Problem 9 - University Physics with Modern Physics

Question

A 0.160 -kg hockey puck is moving on an icy, frictionless, horizontal surface. At \(t=0,\) the puck is moving to the right at 3.00 \(\mathrm{m} / \mathrm{s} .\) (a) Calculate the velocity of the puck (magnitude and direction) after a force of 25.0 \(\mathrm{N}\) directed to the right has been applied for 0.050 s. (b) If, instead, a force of 12.0 \(\mathrm{N}\) directed to the left is applied from \(t=0\) to \(t=0.050 \mathrm{s}\) , what is the final velocity of the puck?

Step-by-Step Solution

Answer

(a) 10.81 m/s right; (b) 0.75 m/s left.

Step 1: Identify Known Values

For both scenarios, we know the initial mass of the hockey puck is \( m = 0.160 \) kg, and its initial velocity \( v_i = 3.00 \) m/s to the right at \( t = 0 \). Also, the time period of the force application \( \Delta t = 0.050 \) s.

Step 2: Calculate Final Velocity for Force to the Right

To find the final velocity when a force of 25.0 N is applied to the right, we use the formula for acceleration \( a = \frac{F}{m} \), where \( F = 25.0 \) N. \[ a = \frac{25.0 \ \text{N}}{0.160 \ \text{kg}} = 156.25 \ \text{m/s}^2 \]Use the formula for final velocity: \( v_f = v_i + a \cdot \Delta t \).Substituting in the known values:\[ v_f = 3.00 \ \text{m/s} + (156.25 \ \text{m/s}^2 \cdot 0.050 \ \text{s}) \]\[ v_f = 3.00 \ \text{m/s} + 7.8125 \ \text{m/s} \]\[ v_f = 10.8125 \ \text{m/s} \]Thus, the velocity of the puck is 10.81 m/s to the right.

Step 3: Calculate Final Velocity for Force to the Left

Now consider a force of 12.0 N to the left. First, determine the acceleration: \[ a = \frac{-12.0 \ \text{N}}{0.160 \ \text{kg}} = -75.0 \ \text{m/s}^2 \]Next, use the formula for final velocity: \( v_f = v_i + a \cdot \Delta t \).Substitute the known values:\[ v_f = 3.00 \ \text{m/s} + (-75.0 \ \text{m/s}^2 \cdot 0.050 \ \text{s}) \]\[ v_f = 3.00 \ \text{m/s} - 3.75 \ \text{m/s} \]\[ v_f = -0.75 \ \text{m/s} \]Thus, the puck moves at 0.75 m/s to the left.

Key Concepts

Frictionless Surfaces

Imagine a surface where there is absolutely no friction to slow things down. This type of surface is deemed frictionless. On such a frictionless surface, an object will continue to move at a constant speed without being slowed by roughness or resistance.
In real life, ice is one of the closest examples. However, no real surface is truly without friction. But in physics problems, a frictionless surface is an idealized concept that helps us focus on the pure effects of forces and motion without additional factors.
When we say that the hockey puck in our problem is on an icy, frictionless surface, it means the only forces acting on it are those actively applied, like pushes or pulls. This allows us to apply physical laws in a simplified manner without worrying about deceleration due to friction.

Newton's Second Law

One of the most fundamental principles of physics is Newton's Second Law of Motion. This law is vital in understanding how forces cause motion. It states that the force applied to an object is equal to the mass of the object times its acceleration. In formula terms, it’s represented as:

• \( F = m imes a \)

For our hockey puck, which has a mass of 0.160 kg, this means any force applied influences its acceleration directly. The greater the force, the greater the acceleration. By rearranging the formula, we also define acceleration \( a \) as \( \frac{F}{m} \).
In our scenarios, applying Newton's Second Law helps us calculate the acceleration produced by forces of 25.0 N and 12.0 N on the puck. Knowing the acceleration and initial velocity allows us to find the final velocity, using another important relationship, which is velocity calculation.

Velocity Calculation

Velocity is the speed of something in a given direction. To find the final velocity of our moving puck, we rely on its initial velocity and the acceleration due to the applied force. The formula to calculate the final velocity \( v_f \) following a constant acceleration is:

• \( v_f = v_i + a \cdot \Delta t \)

Where \( v_i \) is the initial velocity, \( a \) is acceleration, and \( \Delta t \) is the time duration for which the force is applied.
In the exercise problem, we start with an initial velocity of 3.00 m/s. Using the identified acceleration from Newton's Second Law, we compute final velocities for forces applied both in the right and left directions. It demonstrates how quickly the puck accelerates or decelerates based on direction and magnitude of the force applied.

Force and Motion

The relationship between force and motion is a cornerstone of classical mechanics. Forces cause objects to change their state of motion—they either speed up (accelerate), slow down, or change direction. How they change depends greatly on the amount of force and the object's mass.
In our problem, we observe two different forces applied to the puck. A force of 25.0 N directed to the right causes the puck to speed up, whereas a force of 12.0 N to the left makes the puck slow down and eventually move in the opposite direction. These scenarios underline how forces affect velocity and direction.
This study of forces and how they affect motion helps us predict future states of moving objects, which is crucial in diverse applications from sports to engineering and beyond.