Measurement, Vectors & Kinematics

Q1

(a) From $T = 2\pi\sqrt{L/g}$, square: $T^2 = 4\pi^2 L/g \Rightarrow g = \dfrac{4\pi^2 L}{T^2}$. (1 mark)

Period: $T = 96.5/50 = 1.93\ \text{s}$. (1 mark)

$g = \dfrac{4\pi^2 (0.925)}{(1.93)^2} = \dfrac{36.518}{3.7249} = 9.803\ \text{m/s}^2$. (1 mark)

(b) Fractional uncertainties:

• $\Delta L/L = 0.002/0.925 = 0.00216$
• Since $T = t_{total}/50$, $\Delta T/T = \Delta t/t = 0.5/96.5 = 0.00518$. (1 mark)

$g \propto L\, T^{-2}$, so $\dfrac{\Delta g}{g} = \dfrac{\Delta L}{L} + 2\dfrac{\Delta T}{T}$. (1 mark)

$= 0.00216 + 2(0.00518) = 0.00216 + 0.01036 = 0.01253 = 1.25\%$. (1 mark)

$\Delta g = 0.01253 \times 9.803 = 0.123\ \text{m/s}^2$. (1 mark)

Report: $g = (9.80 \pm 0.12)\ \text{m/s}^2$ (uncertainty to 2 s.f., value rounded to match). (1 mark)

Q2

(a) Resultant $\vec R = 10\hat i$. $\vec F_2 = \vec R - \vec F_1 = (10-3)\hat i + (0-4)\hat j = 7\hat i - 4\hat j\ \text{N}$. (3 marks) (1 setup, 2 components)

(b) $\vec F_1\cdot\vec F_2 = (3)(7)+(4)(-4) = 21-16 = 5$. (1) $|\vec F_1| = 5$, $|\vec F_2| = \sqrt{49+16}=\sqrt{65}=8.062$. (1) $\cos\theta = \dfrac{5}{5\times 8.062} = 0.1240 \Rightarrow \theta = 82.9^\circ$. (1)

(c) $\vec F_1\times\vec F_2 = (F_{1x}F_{2y}-F_{1y}F_{2x})\hat k = (3\cdot(-4) - 4\cdot 7)\hat k = (-12-28)\hat k = -40\hat k$. (1) $\tfrac12(\vec F_1\times\vec F_2) = -20\hat k$. (1) Direction: $-\hat k$ (into the page); by right-hand rule, curling from $\vec F_1$ to $\vec F_2$ (clockwise in the $xy$-plane) points into the page. Magnitude $20$ = area of triangle formed by the vectors. (1)

Q3

(a) To go straight across, the upstream component of boat velocity must cancel the current: $5\sin\theta = 3 \Rightarrow \sin\theta = 0.6 \Rightarrow \theta = 36.87^\circ$ upstream from the perpendicular...

Let angle to the bank be measured such that cross-component $= 5\cos\theta$ where $\theta$ is from perpendicular. Cleanest: heading angle upstream from straight-across is $\alpha$ with $\sin\alpha = 3/5$, $\alpha = 36.87^\circ$. (1) So heading makes $90-36.87 = 53.13^\circ$ with the bank. (1) Cross velocity $= 5\cos\alpha = 5(0.8) = 4\ \text{m/s}$. (1) Time $= 200/4 = 50\ \text{s}$. (1)

(b) Cross velocity $= 5\ \text{m/s}$ (full speed perpendicular). Time $= 200/5 = 40\ \text{s}$. (1) Drift $=$ current $\times$ time $= 3\times 40 = 120\ \text{m}$ downstream. (2)

(c) Strategy (b) is faster: $40\ \text{s}$ vs $50\ \text{s}$, shorter by $10\ \text{s}$ (but with drift). (2)

Q4

Components: $v_{0x} = 20\cos30 = 17.32\ \text{m/s}$, $v_{0y} = 20\sin30 = 10\ \text{m/s}$.

