Question

An air-hockey puck has a model rocket rigidly attached to it. The puck is pushed from one corner along the long side of the 2.00 -m-long air-hockey table, with the rocket pointing along the short side of the table, and at the same time the rocket is fired. If the rocket thrust imparts an acceleration of \(2.00 \mathrm{~m} / \mathrm{s}^{2}\) to the puck, and the table is \(1.00 \mathrm{~m}\) wide, with what minimum initial velocity should the puck be pushed to make it to the opposite short side of the table without bouncing off either long side? Draw the trajectory of the puck for three initial velocities: \(vv_{\min } .\) Neglect friction and air resistance.

Short answer

Answer: The minimum initial velocity required is 4 m/s.

Step-by-step solution

Calculate the time required to travel the long side of the table.

We first need to find the time it takes for the puck to travel the 2-meter length of the table. To do this, we will use the following kinematic equation: $$ s = vt + \frac{1}{2}at^2 $$ Here, s is the distance, v is the initial velocity, a is the acceleration, and t is the time. In this problem, we know the distance s (2 meters), the acceleration a (0 m/s^2, since we are considering the vertical distance), and the initial velocity v (unknown, since that is what we need to find). We need to solve for t: $$ 2 = vt + \frac{1}{2} \cdot 0 \cdot t^2 \Rightarrow t = \frac{2}{v} $$

Determine how much distance the puck covers on the short side, considering the rocket's acceleration.

Now, let's find out how much distance the puck covers on the short side (width) of the table, given its rocket-induced acceleration of 2 m/s^2. We will use the same kinematic equation as before but this time considering the horizontal distance. We know the acceleration a (2 m/s^2) and the time t (found in Step 1: \(\frac{2}{v}\)). The initial velocity in this case is 0 m/s since the rocket determines the horizontal motion. We need to solve for s: $$ s = 0 \cdot \frac{2}{v} + \frac{1}{2} \cdot 2 \cdot \left(\frac{2}{v}\right)^2 \Rightarrow s = \frac{4}{v} $$

Find the minimum initial velocity required to travel 1 meter horizontally within that time.

Finally, we want the puck to travel exactly 1 meter horizontally within the time it takes to travel the 2-meter length of the table. Based on our calculations in Step 2, the horizontal distance s is given by: $$ s = \frac{4}{v} $$ Now, we need to find the initial velocity required to travel 1 meter horizontally in the given time: $$ 1 = \frac{4}{v} \Rightarrow v = 4 \mathrm{~m/s} $$ So, the minimum initial velocity required for the puck to make it to the opposite short side without bouncing off either long side is 4 m/s. As for the trajectories, consider the following cases: 1. Initital velocity \(v < 4 \mathrm{~m/s}\): The puck will not reach the opposite short side, and will collide with one of the long sides due to the rocket's acceleration. 2. Initital velocity \(v = 4 \mathrm{~m/s}\): The puck will follow a trajectory that just makes it to the opposite short side without bouncing off either long side 3. Initital velocity \(v > 4 \mathrm{~m/s}\): The puck will move too fast, hit the first long side and bounce back before reaching the opposite short side.

Key concepts

Uniformly Accelerated Motion

Uniformly accelerated motion is a key concept in classical mechanics. Understanding this concept is essential for solving a variety of physics problems, including those involving objects with a constant acceleration.

Imagine you're in a car that accelerates out of a traffic stop; the increase in the car's velocity occurs at a constant rate, that's uniformly accelerated motion. In this case, the rocket attached to the air-hockey puck provides a constant horizontal acceleration, pushing the puck across the table without friction or air resistance. This is a practical example of such motion.

The kinematic equations, as used in this problem, allow us to predict the future position and velocity of an object under uniformly accelerated motion given certain initial conditions. For the puck, with a known acceleration and time, we can calculate the distance it will travel on the short side of the table. This is crucial because if the puck doesn't travel the width of the table (1 meter) before it reaches the opposite long side, it won't accomplish its intended trajectory.

Projectile Motion

Projectile motion is an integral part of physics that involves objects moving through space while being affected by gravity. However, in this particular case of the air-hockey puck, we are neglecting the effects of gravity as we are dealing with a horizontal motion.

Normally, when studying projectile motion, we think of objects being launched into the air and moving along a curved path under the influence of gravity. But the principles behind it also apply to any two-dimensional motion where one component is constant, like the horizontal movement of the puck, and another is changed or influenced by an external force, like the thrust from the rocket.

By treating the hockey puck's path as a product of two independent motions - one without acceleration (along the table's length) and one with a uniform acceleration (provided by the rocket) - we can analyze its trajectory. The puck's initial velocity dictates whether it will fall short, make it precisely, or overshoot the table's width.

Classical Mechanics

Classical mechanics is the branch of physics dealing with the motion of macroscopic objects from projectiles to parts in machinery. It provides us with the tools to describe an object's motion, understand the forces acting on it, and predict its future state.

In the context of the air-hockey puck problem, we're employing concepts from classical mechanics like kinematic equations to resolve the motion of the puck as it glides across the table. The first step identifies the time it takes to traverse the length thanks to its initial velocity, a concept known as linear motion. Then, we regard the rocket's thrust as a force causing constant horizontal acceleration, allowing us to calculate the puck's horizontal displacement.

All these calculations are fundamental applications of Newton’s laws of motion, which are cornerstones of classical mechanics. Overall, a firm grasp of classical mechanics is what enables us to solve complicated real-world problems using simplified models where we can neglect external forces like friction or air resistance in order to focus on the primary factors affecting motion.