Mastering Mechanics: Algebra, Logarithms, and Trigonometry

Problem Statement

In HSC Mathematics Extension 1 and Extension 2, mechanics is not just about understanding physics principles like Newton's Laws of Motion; it is heavily reliant on your mathematical toolkit. Solving mechanics problems frequently involves complex algebraic manipulations, rearranging non-linear equations, using logarithms to solve equations involving time and velocity, and applying exact values of trigonometric ratios and trigonometric identities to resolve forces in multiple dimensions.

Consider the following mechanics problem that tests all these mathematical skills:

A particle of mass $m$ is projected up a smooth plane inclined at an angle $\theta$ to the horizontal, where $\sin \theta = \frac{3}{5}$. The particle experiences air resistance proportional to its velocity, given by $R = mkv$, where $k$ is a positive constant and $v$ is the velocity at time $t$. The initial velocity of the particle is $u$.

(a) By resolving forces parallel to the inclined plane, show that the equation of motion is given by: $\frac{dv}{dt} = -\left(\frac{3g}{5} + kv\right)$

(b) Using integration and the properties of logarithms, find an expression for the time $t$ it takes for the particle to come to instantaneous rest.

(c) Given that the particle comes to rest in $t = \frac{1}{k} \ln 2$ seconds, find the initial velocity $u$ in terms of $g$ and $k$.

Hints

• Part (a): Draw a force diagram. Identify the forces acting parallel to the plane: the component of gravity down the plane and the air resistance down the plane. Use Newton's Second Law ($F = ma$) and the exact trigonometric ratio for $\sin \theta$.
• Part (b): Rearrange the differential equation to separate the variables $v$ and $t$. Integrate both sides. Remember that $\int \frac{f'(x)}{f(x)} dx = \ln |f(x)| + C$. Apply the initial condition $v = u$ when $t = 0$ to find the constant of integration.
• Part (c): Set $v = 0$ in your velocity-time equation to represent instantaneous rest. Substitute the given time $t = \frac{1}{k} \ln 2$ and use logarithmic identities to solve for $u$.

Solutions

Part (a): Resolving Forces

1. Let the direction up the inclined plane be the positive $x$-direction.
2. The forces acting parallel to the plane are:
   • The component of weight acting down the plane: $W_x = -mg \sin \theta$
   • The air resistance acting against the motion (down the plane): $R_x = -mkv$
3. According to Newton's Second Law, the resultant force is $F = m \frac{dv}{dt}$: $m \frac{dv}{dt} = -mg \sin \theta - mkv$
4. Substitute the exact trigonometric value $\sin \theta = \frac{3}{5}$: $m \frac{dv}{dt} = -mg \left(\frac{3}{5}\right) - mkv$
5. Divide the entire equation by $m$: $\frac{dv}{dt} = -\frac{3g}{5} - kv = -\left(\frac{3g}{5} + kv\right)$

Part (b): Using Logarithms to find Time

1. We have the differential equation: $\frac{dv}{dt} = -\left(\frac{3g}{5} + kv\right)$
2. Rearrange to separate variables: $\frac{dt}{dv} = -\frac{1}{\frac{3g}{5} + kv}$
3. Integrate with respect to $v$: $t = -\int \frac{1}{\frac{3g}{5} + kv} dv$
4. To integrate this, we need the numerator to be the derivative of the denominator (which is $k$). So, multiply by $\frac{k}{k}$: $t = -\frac{1}{k} \int \frac{k}{\frac{3g}{5} + kv} dv$ $t = -\frac{1}{k} \ln \left| \frac{3g}{5} + kv \right| + C$
5. Apply initial conditions to find $C$: When $t = 0$, $v = u$. $0 = -\frac{1}{k} \ln \left( \frac{3g}{5} + ku \right) + C$ $C = \frac{1}{k} \ln \left( \frac{3g}{5} + ku \right)$
6. Substitute $C$ back into the equation for $t$: $t = -\frac{1}{k} \ln \left( \frac{3g}{5} + kv \right) + \frac{1}{k} \ln \left( \frac{3g}{5} + ku \right)$
7. Use the logarithmic identity $\ln A - \ln B = \ln(\frac{A}{B})$: $t = \frac{1}{k} \left[ \ln \left( \frac{3g}{5} + ku \right) - \ln \left( \frac{3g}{5} + kv \right) \right]$ $t = \frac{1}{k} \ln \left( \frac{\frac{3g}{5} + ku}{\frac{3g}{5} + kv} \right)$

Part (c): Solving for Initial Velocity

1. The particle comes to instantaneous rest when $v = 0$. Substitute $v = 0$ into our time equation: $t_{rest} = \frac{1}{k} \ln \left( \frac{\frac{3g}{5} + ku}{\frac{3g}{5} + 0} \right)$ $t_{rest} = \frac{1}{k} \ln \left( \frac{3g + 5ku}{3g} \right)$ $t_{rest} = \frac{1}{k} \ln \left( 1 + \frac{5ku}{3g} \right)$
2. We are given that $t_{rest} = \frac{1}{k} \ln 2$. Equate the two expressions: $\frac{1}{k} \ln 2 = \frac{1}{k} \ln \left( 1 + \frac{5ku}{3g} \right)$
3. Cancel $\frac{1}{k}$ from both sides and equate the arguments of the natural logarithm: $2 = 1 + \frac{5ku}{3g}$
4. Rearrange algebraically to solve for $u$: $1 = \frac{5ku}{3g}$ $3g = 5ku$ $u = \frac{3g}{5k}$

Takeaways

• Exact Trigonometric Ratios: When resolving forces on an inclined plane, exact ratios (like those derived from Pythagorean triples, e.g., 3-4-5 triangles) simplify the differential equation.
• Calculus and Logarithms: Resisted motion almost always leads to integrals of the form $\int \frac{1}{ax+b} dx$, yielding logarithmic functions. Fluency in logarithmic laws ($\ln A - \ln B = \ln \frac{A}{B}$) is mandatory for simplifying these expressions.
• Algebraic Manipulation: Setting up the equation is only half the battle. Rearranging equations to make $t$, $v$, or $x$ the subject requires strong algebraic fundamentals, particularly when dealing with constants like $k$ and $g$.

Further Readings

• HSC Mechanics: https://vumaths.com/booklets/hsc-mechanics/
• HSC Integrals: https://vumaths.com/booklets/hsc-integrals/
• HSC Trigonometry: https://vumaths.com/booklets/hsc-trigonometry/

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