A ball of mass 'm' moving with speed 'u' undergoes a head-on elastic collision with a ball of mass 'nm' initially at rest. Find the fraction of the incident energy transferred to the second ball.

Options

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

Explanation

To solve the problem of finding the fraction of the incident energy transferred to the second ball during a head-on elastic collision, we can follow these steps:

### Step 1: Apply Conservation of Momentum
In a head-on elastic collision, the total momentum before the collision is equal to the total momentum after the collision.

Let:
- Mass of the first ball = \( m \)
- Initial speed of the first ball = \( u \)
- Mass of the second ball = \( nm \) (initially at rest)
- Final speed of the first ball = \( v_1 \)
- Final speed of the second ball = \( v_2 \)

Using the conservation of momentum:
\[ m \cdot u + nm \cdot 0 = m \cdot v_1 + nm \cdot v_2 \]
This simplifies to:
\[ u = v_1 + n v_2 \quad \text{(Equation 1)} \]

### Step 2: Apply the Coefficient of Restitution
For an elastic collision, the coefficient of restitution \( e \) is equal to 1. This means that the relative velocity of separation is equal to the relative velocity of approach.

Using the definition of the coefficient of restitution:
\[ e = \frac{\text{Velocity of separation}}{\text{Velocity of approach}} = 1 \]
Thus, we have:
\[ v_2 - v_1 = u \quad \text{(Equation 2)} \]

### Step 3: Solve the Equations Simultaneously
Now we can solve Equations 1 and 2 simultaneously.

From Equation 2:
\[ v_2 = u + v_1 \]

Substituting this expression for \( v_2 \) into Equation 1:
\[ u = v_1 + n(u + v_1) \]
\[ u = v_1 + nu + n v_1 \]
\[ u = (1 + n)v_1 + nu \]
Rearranging gives:
\[ u - nu = (1 + n)v_1 \]
\[ u(1 - n) = (1 + n)v_1 \]
Thus, we find:
\[ v_1 = \frac{u(1 - n)}{1 + n} \quad \text{(Equation 3)} \]

### Step 4: Substitute \( v_1 \) back to find \( v_2 \)
Now substitute \( v_1 \) back into Equation 2 to find \( v_2 \):
\[ v_2 = u + \frac{u(1 - n)}{1 + n} \]
\[ v_2 = u\left(1 + \frac{1 - n}{1 + n}\right) \]
\[ v_2 = u\left(\frac{(1 + n) + (1 - n)}{1 + n}\right) \]
\[ v_2 = u\left(\frac{2}{1 + n}\right) \]

### Step 5: Calculate the Kinetic Energies
Now, we can calculate the kinetic energies before and after the collision.

- Initial kinetic energy of the first ball:
\[ KE_{\text{initial}} = \frac{1}{2} m u^2 \]

- Final kinetic energy of the second ball:
\[ KE_{\text{final}} = \frac{1}{2} nm v_2^2 = \frac{1}{2} nm \left(u \cdot \frac{2}{1 + n}\right)^2 \]
\[ KE_{\text{final}} = \frac{1}{2} nm \cdot \frac{4u^2}{(1 + n)^2} \]
\[ KE_{\text{final}} = \frac{2nm u^2}{(1 + n)^2} \]

### Step 6: Find the Fraction of Energy Transferred
The fraction of the incident energy transferred to the second ball is given by:
\[ \text{Fraction} = \frac{KE_{\text{final}}}{KE_{\text{initial}}} = \frac{\frac{2nm u^2}{(1 + n)^2}}{\frac{1}{2} m u^2} \]
\[ = \frac{2nm u^2}{(1 + n)^2} \cdot \frac{2}{m u^2} \]
\[ = \frac{4n}{(1 + n)^2} \]

### Final Answer
Thus, the fraction of the incident energy transferred to the second ball is:
\[ \frac{4n}{(1 + n)^2} \]