Measurement, Vectors & Kinematics

Q1 [10]

(a) $a = \dfrac{dv}{dt}$ (definition of instantaneous acceleration) [0.5]. Constant $a$ ⇒ $\int_u^v dv = \int_0^t a\,dt$ [1] ⇒ $v-u = at$ ⇒ $v = u+at$ [1.5].

(b) $v = \dfrac{ds}{dt} = u+at$ [0.5]. $\int_0^s ds = \int_0^t (u+at)\,dt$ [1] ⇒ $s = ut + \tfrac12 at^2$ [1.5].

(c) From (a) $t = (v-u)/a$ [0.5]; substitute into (b): $s = u\frac{v-u}{a} + \tfrac12 a\frac{(v-u)^2}{a^2}$ [1] ⇒ simplify $2as = 2u(v-u)+(v-u)^2 = v^2-u^2$ ⇒ $v^2 = u^2+2as$ [1.5]. (Alt: chain rule $a = v\,dv/ds$.)

(d) Area under $v$–$t$ = displacement; slope of $x$–$t$ = instantaneous velocity. [1]

Q2 [12]

(a) Horizontal: $u_x = u\cos\theta$ (no acceleration). Vertical: $u_y = u\sin\theta$, $a_y=-g$ [1].

• Time of flight: $y=u\sin\theta\,t - \tfrac12 g t^2 = 0$ at landing ⇒ $T = \dfrac{2u\sin\theta}{g}$ [2].
• Max height: at top $v_y=0$: $H = \dfrac{u^2\sin^2\theta}{2g}$ [1.5].
• Range: $R = u\cos\theta\cdot T = \dfrac{u^2\sin 2\theta}{g}$ [1.5].

(b) $R$ max when $\sin2\theta=1 ⇒ 2\theta=90^\circ ⇒ \theta=45^\circ$ [1]; $R_{\max}=u^2/g$ [1].

(c) $u=20,\theta=30^\circ$: $\sin30=0.5,\cos30=0.866,\sin60=0.866$.

• $T = \dfrac{2(20)(0.5)}{9.8} = 2.04\ \text{s}$ [1.5]
• $H = \dfrac{400(0.25)}{19.6} = 5.10\ \text{m}$ [1.5]
• $R = \dfrac{400(0.866)}{9.8} = 35.3\ \text{m}$ [1]

Q3 [10]

(a) Assume $T = k\,\ell^a m^b g^c$ [1]. Dimensions: $[T]=T$, $[\ell]=L$, $[m]=M$, $[g]=LT^{-2}$. $T = L^a M^b (LT^{-2})^c = M^b L^{a+c} T^{-2c}$ [1]. Match: $M: b=0$; $T: -2c=1 ⇒ c=-\tfrac12$; $L: a+c=0 ⇒ a=\tfrac12$ [2]. So $T = k\sqrt{\ell/g}$ — independent of mass [1].

(b) $[u]=[v]=LT^{-1}$; $[at]=LT^{-2}\cdot T = LT^{-1}$. All terms $LT^{-1}$ ⇒ consistent [2].

(c) $A = 2.31\times1.4 = 3.234\ \text{m}^2$ [1]. Least sig figs = 2 (from 1.4) [1] ⇒ report $A = 3.2\ \text{m}^2$ [1].

Q4 [10]

(a) $|\vec A|=\sqrt{9+16+1}=\sqrt{26}\approx 5.10$ [1.5]; $\hat A = \dfrac{1}{\sqrt{26}}(3\hat i+4\hat j-\hat k)$ [1.5].

(b) $\vec A\cdot\vec B = 3(1)+4(-2)+(-1)(2)= 3-8-2 = -7$ [2]. $|\vec B|=\sqrt{1+4+4}=3$. $\cos\theta = \dfrac{-7}{3\sqrt{26}} = -0.4577$ [1] ⇒ $\theta \approx 117.2^\circ$ [1].

(c) $\vec A\times\vec B = \begin{vmatrix}\hat i&\hat j&\hat k\\3&4&-1\\1&-2&2\end{vmatrix}$ $=\hat i(4\cdot2 -(-1)(-2)) - \hat j(3\cdot2-(-1)(1)) + \hat k(3(-2)-4(1))$ $=\hat i(8-2) - \hat j(6+1)+\hat k(-6-4) = 6\hat i -7\hat j -10\hat k$ [2]. Physical: torque ($\vec r\times\vec F$) or area of parallelogram / angular momentum [1].

Q5 [10]

(a) Shortest time: head straight across (perpendicular to bank) so full $5\ \text{m/s}$ crosses [1]. $t = 200/5 = 40\ \text{s}$ [1.5]. Downstream drift $= 3\times40 = 120\ \text{m}$ [1.5].

(b) Shortest path (straight across): cross-stream component of boat velocity must cancel current. Head upstream at angle $\phi$ from the direct-crossing line such that $5\sin\phi = 3 ⇒ \sin\phi=0.6 ⇒ \phi=36.87^\circ$ upstream from perpendicular (i.e. $53.13^\circ$ to the bank) [2]. Effective cross speed $=5\cos\phi=5(0.8)=4\ \text{m/s}$ [1] ⇒ $t = 200/4 = 50\ \text{s}$ [1].

(c) Shortest-time maximises the cross-stream velocity component (all speed used to cross); shortest-path minimises distance by cancelling drift, at the cost of a longer time. [2]

Q6 [8]

(a) $\dfrac{\Delta P}{P} = \dfrac{\Delta a}{a} + 2\dfrac{\Delta b}{b} + \dfrac{\Delta c}{c}$ (powers add multiplied by exponent) [2] $= 1\% + 2(2\%) + 3\% = 8\%$ [3].

(b) Systematic: consistent bias in one direction (e.g. zero error of instrument / calibration); reduced by calibration/correction. Random: unpredictable scatter (e.g. reaction-time fluctuations in timing); reduced by averaging many readings. [3]

[
 {"claim":"Q2c T=2.04s, H=5.10m, R=35.3m for u=20 theta=30",
  "code":"import math\nu=20;th=math.radians(30);g=9.8\nT=2*u*math.sin(th)/g\nH=(u**2*math.sin(th)**2)/(2*g)\nR=(u**2*math.sin(2*th))/g\nresult = abs(T-2.04)<0.02 and abs(H-5.10)<0.05 and abs(R-35.3)<0.3"},
 {"claim":"Q4b dot=-7 and angle ~117.2 deg",
  "code":"import math\nA=[3,4,-1];B=[1,-2,2]\ndot=sum(a*b for a,b in zip(A,B))\nmA=math.sqrt(sum(x*x for x in A));mB=math.sqrt(sum(x*x for x in B))\nang=math.degrees(math.acos(dot/(mA*mB)))\nresult = dot==-7 and abs(ang-117.2)<0.3"},
 {"claim":"Q4c cross product = (6,-7,-10)",
  "code":"A=Matrix([3,4,-1]);B=Matrix([1,-2,2])\nc=A.cross(B)\nresult = list(c)==[6,-7,-10]"},
 {"claim":"Q5 shortest-time drift 120m t40s; shortest-path t50s",
  "code":"import math\nw=200;vc=3;vb=5\nt1=w/vb;drift=vc*t1\nphi=math.asin(vc/vb);t2=w/(vb*math.cos(phi))\nresult = t1==40 and drift==120 and abs(t2-50)<0.01"},
 {"claim":"Q6 max percent error = 8%",
  "code":"e=1+2*2+3\nresult = e==8"}
]