A particle of mass `m_(1)` makes an elastic one dimensional collision with a stationary particle of mass `m_(2)`.What fraction of the kinetic energy of `m_(1)` is carried away by `m_(2)`?

Options

A. `m_(1)/m_(2)`
B. `m_(2)/m_(1)`
C. `(2m_(1)m_(2))/((m_(1)+m_(2))^(2))`
D. `(4m_(1)m_(2))/((m_(1)+m_(2))^(2))`

Answer

To solve the problem of finding the fraction of kinetic energy of mass \( m_1 \) that is carried away by mass \( m_2 \) after an elastic collision, we can follow these steps:

### Step 1: Understand the Initial Conditions
- Let the mass of the first particle be \( m_1 \) and its initial velocity be \( v \).
- The second particle has a mass \( m_2 \) and is initially stationary, so its initial velocity \( u_2 = 0 \).

### Step 2: Calculate Initial Kinetic Energy of \( m_1 \)
- The initial kinetic energy \( KE_1 \) of mass \( m_1 \) is given by:
\[ KE_1 = \frac{1}{2} m_1 v^2 \]

### Step 3: Use the Formula for Final Velocities in Elastic Collisions
- For an elastic collision, the final velocities can be calculated using the following formulas:
- Final velocity of \( m_1 \):
\[ v_1' = \frac{(m_1 - m_2)}{(m_1 + m_2)} v \]
- Final velocity of \( m_2 \):
\[ v_2' = \frac{2 m_1}{(m_1 + m_2)} v \]

### Step 4: Calculate Final Kinetic Energy of \( m_2 \)
- The final kinetic energy \( KE_2 \) of mass \( m_2 \) after the collision is given by:
\[ KE_2 = \frac{1}{2} m_2 (v_2')^2 \]
- Substitute \( v_2' \):
\[ KE_2 = \frac{1}{2} m_2 \left(\frac{2 m_1}{(m_1 + m_2)} v\right)^2 \]
- Simplifying this gives:
\[ KE_2 = \frac{1}{2} m_2 \cdot \frac{4 m_1^2}{(m_1 + m_2)^2} v^2 = \frac{2 m_1^2 m_2}{(m_1 + m_2)^2} v^2 \]

### Step 5: Calculate the Fraction of Kinetic Energy Transferred
- The fraction of kinetic energy transferred to \( m_2 \) is given by:
\[ \text{Fraction} = \frac{KE_2}{KE_1} \]
- Substitute the expressions for \( KE_2 \) and \( KE_1 \):
\[ \text{Fraction} = \frac{\frac{2 m_1^2 m_2}{(m_1 + m_2)^2} v^2}{\frac{1}{2} m_1 v^2} \]
- Simplifying this gives:
\[ \text{Fraction} = \frac{2 m_1^2 m_2}{(m_1 + m_2)^2} \cdot \frac{2}{m_1} = \frac{4 m_1 m_2}{(m_1 + m_2)^2} \]

### Final Answer
The fraction of the kinetic energy of \( m_1 \) that is carried away by \( m_2 \) is:
\[ \frac{4 m_1 m_2}{(m_1 + m_2)^2} \]

---