Physics interleaved practice

1. (Tests 1.1.11 — Dot product / work) Work $W = \vec{F}\cdot\vec{d} = (3)(2)+(4)(-1) = 6-4 = 2\,\text{J}$. $|\vec{F}| = \sqrt{9+16}=5$, $|\vec{d}|=\sqrt{4+1}=\sqrt5$. $\cos\theta = \dfrac{2}{5\sqrt5} = 0.1789 \Rightarrow \theta \approx 79.7^\circ$. Why: Work asks for a scalar projection → dot product, not cross.

2. (Tests 1.1.3 — Dimensional analysis, checking) $[v^2] = (\text{LT}^{-1})^2 = \text{L}^2\text{T}^{-2}$. $[u^2]=\text{L}^2\text{T}^{-2}$. $[2as] = (\text{LT}^{-2})(\text{L}) = \text{L}^2\text{T}^{-2}$. All terms match → dimensionally consistent. Why: "Check equation" cues dimensional consistency, not solving.

3. (Tests 1.1.17 — Free fall / SUVAT) (a) At top $v=0$: $0 = 19.6 - 9.8t \Rightarrow t = 2\,\text{s}$. (b) $h = 19.6(2) - \tfrac12(9.8)(4) = 39.2 - 19.6 = 19.6\,\text{m}$. (c) By symmetry $v = -19.6\,\text{m/s}$ (downward, same magnitude). Why: Vertical single-axis motion with $g$ → free-fall SUVAT.

4. (Tests 1.1.3 — Dimensional analysis, deriving) $[T] = [L]^a[g]^b = \text{L}^a(\text{LT}^{-2})^b = \text{L}^{a+b}\text{T}^{-2b}$. Match: $\text{T}^1$: $-2b = 1 \Rightarrow b = -\tfrac12$. $\text{L}^0$: $a+b=0 \Rightarrow a=\tfrac12$. So $T \propto \sqrt{L/g}$. (Mass drops out — cannot appear.) Why: "Derive relation" cues dimensional method; mass has no way to enter.

5. (Tests 1.1.21 — River-boat relative motion) Heading straight across, crossing speed $=2\,\text{m/s}$. (a) $t = 60/2 = 30\,\text{s}$. (b) Drift $= 1 \times 30 = 30\,\text{m}$. (c) Ground speed $=\sqrt{2^2+1^2}=\sqrt5 \approx 2.24\,\text{m/s}$. Why: Perpendicular components are independent → cross-time uses only swim speed.

6. (Tests 1.1.4 — Significant figures) Area $= 4.52 \times 2.1 = 9.492 \to$ round to 2 sig figs (limited by $2.1$): $\boxed{9.5\,\text{cm}^2}$. Why: Multiplication → fewest sig figs of the factors governs.

7. (Tests 1.1.20 — Range/height/time of flight) $u_x = 20\cos30 = 17.32$, $u_y = 20\sin30 = 10\,\text{m/s}$. $T = \dfrac{2u_y}{g} = \dfrac{20}{9.8} = 2.04\,\text{s}$. $H = \dfrac{u_y^2}{2g} = \dfrac{100}{19.6} = 5.10\,\text{m}$. $R = u_x T = 17.32 \times 2.04 = 35.3\,\text{m}$. Why: Full projectile → apply derived formulas after resolving.

8. (Tests 1.1.12 — Cross product / area) $\vec{A}\times\vec{B} = \begin{vmatrix}\hat i&\hat j&\hat k\\2&3&-1\\1&-1&2\end{vmatrix}$ $= \hat i(3\cdot2 - (-1)(-1)) - \hat j(2\cdot2-(-1)(1)) + \hat k(2(-1)-3(1))$ $= \hat i(6-1) - \hat j(4+1) + \hat k(-2-3) = 5\hat i - 5\hat j - 5\hat k$. Area $= |\vec A\times\vec B| = \sqrt{25+25+25} = 5\sqrt3 \approx 8.66$ (units²). Why: Parallelogram area = magnitude of cross product, not dot.

9. (Tests 1.1.5 — Errors / relative error) $V = \tfrac43\pi r^3 \Rightarrow \dfrac{\Delta V}{V} = 3\dfrac{\Delta r}{r} = 3\times\dfrac{0.1}{5.0} = 0.06 = 6\%$. Why: Power in formula multiplies fractional error by the exponent.

10. (Tests 1.1.14 & 1.1.15 — Avg vs instantaneous velocity/accel) $\vec v = \dfrac{d\vec r}{dt} = 2t\,\hat i + 3\hat j$. At $t=2$: $\vec v = 4\hat i + 3\hat j$ (mag $5\,\text{m/s}$). (b) $\vec r(0) = 0$, $\vec r(2) = 4\hat i + 6\hat j$. Avg $\vec v = \dfrac{4\hat i+6\hat j}{2} = 2\hat i + 3\hat j\,\text{m/s}$. (c) $\vec a = \dfrac{d\vec v}{dt} = 2\hat i\,\text{m/s}^2$ (constant). Why: Instantaneous → differentiate; average → displacement/time.

[
  {"claim":"Q1 work = 2 J and angle approx 79.7 deg",
   "code":"import sympy as sp\nF=sp.Matrix([3,4]); d=sp.Matrix([2,-1])\nW=F.dot(d)\nctheta=W/(F.norm()*d.norm())\nang=sp.deg(sp.acos(ctheta))\nresult = (W==2) and abs(float(ang)-79.695)<0.1"},
  {"claim":"Q4 exponents a=1/2, b=-1/2",
   "code":"import sympy as sp\na,b=sp.symbols('a b')\nsol=sp.solve([a+b,-2*b-1],[a,b])\nresult = (sol[a]==sp.Rational(1,2)) and (sol[b]==-sp.Rational(1,2))"},
  {"claim":"Q8 cross product = (5,-5,-5), area = 5*sqrt3",
   "code":"import sympy as sp\nA=sp.Matrix([2,3,-1]); B=sp.Matrix([1,-1,2])\nc=A.cross(B)\nresult = (c==sp.Matrix([5,-5,-5])) and sp.simplify(c.norm()-5*sp.sqrt(3))==0"}
]