Question

I kick a puck of mass $m$ up an incline (angle of slope $=\theta$ ) with initial speed $v_{\mathrm{o}}$. There is no friction between the puck and the incline, but there is air resistance with magnitude $f(v)=c{v}^2.$ Write down and solve Newton's second law for the puck's velocity as a function of $t$ on the upward journey. How long does the upward journey last?

Short answer

The upward journey lasts until the velocity becomes zero, determined by solving the integrated function for $t$.

Step-by-step solution

Identify the Forces

We need to identify the forces acting on the puck. The gravitational force acts downward, and air resistance acts opposite to the direction of motion. The gravitational force is $mg$, where $g$ is the gravitational acceleration. For an inclined plane, the component of gravitational force acting parallel to the incline is $mg\sin \theta$. Air resistance is given by $f(v)=c{v}^2$.

Apply Newton's Second Law

Newton's second law states that $F=ma$. The net force on the puck along the incline is $-mg\sin \theta -c{v}^2$ (negative because both forces oppose the motion). Thus, $ma=-mg\sin \theta -c{v}^2$.

Set Up the Differential Equation

We have the equation $m\frac{dv}{dt}=-mg\sin \theta -c{v}^2$. Simplifying, we get $\frac{dv}{dt}=-g\sin \theta -\frac{c}{m}{v}^2$.

Solve the Differential Equation

This is a separable differential equation. Rearrange terms to separate variables: $\frac{dv}{g\sin \theta +\frac{c}{m}{v}^2}=-dt$. Integrate both sides. The left side requires partial fraction decomposition and results in an integration that involves an arctan function. Integrating both sides: $\int \frac{dv}{g\sin \theta +\frac{c}{m}{v}^2}=-\int dt+C$.

Solve for the Time of the Upward Journey

Evaluate the integral to find $v(t)$. The upward journey lasts until the velocity becomes zero. Solve the equation for $t$ when $v=0$. Let $v(0)=v_0$ be the initial condition. After performing integration and algebra, obtain $t=\frac{m}{c}(\frac{1}{v_0}-\frac{1}{v(t)})-\frac{g\sin \theta }{c}$. Set $v(t)=0$ and solve for $t$, you get the time it takes for the puck to come to a stop.

Key concepts

Air Resistance

Air resistance, often referred to as drag, is a force that opposes the motion of an object through the air. In our scenario with the puck, this force is given by the formula $f(v)=c{v}^2$, where $c$ is a constant coefficient that depends on factors like the air density and shape of the object, and $v$ is the velocity of the puck.
Air resistance varies with respect to velocity. Since it is proportional to the square of the velocity, as the puck moves faster, the air resistance significantly increases. This quadratic relationship can cause dramatic oppositional forces during high-speed motion.
Understanding air resistance is crucial as it plays a vital role in moderating speed. It not only slows the object down but also alters the overall energy dynamics of the object's motion. When solving problems involving air resistance, one must consider how it interacts with other forces to affect the object's velocity and motion trajectory.

Inclined Plane

An inclined plane is a flat surface that is tilted at an angle, $\theta$, to the horizontal. It is a simple mechanical device used to facilitate the raising or lowering of a load. In physics problems, an inclined plane helps illustrate how forces resolve and interact differently compared to horizontal motion.
In this problem, the gravitational force acting down the plane can be calculated using the component $mg\sin \theta$, where $m$ is the mass of the puck, and $g$ is the acceleration due to gravity. This component represents the force pulling the puck back as it slides upwards, working against the initial push and effectively lowering the puck's speed along with air resistance.
Considering inclined plane mechanics along with air resistance, like in this exercise, provides a layered understanding of how objects behave on slopes. The forces at play determine how efficiently an object can ascend the incline and how quickly it slows down when moving against gravity.

Separable Differential Equations

A separable differential equation is a type of differential equation that can be written such that all terms involving one variable and its differential are on one side of the equation, and all terms involving the other variable and its differential are on the other side. This allows the equation to be solved in a step-by-step manner by means of integration.
In the exercise, the equation $m\frac{dv}{dt}=-mg\sin \theta -c{v}^2$ is a separable differential equation. By rearranging terms, we get:$$\frac{dv}{g\sin \theta +\frac{c}{m}{v}^2}=-dt$$
Solving this involves integrating both sides of the equation. This integration helps find the expression for velocity, $v(t)$, as a function of time. The process of solving these equations typically involves techniques like partial fraction decomposition or substitution to simplify the integration tasks.
Separable differential equations are foundational in understanding motion dynamics, where forces interact in complex ways. Mastering these techniques is critical for accurately predicting the velocity or position of objects over time in physics problems, particularly in cases involving mixed forces like gravity and resistance.