A 0.180 kg hockey puck is moving on a frictionless, horizontal icy surface. At t = 0, the puck is moving to the right at 3.00 m/s. What is the velocity of the puck (magnitude and direction) after a force of 30.0 N directed to the right has been applied for 0.07 s. 4.8 m/s to the right 22.1 m/s to the right 14.7 m/s to the right 9.3 m/s to the left.

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### Physics Problem: Frictionless Motion of a Hockey Puck

#### Problem Statement:
A 0.180 kg hockey puck is moving on a frictionless, horizontal icy surface. At \( t = 0 \), the puck is moving to the right at 3.00 m/s. What is the velocity of the puck (magnitude and direction) after a force of 30.0 N directed to the right has been applied for 0.07 s?

#### Multiple Choice Answers:
- A) 4.8 m/s to the right
- B) 22.1 m/s to the right
- C) 14.7 m/s to the right
- D) 9.3 m/s to the left

To solve this problem, you need to use the principles of Newton's second law and the kinematic equations for motion in one dimension. 

##### Solution Steps:
1. **Determine the acceleration**:
   - Force (\( F \)) is given as 30 N.
   - Mass (\( m \)) of the puck is 0.180 kg.
   - By Newton's second law, \( F = ma \) or \( a = \frac{F}{m} \).
   - \( a = \frac{30.0 \, \text{N}}{0.180 \, \text{kg}} = 166.67 \, \text{m/s}^2 \).

2. **Calculate the change in velocity (\( \Delta v \))**:
   - The acceleration is applied for \( t = 0.07 \) s.
   - Change in velocity, \( \Delta v = a \times t = 166.67 \, \text{m/s}^2 \times 0.07 \, \text{s} \).
   - \( \Delta v = 11.67 \, \text{m/s} \).

3. **Determine the final velocity**:
   - Initial velocity (\( v_0 \)) is 3.00 m/s to the right.
   - Final velocity, \( v = v_0 + \Delta v \).
   - \( v = 3.00 \, \text{m/s} + 11.67 \, \text{m/s} = 14.67 \, \text{m/s} \).

Therefore, the correct answer is:
-
Transcribed Image Text:### Physics Problem: Frictionless Motion of a Hockey Puck #### Problem Statement: A 0.180 kg hockey puck is moving on a frictionless, horizontal icy surface. At \( t = 0 \), the puck is moving to the right at 3.00 m/s. What is the velocity of the puck (magnitude and direction) after a force of 30.0 N directed to the right has been applied for 0.07 s? #### Multiple Choice Answers: - A) 4.8 m/s to the right - B) 22.1 m/s to the right - C) 14.7 m/s to the right - D) 9.3 m/s to the left To solve this problem, you need to use the principles of Newton's second law and the kinematic equations for motion in one dimension. ##### Solution Steps: 1. **Determine the acceleration**: - Force (\( F \)) is given as 30 N. - Mass (\( m \)) of the puck is 0.180 kg. - By Newton's second law, \( F = ma \) or \( a = \frac{F}{m} \). - \( a = \frac{30.0 \, \text{N}}{0.180 \, \text{kg}} = 166.67 \, \text{m/s}^2 \). 2. **Calculate the change in velocity (\( \Delta v \))**: - The acceleration is applied for \( t = 0.07 \) s. - Change in velocity, \( \Delta v = a \times t = 166.67 \, \text{m/s}^2 \times 0.07 \, \text{s} \). - \( \Delta v = 11.67 \, \text{m/s} \). 3. **Determine the final velocity**: - Initial velocity (\( v_0 \)) is 3.00 m/s to the right. - Final velocity, \( v = v_0 + \Delta v \). - \( v = 3.00 \, \text{m/s} + 11.67 \, \text{m/s} = 14.67 \, \text{m/s} \). Therefore, the correct answer is: -
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