9231 · 3.5
Linear motion under a variable force flashcards
Revision flashcards for Cambridge 9231 Linear motion under a variable force (syllabus 3.5). Flip, recall, then mark a real past-paper question.
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What is the fundamental equation for motion under a variable force?
Newton's Second Law, $F = ma$. Since the force $F$ is variable, the acceleration $a$ must also be variable.
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When should you express acceleration as $a = \frac{dv}{dt}$?
When the net force is given as a function of time, $F(t)$, or as a function of velocity, $F(v)$, and you want to find velocity as a function of time.
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When should you express acceleration as $a = v\frac{dv}{dx}$?
When the net force is given as a function of displacement (position), $F(x)$, or as a function of velocity, $F(v)$, and you want to find a relationship between velocity and displacement.
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Why can't 'suvat' equations be used for variable force problems?
The 'suvat' equations ($v=u+at$, $s=ut+\frac{1}{2}at^2$, etc.) are derived on the assumption that acceleration, $a$, is constant. If force is variable, acceleration is variable, so this assumption is violated.
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How is the work done by a variable force $F(x)$ from $x=a$ to $x=b$ calculated?
The work done is calculated by the integral: $W = \int_{a}^{b} F(x) \,dx$. This is also equal to the change in kinetic energy.
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What is the equation of motion if force $F$ is a function of displacement $x$?
$F(x) = m v \frac{dv}{dx}$. This sets up a differential equation that can be solved by separating variables.
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What is the equation of motion if force $F$ is a function of time $t$?
$F(t) = m \frac{dv}{dt}$. This can be integrated with respect to time to find velocity.
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What is the role of initial conditions?
When you integrate to solve the differential equation, you get a constant of integration (e.g., '+ C'). Initial conditions (like velocity at $t=0$ or position at $t=0$) allow you to find the specific value of this constant.
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A common trap is forgetting the constant of integration. What's another one?
Incorrectly separating variables. For example, when solving $m v \frac{dv}{dx} = F(x)$, ensure all $v$ terms are with $dv$ and all $x$ terms are with $dx$ before integrating: $\int mv \,dv = \int F(x) \,dx$.
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How do you derive $a = v\frac{dv}{dx}$?
Using the chain rule: $a = \frac{dv}{dt} = \frac{dv}{dx} \times \frac{dx}{dt}$. Since $v = \frac{dx}{dt}$, this becomes $a = \frac{dv}{dx} \times v = v\frac{dv}{dx}$.
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What is the relationship between power ($P$), force ($F$) and velocity ($v$)?
$P = Fv$. If power is constant, then force is inversely proportional to velocity: $F = P/v$. This is a common source of variable force problems.
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Variable force
A force whose size changes with time, displacement or velocity, so that the acceleration it produces is not constant and Newton's second law $F = ma$ becomes a differential equation rather than a constant-acceleration formula.
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Equation of motion
Newton's second law written for the particle in hand, $F = ma$, with the acceleration expressed as $\dfrac{dv}{dt}$, $\dfrac{d^2x}{dt^2}$ or $v\dfrac{dv}{dx}$ according to whether the force depends on time, displacement or velocity.
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Separation of variables
The method of solving a first-order differential equation by collecting each variable on its own side, for example $m v\,dv = F(x)\,dx$, and integrating both sides, usually with a boundary condition to fix the constant.
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Force dependent on displacement
A force given as a function of position, $F(x)$, such as a spring force or gravitational attraction, for which the acceleration is written as $a = v\dfrac{dv}{dx}$ so that the equation of motion $mv\dfrac{dv}{dx} = F(x)$ can be separated and integrated.
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Force dependent on velocity
A force given as a function of speed, $F(v)$, such as air resistance. Either form of acceleration works: use $a = \dfrac{dv}{dt}$ to find $v$ as a function of time, or $a = v\dfrac{dv}{dx}$ to find $v$ as a function of displacement; the question's wording decides which.