A projectile (mass = 0.20 kg) is fired at and embeds itself in a target (mass = 2.50 kg). The target (with the projectile in it) flies off after being struck. What percentage of the projectile's incident kinetic energy does the target (with the projectile in it) carry off after being struck?

Options

A. `2.1%`
B. `5.0%`
C. `3.8%`
D. `7.4%`

Answer

To solve the problem step by step, we will use the principles of conservation of momentum and kinetic energy.

### Step 1: Understand the System
We have a projectile with mass \( m_p = 0.20 \, \text{kg} \) that embeds itself into a target with mass \( m_t = 2.50 \, \text{kg} \). After the collision, they move together with a common velocity \( V' \).

### Step 2: Conservation of Momentum
Since no external forces are acting on the system, momentum is conserved. The initial momentum of the system is equal to the final momentum.

\[ m_p \cdot V = (m_p + m_t) \cdot V' \]

Where \( V \) is the initial velocity of the projectile.

### Step 3: Solve for Final Velocity \( V' \)
Rearranging the momentum conservation equation gives us:

\[ V' = \frac{m_p \cdot V}{m_p + m_t} \]

### Step 4: Calculate Initial Kinetic Energy of the Projectile
The initial kinetic energy \( KE_i \) of the projectile is given by:

\[ KE_i = \frac{1}{2} m_p V^2 \]

### Step 5: Calculate Final Kinetic Energy of the Combined Mass
The final kinetic energy \( KE_f \) of the target and projectile together is:

\[ KE_f = \frac{1}{2} (m_p + m_t) V'^2 \]

Substituting \( V' \) from Step 3 into this equation:

\[ KE_f = \frac{1}{2} (m_p + m_t) \left( \frac{m_p \cdot V}{m_p + m_t} \right)^2 \]

### Step 6: Simplify the Final Kinetic Energy Expression
\[ KE_f = \frac{1}{2} (m_p + m_t) \cdot \frac{m_p^2 \cdot V^2}{(m_p + m_t)^2} \]

\[ KE_f = \frac{1}{2} \cdot \frac{m_p^2 \cdot V^2}{m_p + m_t} \]

### Step 7: Calculate the Percentage of Kinetic Energy Carried Off
To find the percentage of the initial kinetic energy that the target carries off after being struck, we calculate:

\[ \text{Percentage} = \left( \frac{KE_f}{KE_i} \right) \times 100 \]

Substituting the expressions for \( KE_f \) and \( KE_i \):

\[ \text{Percentage} = \left( \frac{\frac{1}{2} \cdot \frac{m_p^2 \cdot V^2}{m_p + m_t}}{\frac{1}{2} m_p V^2} \right) \times 100 \]

### Step 8: Simplify the Percentage Expression
\[ \text{Percentage} = \left( \frac{m_p^2}{m_p(m_p + m_t)} \right) \times 100 \]

\[ \text{Percentage} = \left( \frac{m_p}{m_p + m_t} \right) \times 100 \]

### Step 9: Substitute the Values
Substituting \( m_p = 0.20 \, \text{kg} \) and \( m_t = 2.50 \, \text{kg} \):

\[ \text{Percentage} = \left( \frac{0.20}{0.20 + 2.50} \right) \times 100 \]

\[ \text{Percentage} = \left( \frac{0.20}{2.70} \right) \times 100 \approx 7.41\% \]

### Final Answer
The target (with the projectile in it) carries off approximately **7.41%** of the projectile's incident kinetic energy after being struck.

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