STEM Master Solver Guide
Module 01 Applied Mathematics
Worked Solutions Series
Quadratic Optimization & Extrema
Analytical modeling, vertex calculation, and critical point derivation with full verification.
Benchmark Problem
A parabolic arch bridge has height \(h(x) = -0.05x^2 + 3.2x + 4.8\) (meters), where \(x\) is distance from the left abutment. Calculate: (A) the maximum clearance height and horizontal location, and (B) the total horizontal span between ground footings (\(h = 0\)).
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Function Identification & Symmetry Analysis
Setup
\(a = -0.05\)
\(b = 3.2\)
\(c = 4.8\)
Since \(a < 0\), the parabola opens downward; the vertex is guaranteed to be a global maximum.
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Part (A): Maximum Height (Vertex Coordinates)
Extremum
Method 1: Axis of Symmetry
\(x_v = -\frac{b}{2a} = -\frac{3.2}{2(-0.05)} = \mathbf{32\text{ m}}\)
Method 2: Calculus Derivative
\(h'(x) = -0.10x + 3.2 = 0 \implies x = \mathbf{32\text{ m}}\)
Evaluate Maximum Height: \(h(32) = -0.05(32)^2 + 3.2(32) + 4.8 = -51.2 + 102.4 + 4.8 = \mathbf{56.0\text{ m}}\)
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Part (B): Footing Coordinates & Span Width
Roots
Set \(h(x) = 0\) and apply quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
\(\Delta = b^2 - 4ac = (3.2)^2 - 4(-0.05)(4.8) = 10.24 + 0.96 = 11.20 \implies \sqrt{11.20} \approx 3.3466\)
\(x = \frac{-3.2 \pm 3.3466}{-0.10} \implies x_1 = \mathbf{-1.47\text{ m}}, \quad x_2 = \mathbf{65.47\text{ m}}\)
Total Span: \(\Delta x = x_2 - x_1 = 65.47 - (-1.47) = \mathbf{66.94\text{ m}}\) (Terrain at continuous \(y=0\))
Sanity Verification
Midpoint between roots: \(\frac{-1.47 + 65.47}{2} = 32.00\text{ m}\). Perfectly matches vertex \(x_v = 32\text{ m}\), verifying symmetry.
Common Trap
Do not assume left footing is at \(x = 0\). At \(x=0\), \(h(0) = 4.8\text{ m}\) (abutment height), so the footing extends back to \(x = -1.47\text{ m}\).
STEM Problem Solver Master Series • Mathematics Page 1 of 3
Module 02 Classical Mechanics
Worked Solutions Series
Two-Dimensional Projectile Motion
Vector decomposition, decoupled kinematic equations, and impact state determination.
Benchmark Problem
A rescue canister is launched from a cliff \(y_0 = 45.0\text{ m}\) at speed \(v_0 = 25.0\text{ m/s}\) angled \(\theta = 35.0^\circ\) above horizontal (\(g = 9.80\text{ m/s}^2\)). Determine: (A) flight time \(t_{\text{land}}\), (B) horizontal impact range \(R\), and (C) impact velocity vector (magnitude & angle).
1
Vector Decomposition & Coordinate Frame
Vectors
Horizontal Component (Uniform Motion)
\(v_{0x} = v_0 \cos(35^\circ) = 25.0(0.8192) = \mathbf{20.48\text{ m/s}}\)
\(a_x = 0 \implies v_x(t) = v_{0x}\) (constant)
Vertical Component (Free Fall)
\(v_{0y} = v_0 \sin(35^\circ) = 25.0(0.5736) = \mathbf{14.34\text{ m/s}}\)
\(a_y = -g = -9.80\text{ m/s}^2\)
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Part (A): Total Flight Time (\(t_{\text{land}}\))
Kinematics
Vertical position equation with landing condition \(y(t) = 0\):
\(y(t) = y_0 + v_{0y}t - \tfrac{1}{2}gt^2 \implies 0 = 45.0 + 14.34t - 4.90t^2\)
Standard form: \(4.90t^2 - 14.34t - 45.0 = 0\)
\(t = \frac{14.34 \pm \sqrt{(-14.34)^2 - 4(4.90)(-45.0)}}{9.80} = \frac{14.34 \pm \sqrt{1087.64}}{9.80} = \frac{14.34 \pm 32.98}{9.80}\)
Rejecting negative root: \(t_{\text{land}} = \frac{14.34 + 32.98}{9.80} = \frac{47.32}{9.80} = \mathbf{4.83\text{ s}}\).
