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Answer:
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Explanation:
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The period of the simple harmonic motion (SHM) is given by the formula:
\[ T = 2π\sqrt{\frac{m}{k}} \]
Where:
- \( T \) = period of the motion
- \( m \) = mass of the block
- \( k \) = spring constant
To find the period, we first need to find the spring constant. The potential energy stored in the spring at maximum compression is equal to the kinetic energy of the bullet just before it embeds itself in the block.
The potential energy stored in the spring (\( U \)) is given by:
\[ U = \frac{1}{2} k x^2 \]
Where:
- \( k \) = spring constant
- \( x \) = maximum compression distance
The kinetic energy of the bullet (\( KE \)) just before embedding itself in the block is given by:
\[ KE = \frac{1}{2} m v^2 \]
Where:
- \( m \) = mass of the bullet
- \( v \) = initial velocity of the bullet
Setting the potential energy equal to the kinetic energy, we can solve for \( k \):
\[ \frac{1}{2} k x^2 = \frac{1}{2} m v^2 \]
Solving for \( k \) gives:
\[ k = \frac{m v^2}{x^2} \]
Now that we have \( k \), we can find the period \( T \).
Given:
- Mass of the block, \( m = 0.992 \, \text{kg} \)
- Initial velocity of the bullet, \( v = 280 \, \text{m/s} \)
- Maximum compression distance, \( x = 17.0 \, \text{cm} = 0.17 \, \text{m} \)
- Mass of the bullet, \( 8.00 \, \text{g} = 0.008 \, \text{kg} \)
After calculating \( k \), we can plug it into the formula for the period of the SHM to find the period \( T \).
Using the provided values, we can calculate the spring constant \( k \) as follows:
\[ k = \frac{0.008 \, \text{kg} \times (280 \, \text{m/s})^2}{(0.17 \, \text{m})^2} \]
\[ k \approx 75294.12 \, \text{N/m} \]
Now, we can use this value to calculate the period of the simple harmonic motion (SHM):
\[ T = 2π\sqrt{\frac{0.992 \, \text{kg}}{75294.12 \, \text{N/m}}} \]
\[ T = 2π\sqrt{1.3191 \times 10^{-5}} \]
\[ T \approx 0.005755 \, \text{s} \]
So, the period of the motion is approximately 0.005755 seconds.
\[ T = 2π\sqrt{\frac{m}{k}} \]
Where:
- \( T \) = period of the motion
- \( m \) = mass of the block
- \( k \) = spring constant
To find the period, we first need to find the spring constant. The potential energy stored in the spring at maximum compression is equal to the kinetic energy of the bullet just before it embeds itself in the block.
The potential energy stored in the spring (\( U \)) is given by:
\[ U = \frac{1}{2} k x^2 \]
Where:
- \( k \) = spring constant
- \( x \) = maximum compression distance
The kinetic energy of the bullet (\( KE \)) just before embedding itself in the block is given by:
\[ KE = \frac{1}{2} m v^2 \]
Where:
- \( m \) = mass of the bullet
- \( v \) = initial velocity of the bullet
Setting the potential energy equal to the kinetic energy, we can solve for \( k \):
\[ \frac{1}{2} k x^2 = \frac{1}{2} m v^2 \]
Solving for \( k \) gives:
\[ k = \frac{m v^2}{x^2} \]
Now that we have \( k \), we can find the period \( T \).
Given:
- Mass of the block, \( m = 0.992 \, \text{kg} \)
- Initial velocity of the bullet, \( v = 280 \, \text{m/s} \)
- Maximum compression distance, \( x = 17.0 \, \text{cm} = 0.17 \, \text{m} \)
- Mass of the bullet, \( 8.00 \, \text{g} = 0.008 \, \text{kg} \)
After calculating \( k \), we can plug it into the formula for the period of the SHM to find the period \( T \).
Using the provided values, we can calculate the spring constant \( k \) as follows:
\[ k = \frac{0.008 \, \text{kg} \times (280 \, \text{m/s})^2}{(0.17 \, \text{m})^2} \]
\[ k \approx 75294.12 \, \text{N/m} \]
Now, we can use this value to calculate the period of the simple harmonic motion (SHM):
\[ T = 2π\sqrt{\frac{0.992 \, \text{kg}}{75294.12 \, \text{N/m}}} \]
\[ T = 2π\sqrt{1.3191 \times 10^{-5}} \]
\[ T \approx 0.005755 \, \text{s} \]
So, the period of the motion is approximately 0.005755 seconds.