(a) Take up positive, origin at launch. $y = v_{0y}t - \tfrac12 g t^2 = -45$ at landing. $-45 = 10t - 4.9t^2 \Rightarrow 4.9t^2 - 10t - 45 = 0$. (2) $t = \dfrac{10 \pm \sqrt{100 + 4(4.9)(45)}}{2(4.9)} = \dfrac{10 \pm \sqrt{982}}{9.8} = \dfrac{10 \pm 31.34}{9.8}$. (1) Positive root: $t = 41.34/9.8 = 4.218\ \text{s}$. (1)

(b) $x = v_{0x}t = 17.32\times 4.218 = 73.06\ \text{m} \approx 73.1\ \text{m}$. (2)

(c) $v_x = 17.32\ \text{m/s}$; $v_y = v_{0y} - gt = 10 - 9.8(4.218) = 10 - 41.34 = -31.34\ \text{m/s}$. (2) Speed $= \sqrt{17.32^2 + 31.34^2} = \sqrt{299.9 + 982.2} = \sqrt{1282.1} = 35.81\ \text{m/s}$. (1) Direction: $\theta = \arctan(31.34/17.32) = 61.1^\circ$ below horizontal. (1)

(d) Max height above launch: $H = \dfrac{v_{0y}^2}{2g} = \dfrac{100}{19.6} = 5.10\ \text{m}$. (2)

Q5

(a) $a = dv/dt = 6 - 6t$. At $t=2$: $a = 6-12 = -6\ \text{m/s}^2$. (2)

(b) $x(t) = x_0 + \int_0^t (6t-3t^2)\,dt = 4 + 3t^2 - t^3$. (2) $x(3) = 4 + 27 - 27 = 4\ \text{m}$; $x(0)=4$. Displacement $= 4-4 = 0\ \text{m}$. (1)

(c) $v=0$ when $6t-3t^2 = 3t(2-t)=0 \Rightarrow t=0, 2$. Particle reverses at $t=2$. (1) $x(2) = 4 + 12 - 8 = 8\ \text{m}$; so from $0\to2$: $\Delta = 8-4 = +4\ \text{m}$. (1) From $2\to3$: $x(3)-x(2) = 4-8 = -4\ \text{m}$. (1) Total distance $= |4| + |-4| = 8\ \text{m}$. (1)

(d) $v$ has units m/s. In $6t$, $t$ has units s, so $6$ must have units $\dfrac{\text{m/s}}{\text{s}} = \text{m/s}^2$ = acceleration. ✓ (1)

[
  {"claim": "Q1: g = 4*pi^2*L/T^2 with L=0.925, T=1.93 gives ~9.803", "code": "import sympy as sp; g = 4*sp.pi**2*sp.Rational(925,1000)/sp.Rational(193,100)**2; result = abs(float(g) - 9.803) < 0.01"},
  {"claim": "Q1: percentage uncertainty ~1.25%", "code": "dL=0.002/0.925; dT=0.5/96.5; frac=dL+2*dT; result = abs(frac-0.01253) < 0.001"},
  {"claim": "Q2: F2 = 7i-4j and F1xF2 z-component = -40", "code": "F1=(3,4); F2=(7,-4); cross=F1[0]*F2[1]-F1[1]*F2[0]; result = cross == -40"},
  {"claim": "Q3: strategy b faster by 10s (40 vs 50)", "code": "import sympy as sp; ta=200/(5*sp.cos(sp.asin(sp.Rational(3,5)))); tb=200/5; result = abs(float(ta)-50)<0.01 and tb==40 and float(ta)-tb==10"},
  {"claim": "Q4: time of flight ~4.218s", "code": "import sympy as sp; t=sp.symbols('t',positive=True); sol=sp.solve(sp.Eq(10*t-sp.Rational(49,10)*t**2,-45),t); result = any(abs(float(s)-4.218)<0.01 for s in sol)"},
  {"claim": "Q5: total distance = 8m, displacement = 0", "code": "x=lambda t:4+3*t**2-t**3; disp=x(3)-x(0); dist=abs(x(2)-x(0))+abs(x(3)-x(2)); result = disp==0 and dist==8"}
]