Problem 37 - Classical Mechanics

Question

A student kicks a frictionless puck with initial speed \(v_{\mathrm{o}},\) so that it slides straight up a plane that is inclined at an angle \(\theta\) above the horizontal. (a) Write down Newton's second law for the puck and solve to give its position as a function of time. (b) How long will the puck take to return to its starting point?

Step-by-Step Solution

Answer

The position is \(x(t) = v_0 t - \frac{1}{2} g \sin \theta \cdot t^2\), and the return time is \(\frac{2v_0}{g \sin \theta}\).

Step 1: Understanding Forces

First, analyze the forces acting on the puck as it moves up the inclined plane. The primary force acting on it is gravity, which can be decomposed into components parallel and perpendicular to the incline. The component of gravitational force along the incline is \( mg \sin \theta \), where \( m \) is the mass of the puck, \( g \) is the acceleration due to gravity, and \( \theta \) is the angle of inclination.

Step 2: Applying Newton's Second Law

Next, apply Newton's second law along the direction of the incline. Since there is no friction, the net force along the incline is \[ F = ma = -mg \sin \theta \] Thus, the acceleration of the puck along the incline is \[ a = -g \sin \theta \].

Step 3: Deriving the Position Function

Knowing the initial velocity \( v_0 \) and the acceleration, we can use the kinematic equation to describe the position \( x(t) \) as a function of time: \[ x(t) = v_0 t + \frac{1}{2} a t^2 \]Substitute \( a = -g \sin \theta \) into the equation:\[ x(t) = v_0 t - \frac{1}{2} g \sin \theta \cdot t^2 \].

Step 4: Finding the Ascend Time

To determine when the puck reaches the highest point (where its velocity is zero), use the velocity equation:\[ v(t) = v_0 + at \]Set \( v(t) = 0 \):\[ 0 = v_0 - g \sin \theta \cdot t \]Solving for \( t \), we find:\[ t = \frac{v_0}{g \sin \theta} \] is the time to reach the highest point.

Step 5: Finding the Total Time to Return

The time to ascend and return is twice the ascend time because the motion is symmetric. Thus the total time to return to the starting point is:\[ T = 2 \times \frac{v_0}{g \sin \theta} = \frac{2v_0}{g \sin \theta} \].

Key Concepts

Inclined Plane Motion

When a puck slides on an inclined plane, it experiences a unique type of motion influenced by gravity and the angle of the incline. Let's break down what happens:

Imagine an inclined plane like a ramp, sloping upwards at an angle \( \theta \) to the horizontal. When you slide an object, like a puck, up this ramp, its motion is affected by gravity pulling it downwards.
Gravity acts in two directions here: parallel to the plane and perpendicular to it.
The component of gravity that affects the motion parallel to the incline determines how the puck slows down as it ascends.
The frictionless nature of the plane means there's no opposing force to slow it down other than this gravity component.

Real-world example: Think of trying to push a cart up a steep hill. The steeper the incline, the harder you have to push because gravity pulls more strongly against the direction you're moving the cart.

Kinematic Equations

Kinematics is the branch of physics that deals with the motion of objects. On our inclined plane, kinematic equations help describe how the puck moves over time under the influence of gravity.

A main equation we use is for position as a function of time:

• \( x(t) = v_0 t + \frac{1}{2} a t^2 \)

Here, \( x(t) \) is the position at time \( t \), \( v_0 \) is the initial speed, and \( a \) is the acceleration along the incline.
Substitute \( a = -g \sin \theta \), because of gravity's pull along the incline.
This equation tells us how far along the plane the puck gets at any time \( t \).
Similarly, the velocity at any point in time can be found by:

• \( v(t) = v_0 + at \)

These equations are crucial for predicting how the puck behaves; from its highest point to when it slides back down, they provide the complete picture.

Gravitational Force Decomposition

Gravity is constantly working to pull objects down towards the Earth. When dealing with an inclined plane, we can break down, or "decompose," this gravitational pull into components to better understand its effects.

• Parallel Component: The force pulling the puck down the incline. This is given by \( mg \sin \theta \).
• Perpendicular Component: This is the force pressing the puck into the plane and is given by \( mg \cos \theta \). It doesn't affect the sliding motion, but it's crucial for friction calculations, if friction were present.

Why do we decompose? It's simpler to analyze motion using these components instead of the whole force of gravity at an angle.
It helps us to apply Newton's Second Law just along the inclined plane.
This decomposition is an essential concept when studying physics problems involving slopes or angles.