In a ballistic pendulum experiment a ball of mass 46 g is fired into a pendulum of mass 199 g. Calculate the fractional loss of energy.

In a ballistic pendulum experiment, the fractional loss of energy refers to the ratio of the difference in kinetic energy before and after the collision to the initial kinetic energy. This loss happens because the collision between the ball and the pendulum is inelastic.

### Step 1: Understanding the process
- **Mass of the ball** \( m_b = 46 \, \text{g} = 0.046 \, \text{kg} \)
- **Mass of the pendulum** \( m_p = 199 \, \text{g} = 0.199 \, \text{kg} \)
- **Initial velocity of the ball**: Let's denote this as \( v_b \).
- The pendulum is initially at rest, so its velocity \( v_p = 0 \).

### Step 2: Conservation of momentum (during the collision)
Since momentum is conserved during the inelastic collision, the total momentum before and after the collision must be equal. After the collision, the ball and pendulum move together with a common velocity \( v_f \).

\[
m_b v_b = (m_b + m_p) v_f
\]

Solving for \( v_f \):

\[
v_f = \frac{m_b v_b}{m_b + m_p}
\]

### Step 3: Energy before and after the collision

1. **Initial kinetic energy** (before the collision):
The kinetic energy of the ball before the collision is given by:

\[
KE_{\text{initial}} = \frac{1}{2} m_b v_b^2
\]

2. **Final kinetic energy** (after the collision):
After the collision, both the ball and the pendulum move together, and the final kinetic energy is:

\[
KE_{\text{final}} = \frac{1}{2} (m_b + m_p) v_f^2
\]

Substituting \( v_f = \frac{m_b v_b}{m_b + m_p} \):

\[
KE_{\text{final}} = \frac{1}{2} (m_b + m_p) \left( \frac{m_b v_b}{m_b + m_p} \right)^2 = \frac{1}{2} \frac{m_b^2 v_b^2}{m_b + m_p}
\]

### Step 4: Fractional loss of energy

The fractional loss of energy is the difference between the initial and final kinetic energy divided by the initial kinetic energy:

\[
\text{Fractional Loss} = \frac{KE_{\text{initial}} - KE_{\text{final}}}{KE_{\text{initial}}}
\]

Substituting the expressions for \( KE_{\text{initial}} \) and \( KE_{\text{final}} \):

\[
\text{Fractional Loss} = \frac{\frac{1}{2} m_b v_b^2 - \frac{1}{2} \frac{m_b^2 v_b^2}{m_b + m_p}}{\frac{1}{2} m_b v_b^2}
\]

Simplifying:

\[
\text{Fractional Loss} = 1 - \frac{m_b}{m_b + m_p}
\]

Substitute the values for \( m_b = 0.046 \, \text{kg} \) and \( m_p = 0.199 \, \text{kg} \):

\[
\text{Fractional Loss} = 1 - \frac{0.046}{0.046 + 0.199} = 1 - \frac{0.046}{0.245} \approx 1 - 0.1878 = 0.8122
\]

Thus, the **fractional loss of energy** is approximately **0.812**, or **81.2%**.

### Conclusion:
The ballistic pendulum experiment results in a fractional energy loss of about 81.2%. This loss occurs due to the inelastic nature of the collision.

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Do you want further clarification or details on any step?

Here are 5 relative questions:
1. What factors contribute to the loss of kinetic energy in an inelastic collision?
2. How would the result change if the mass of the ball was doubled?
3. How is momentum conserved in inelastic collisions?
4. What is the difference between elastic and inelastic collisions in terms of energy conservation?
5. How could this experiment be used to calculate the initial velocity of the ball?

**Tip:** In an inelastic collision, kinetic energy is not conserved, but momentum always is.

Learn how to calculate the fractional loss of energy in a ballistic pendulum experiment where a 46 g ball is fired into a 199 g pendulum. The solution uses concepts of conservation of momentum and kinetic energy in inelastic collisions, with an 81.2% energy loss.

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