Chapter 8: Problem 9
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
Step by step solution
Understanding the problem
Identify forces and known quantities
Apply Newton's Second Law
Step 4(a): Calculate acceleration for a force to the right
Step 5(a): Find the final velocity for part (a)
Step 6(b): Calculate acceleration for a force to the left
Step 7(b): Find the final velocity for part (b)
Final Thoughts
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Kinematics
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.
Force and Acceleration
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.
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
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.