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Parts (B) & (C): Impact Range & Velocity Vector
Impact State
Horizontal Range:
\(R = v_{0x} \cdot t_{\text{land}} = (20.48)(4.83) = \mathbf{98.92\text{ m}}\)
Vertical Impact Velocity:
\(v_y = v_{0y} - gt = 14.34 - (9.80)(4.83) = \mathbf{-32.99\text{ m/s}}\)
Magnitude: \(v = \sqrt{v_x^2 + v_y^2} = \sqrt{(20.48)^2 + (-32.99)^2} = \mathbf{38.83\text{ m/s}}\)
Impact Angle: \(\theta = \arctan\left(\tfrac{-32.99}{20.48}\right) = \mathbf{-58.2^\circ}\)
Conservation of Energy Check
\(v_f = \sqrt{v_0^2 + 2gy_0} = \sqrt{25.0^2 + 2(9.80)(45.0)} = \sqrt{1507} \approx \mathbf{38.82\text{ m/s}}\). Matches kinematics precisely!
Formula Limitation Note
Never use \(R = \frac{v_0^2\sin(2\theta)}{g}\) when launch and landing heights differ (\(\Delta y \neq 0\)). It assumes level ground.
STEM Problem Solver Master Series • Physics Page 2 of 3
Module 03 Stoichiometry & Reactions
Worked Solutions Series
Limiting Reactants & Percent Yield
Dimensional analysis, BCA molar accounting, theoretical yield, and mass conservation.
Benchmark Problem
Thermite: \(2\text{Al}(s) + \text{Fe}_2\text{O}_3(s) \longrightarrow 2\text{Fe}(l) + \text{Al}_2\text{O}_3(s)\). Given \(54.0\text{ g}\) of \(\text{Al}\) and \(191.6\text{ g}\) of \(\text{Fe}_2\text{O}_3\), producing \(105.0\text{ g}\) of \(\text{Fe}\), find: (A) limiting reactant, (B) theoretical yield of \(\text{Fe}\), (C) excess mass remaining, and (D) percent yield.
1
Molar Mass Reference & Initial Mole Conversion
Moles
\(M(\text{Al}) = 26.98\text{ g/mol}\)
\(n_{\text{Al}} = \frac{54.0\text{ g}}{26.98\text{ g/mol}} = \mathbf{2.001\text{ mol Al}}\)
\(M(\text{Fe}_2\text{O}_3) = 2(55.85) + 3(16.00) = 159.70\text{ g/mol}\)
\(n_{\text{Fe}_2\text{O}_3} = \frac{191.6\text{ g}}{159.70\text{ g/mol}} = \mathbf{1.200\text{ mol Fe}_2\text{O}_3}\)
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Part (A): BCA Molar Table & Limiting Reactant
Stoichiometry
Stoichiometric ratio check: \(\frac{2.001\text{ mol Al}}{2} = 1.000\text{ eq}\) vs \(\frac{1.200\text{ mol Fe}_2\text{O}_3}{1} = 1.200\text{ eq}\). Since \(1.000 < 1.200\), \(\text{Al}\) is the Limiting Reactant.
| State | 2 Al | 1 Fe₂O₃ | 2 Fe | 1 Al₂O₃ |
|---|
| Before | 2.001 mol | 1.200 mol | 0.000 mol | 0.000 mol |
| Change | -2.001 mol | -1.000 mol | +2.001 mol | +1.000 mol |
| After | 0.000 (LR) | 0.200 (XS) | 2.001 mol | 1.000 mol |
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Parts (B), (C) & (D): Yields & Excess Consumption
Yields
Theoretical Yield (\(\text{Fe}\)):
\(2.001 \times 55.85\text{ g/mol}\)
\(= \mathbf{111.76\text{ g Fe}}\)
Excess Remaining:
\(0.200 \times 159.70\text{ g/mol}\)
\(= \mathbf{31.94\text{ g Fe}_2\text{O}_3}\)
Percent Yield (\(\text{Fe}\)):
\(\frac{105.0\text{ g}}{111.76\text{ g}} \times 100\)
\(= \mathbf{93.95\%}\)
Conservation of Mass Check
Reactants: \(54.0 + 191.6 = \mathbf{245.6\text{ g}}\).
Products + excess: \(111.76 (\text{Fe}) + 101.96 (\text{Al}_2\text{O}_3) + 31.94 (\text{XS}) = \mathbf{245.66\text{ g}}\). (Conserved within \(<0.03\%\)).
Key Rule
Never compare initial masses directly to find the limiting reactant. Always convert mass to moles first, then divide by the balanced equation coefficients.
STEM Problem Solver Master Series • Chemistry Page 3 of 3
