Answer :
Let's solve the problem step by step to understand how the momentum of the bowling ball changes when its speed increases from \(1 \, \text{m/s}\) to \(2 \, \text{m/s}\).
### Step 1: Understanding Momentum
Momentum (\(p\)) is calculated using the formula:
[tex]\[ p = m \times v \][/tex]
where:
- \(m\) is the mass of the object (in kilograms, kg),
- \(v\) is the velocity of the object (in meters per second, m/s).
### Step 2: Initial Momentum
First, let's determine the initial momentum of the bowling ball when its speed is \(1 \, \text{m/s}\).
Given:
- Mass \( m = 5 \, \text{kg} \)
- Initial speed \( v_{\text{initial}} = 1 \, \text{m/s} \)
Using the momentum formula:
[tex]\[ \text{Initial Momentum} = m \times v_{\text{initial}} \][/tex]
[tex]\[ \text{Initial Momentum} = 5 \, \text{kg} \times 1 \, \text{m/s} = 5 \, \text{kg} \cdot \text{m/s} \][/tex]
### Step 3: Final Momentum
Next, let's determine the final momentum of the bowling ball when its speed increases to \(2 \, \text{m/s}\).
Given:
- Mass \( m = 5 \, \text{kg} \)
- Final speed \( v_{\text{final}} = 2 \, \text{m/s} \)
Using the momentum formula:
[tex]\[ \text{Final Momentum} = m \times v_{\text{final}} \][/tex]
[tex]\[ \text{Final Momentum} = 5 \, \text{kg} \times 2 \, \text{m/s} = 10 \, \text{kg} \cdot \text{m/s} \][/tex]
### Step 4: Conclusion
The initial momentum of the bowling ball is \(5 \, \text{kg} \cdot \text{m/s}\), and the final momentum after the speed increase is \(10 \, \text{kg} \cdot \text{m/s}\).
Hence, the correct answer is:
- The initial momentum is [tex]\(5 \, \text{kg} \cdot \text{m/s}\)[/tex], and the final momentum is [tex]\(10 \, \text{kg} \cdot \text{m/s}\)[/tex].
### Step 1: Understanding Momentum
Momentum (\(p\)) is calculated using the formula:
[tex]\[ p = m \times v \][/tex]
where:
- \(m\) is the mass of the object (in kilograms, kg),
- \(v\) is the velocity of the object (in meters per second, m/s).
### Step 2: Initial Momentum
First, let's determine the initial momentum of the bowling ball when its speed is \(1 \, \text{m/s}\).
Given:
- Mass \( m = 5 \, \text{kg} \)
- Initial speed \( v_{\text{initial}} = 1 \, \text{m/s} \)
Using the momentum formula:
[tex]\[ \text{Initial Momentum} = m \times v_{\text{initial}} \][/tex]
[tex]\[ \text{Initial Momentum} = 5 \, \text{kg} \times 1 \, \text{m/s} = 5 \, \text{kg} \cdot \text{m/s} \][/tex]
### Step 3: Final Momentum
Next, let's determine the final momentum of the bowling ball when its speed increases to \(2 \, \text{m/s}\).
Given:
- Mass \( m = 5 \, \text{kg} \)
- Final speed \( v_{\text{final}} = 2 \, \text{m/s} \)
Using the momentum formula:
[tex]\[ \text{Final Momentum} = m \times v_{\text{final}} \][/tex]
[tex]\[ \text{Final Momentum} = 5 \, \text{kg} \times 2 \, \text{m/s} = 10 \, \text{kg} \cdot \text{m/s} \][/tex]
### Step 4: Conclusion
The initial momentum of the bowling ball is \(5 \, \text{kg} \cdot \text{m/s}\), and the final momentum after the speed increase is \(10 \, \text{kg} \cdot \text{m/s}\).
Hence, the correct answer is:
- The initial momentum is [tex]\(5 \, \text{kg} \cdot \text{m/s}\)[/tex], and the final momentum is [tex]\(10 \, \text{kg} \cdot \text{m/s}\)[/tex].
