^{-6} = 0.094080. Divide by 9: vT=0.010453 m/sv_T = 0.010453\ \text{m/s}. (2)

Error propagation (vTΔρr2η1v_T\propto \Delta\rho\, r^2 \eta^{-1}):

= \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