20. A hockey puck with mass 0.20 kg is traveling north with a speed of 15 m/s. A Dallas Star forward hits the puck with his hockey stick and after being hit the puck is traveling south with a speed of 25 m/s. What is the magnitude of the impulse that the stick applied to the puck?

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### Physics Problem: Calculating Impulse Applied to a Hockey Puck

**Problem Statement:**

A hockey puck with a mass of 0.20 kg is traveling north with a speed of 15 m/s. A Dallas Star forward hits the puck with his hockey stick and, after being hit, the puck is traveling south with a speed of 25 m/s. 

**Question:**
What is the magnitude of the impulse that the stick applied to the puck?

**Solution:**

When solving this problem, remember that impulse is defined as the change in momentum (Δp), and it can be calculated using the formula:

\[ \vec{J} = \Delta \vec{p} = m \cdot \Delta \vec{v} \]

where:
- \(\vec{J}\) is the impulse,
- \(m\) is the mass of the puck,
- \(\Delta \vec{v}\) is the change in velocity.

First, we need to determine the change in velocity:
1. Before being hit, the puck is traveling north with a velocity of \(15 \, m/s \). 
2. After being hit, the puck is traveling south with a velocity of \(25 \, m/s \).

For the purpose of calculation, let’s consider the north direction as positive and the south direction as negative. Thus:
- Initial velocity (\(v_i\)) = \(+15 \, m/s\) 
- Final velocity (\(v_f\)) = \(-25 \, m/s\)

The change in velocity (\(\Delta \vec{v}\)) is:
\[ \Delta \vec{v} = v_f - v_i \]
\[ \Delta \vec{v} = -25 \, m/s - (+15 \, m/s) \]
\[ \Delta \vec{v} = -25 \, m/s - 15 \, m/s \]
\[ \Delta \vec{v} = -40 \, m/s \]

Now, we use the mass of the puck (0.20 kg) to calculate the impulse:

\[ \vec{J} = m \cdot \Delta \vec{v} \]
\[ \vec{J} = 0.20 \, kg \cdot -40 \, m/s \]
\[ \vec{J} = -8.0 \, kg \cdot m/s \]
Transcribed Image Text:### Physics Problem: Calculating Impulse Applied to a Hockey Puck **Problem Statement:** A hockey puck with a mass of 0.20 kg is traveling north with a speed of 15 m/s. A Dallas Star forward hits the puck with his hockey stick and, after being hit, the puck is traveling south with a speed of 25 m/s. **Question:** What is the magnitude of the impulse that the stick applied to the puck? **Solution:** When solving this problem, remember that impulse is defined as the change in momentum (Δp), and it can be calculated using the formula: \[ \vec{J} = \Delta \vec{p} = m \cdot \Delta \vec{v} \] where: - \(\vec{J}\) is the impulse, - \(m\) is the mass of the puck, - \(\Delta \vec{v}\) is the change in velocity. First, we need to determine the change in velocity: 1. Before being hit, the puck is traveling north with a velocity of \(15 \, m/s \). 2. After being hit, the puck is traveling south with a velocity of \(25 \, m/s \). For the purpose of calculation, let’s consider the north direction as positive and the south direction as negative. Thus: - Initial velocity (\(v_i\)) = \(+15 \, m/s\) - Final velocity (\(v_f\)) = \(-25 \, m/s\) The change in velocity (\(\Delta \vec{v}\)) is: \[ \Delta \vec{v} = v_f - v_i \] \[ \Delta \vec{v} = -25 \, m/s - (+15 \, m/s) \] \[ \Delta \vec{v} = -25 \, m/s - 15 \, m/s \] \[ \Delta \vec{v} = -40 \, m/s \] Now, we use the mass of the puck (0.20 kg) to calculate the impulse: \[ \vec{J} = m \cdot \Delta \vec{v} \] \[ \vec{J} = 0.20 \, kg \cdot -40 \, m/s \] \[ \vec{J} = -8.0 \, kg \cdot m/s \]
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