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 m/s. (a) Calculate the velocity of the puck (magnitude and direction) after a force of 25.0 N directed to the right has been applied for 0.050 s. (b) If, instead, a force of 12.0 N directed to the left is applied from \(t\) = 0 to \(t\) = 0.050 s, what is the final velocity of the puck?

Short answer

(a) Final velocity: 10.8125 m/s to the right; (b) Final velocity: -0.75 m/s to the left.

Step-by-step solution

Understanding the problem

We have a mass (hockey puck) initially moving with a certain velocity and we want to find out how it changes when a force is applied. Two different scenarios are given: (a) a force to the right and (b) a force to the left.

Identify forces and known quantities

The initial velocity, mass, and force values are provided. For both parts of the question, the mass of the puck is 0.160 kg, and the initial velocity is 3.00 m/s to the right. The force and duration of application differ in each scenario.

Apply Newton's Second Law

Newton's second law states: \[ F = ma \]where \( F \) is the force applied, \( m \) is the mass, and \( a \) is the acceleration. From this, we can find acceleration:\[ a = \frac{F}{m} \]

Step 4(a): Calculate acceleration for a force to the right

For part (a), the force \( F = 25.0 \) N is applied to the right. Calculate the acceleration:\[ a = \frac{25.0}{0.160} = 156.25 \text{ m/s}^2 \]

Step 5(a): Find the final velocity for part (a)

Use the equation of motion to find the final velocity:\[ v_f = v_i + at \]where \( v_i = 3.00 \text{ m/s} \) and \( t = 0.050 \text{ s} \):\[ v_f = 3.00 + (156.25 \times 0.050) \approx 10.8125 \text{ m/s} \]

Step 6(b): Calculate acceleration for a force to the left

For part (b), the force \( F = -12.0 \) N (negative because it is to the left) is applied. Calculate the acceleration:\[ a = \frac{-12.0}{0.160} = -75.0 \text{ m/s}^2 \]

Step 7(b): Find the final velocity for part (b)

Using the same formula as in part (a):\[ v_f = v_i + at \]with \( v_i = 3.00 \text{ m/s} \) and \( t = 0.050 \text{ s} \):\[ v_f = 3.00 + (-75.0 \times 0.050) = 3.00 - 3.75 = -0.75 \text{ m/s} \]

Final Thoughts

In part (a), the force increases the puck's speed to the right. In part (b), the force slows it down and briefly sends it backwards.

Key concepts

Kinematics

Kinematics is the branch of physics that describes the motion of objects without considering the forces that cause it. It helps us describe how objects move based on parameters like velocity, acceleration, and time.
In our exercise, kinematics tells us how the hockey puck moves when different forces are applied.

• The initial velocity of the puck is 3.00 m/s to the right.
• An external force alters this motion either speeding it up or slowing it down.
• Understanding the puck's trajectory involves analyzing how acceleration impacts its velocity and direction.

Kinematics does not look into what causes the acceleration; it simply provides the means to predict the future position and velocity based on the current state.

Force and Acceleration

Force and acceleration are intimately connected through Newton's Second Law of Motion. This law states that the acceleration of an object is directly proportional to the force applied, and inversely proportional to its mass.
Newton's Second Law is expressed as the equation: \[ F = ma \]where:

• F is the force applied.
• m is the mass of the object, which in this case is the hockey puck (0.160 kg).
• a is the acceleration that results from the applied force.

The exercise involves two scenarios:

Scenario 1: Force to the Right

- In part (a), a force of 25.0 N to the right causes acceleration.- The puck's acceleration is determined by \( a = \frac{25.0}{0.160} = 156.25 \text{ m/s}^2 \).

Scenario 2: Force to the Left

- In part (b), a force of 12.0 N to the left yields acceleration in the opposite direction.- The resulting acceleration is \( a = \frac{-12.0}{0.160} = -75.0 \text{ m/s}^2 \).These calculations demonstrate how the direction and magnitude of force determine an object's acceleration.

Velocity Calculation

Calculating velocity involves understanding how the initial velocity, acceleration, and time come together to affect an object's final velocity. Using the formula for motion, we can find the final velocity of the hockey puck after a force has been applied.
The equation is:\[ v_f = v_i + at \]- \( v_f \) is the final velocity.- \( v_i \) is the initial velocity of 3.00 m/s.- \( a \) is the acceleration calculated from the force applied.- \( t \) is the time in seconds for which the force is applied (0.050 s).

Applying the Formula

• **For part (a):** - With the acceleration to the right \( (156.25 \text{ m/s}^2) \), - The formula yields \( v_f = 3.00 + (156.25 \,\times\, 0.050) \approx 10.8125 \text{ m/s} \).
• **For part (b):** - With acceleration to the left \( (-75.0 \text{ m/s}^2) \), - The resulting velocity is \( v_f = 3.00 - 3.75 = -0.75 \text{ m/s} \), which indicates a reversal of direction.

These calculations show the pivotal role of velocity analysis in understanding motion, especially when various forces interact over time.