A projectile (mass = 0.227 kg) is fired at and embeds itself in a target (mass = 1.88 kg). (The target is initially stationary). 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?

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### Problem Statement:

A projectile (mass = 0.227 kg) is fired and embeds itself in a target (mass = 1.88 kg). (The target is initially stationary). 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?

### Explanation and Solution:

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

#### Step 1: Conservation of Momentum

The initial momentum of the system is the momentum of the projectile, as the target is stationary. The final momentum is the combined momentum of the target and the projectile after the collision.

Let's denote:
- Mass of the projectile as \( m_p = 0.227 \) kg
- Mass of the target as \( m_t = 1.88 \) kg
- Initial velocity of the projectile as \( v_p \)
- Final combined velocity of the target and projectile as \( v_f \)

Using the conservation of momentum:
\[ m_p \cdot v_p = (m_p + m_t) \cdot v_f \]

From this, we can solve for \( v_f \):
\[ v_f = \frac{m_p \cdot v_p}{m_p + m_t} \]

#### Step 2: Initial Kinetic Energy of the Projectile

The initial kinetic energy (\( KE_{initial} \)) of the projectile is given by:
\[ KE_{initial} = \frac{1}{2} m_p v_p^2 \]

#### Step 3: Final Kinetic Energy of the Combined System

The final kinetic energy (\( KE_{final} \)) of the combined system is:
\[ KE_{final} = \frac{1}{2} (m_p + m_t) v_f^2 \]

Substituting \( v_f \):
\[ KE_{final} = \frac{1}{2} (m_p + m_t) \left( \frac{m_p \cdot v_p}{m_p + m_t} \right)^2 \]
\[ KE_{final} = \frac{1}{2} \frac{m_p^2 \cdot v_p^2}{m_p + m_t} \]

#### Step 4: Percentage of Kinetic Energy Carried Off

To find the percentage of the projectile's
Transcribed Image Text:### Problem Statement: A projectile (mass = 0.227 kg) is fired and embeds itself in a target (mass = 1.88 kg). (The target is initially stationary). 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? ### Explanation and Solution: To solve this problem, we will use the principles of conservation of momentum and kinetic energy. #### Step 1: Conservation of Momentum The initial momentum of the system is the momentum of the projectile, as the target is stationary. The final momentum is the combined momentum of the target and the projectile after the collision. Let's denote: - Mass of the projectile as \( m_p = 0.227 \) kg - Mass of the target as \( m_t = 1.88 \) kg - Initial velocity of the projectile as \( v_p \) - Final combined velocity of the target and projectile as \( v_f \) Using the conservation of momentum: \[ m_p \cdot v_p = (m_p + m_t) \cdot v_f \] From this, we can solve for \( v_f \): \[ v_f = \frac{m_p \cdot v_p}{m_p + m_t} \] #### Step 2: Initial Kinetic Energy of the Projectile The initial kinetic energy (\( KE_{initial} \)) of the projectile is given by: \[ KE_{initial} = \frac{1}{2} m_p v_p^2 \] #### Step 3: Final Kinetic Energy of the Combined System The final kinetic energy (\( KE_{final} \)) of the combined system is: \[ KE_{final} = \frac{1}{2} (m_p + m_t) v_f^2 \] Substituting \( v_f \): \[ KE_{final} = \frac{1}{2} (m_p + m_t) \left( \frac{m_p \cdot v_p}{m_p + m_t} \right)^2 \] \[ KE_{final} = \frac{1}{2} \frac{m_p^2 \cdot v_p^2}{m_p + m_t} \] #### Step 4: Percentage of Kinetic Energy Carried Off To find the percentage of the projectile's
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