A ball of mass `m` makes head-on elastic collision with a ball of mass nm which is initially at rest. Show that the fractional transfer of energy by the first ball is `4n/(1 + n)^(2)`. Deduce the value of `n` for which the transfer is maximum.

Explanation

To solve the problem of a head-on elastic collision between two balls, we will follow these steps:

### Step 1: Define the System
Let:
- Mass of the first ball = \( m \)
- Mass of the second ball = \( nm \) (where \( n \) is a constant)
- Initial velocity of the first ball = \( u \)
- Initial velocity of the second ball = \( 0 \)

### Step 2: Apply Conservation of Momentum
According to the conservation of momentum, the total momentum before the collision is equal to the total momentum after the collision.

Before collision:
\[ \text{Initial Momentum} = mu + nm \cdot 0 = mu \]

After collision, let \( v_1 \) be the final velocity of the first ball and \( v_2 \) be the final velocity of the second ball:
\[ \text{Final Momentum} = mv_1 + nm v_2 \]

Setting initial momentum equal to final momentum:
\[ mu = mv_1 + nm v_2 \quad \text{(1)} \]

### Step 3: Apply Conservation of Kinetic Energy
For elastic collisions, the total kinetic energy before the collision is equal to the total kinetic energy after the collision.

Before collision:
\[ \text{Initial Kinetic Energy} = \frac{1}{2} mu^2 + 0 = \frac{1}{2} mu^2 \]

After collision:
\[ \text{Final Kinetic Energy} = \frac{1}{2} mv_1^2 + \frac{1}{2} nm v_2^2 \]

Setting initial kinetic energy equal to final kinetic energy:
\[ \frac{1}{2} mu^2 = \frac{1}{2} mv_1^2 + \frac{1}{2} nm v_2^2 \quad \text{(2)} \]

### Step 4: Solve the Equations
From equation (1):
\[ mu = mv_1 + nm v_2 \implies u = v_1 + nv_2 \quad \text{(3)} \]

From equation (2):
\[ mu^2 = mv_1^2 + nm v_2^2 \quad \text{(4)} \]

### Step 5: Express \( v_1 \) and \( v_2 \) in Terms of \( u \)
From equation (3), we can express \( v_1 \):
\[ v_1 = u - nv_2 \quad \text{(5)} \]

Substituting equation (5) into equation (4):
\[ mu^2 = m(u - nv_2)^2 + nm v_2^2 \]

Expanding the equation:
\[ mu^2 = m(u^2 - 2unv_2 + n^2v_2^2) + nm v_2^2 \]
\[ mu^2 = mu^2 - 2munv_2 + mn^2v_2^2 + nm v_2^2 \]
\[ 0 = -2munv_2 + mn^2v_2^2 + nm v_2^2 \]
\[ 0 = v_2(-2mu + mn^2v_2 + nm v_2) \]

### Step 6: Solve for \( v_2 \)
Factoring out \( v_2 \):
\[ v_2( -2mu + mn(n+1)v_2) = 0 \]
This gives us:
\[ v_2 = 0 \quad \text{or} \quad -2mu + mn(n+1)v_2 = 0 \]
Solving for \( v_2 \):
\[ v_2 = \frac{2u}{n+1} \]

### Step 7: Substitute \( v_2 \) Back to Find \( v_1 \)
Using equation (5):
\[ v_1 = u - n\left(\frac{2u}{n+1}\right) = u\left(1 - \frac{2n}{n+1}\right) = \frac{u(1-n)}{n+1} \]

### Step 8: Calculate the Kinetic Energies
Initial kinetic energy:
\[ KE_i = \frac{1}{2} mu^2 \]

Final kinetic energy:
\[ KE_f = \frac{1}{2} m\left(\frac{u(1-n)}{n+1}\right)^2 + \frac{1}{2} nm\left(\frac{2u}{n+1}\right)^2 \]
Calculating \( KE_f \):
\[ KE_f = \frac{1}{2} m \frac{u^2(1-n)^2}{(n+1)^2} + \frac{1}{2} nm \frac{4u^2}{(n+1)^2} \]
\[ KE_f = \frac{mu^2}{2(n+1)^2} \left((1-n)^2 + 4n\right) \]

### Step 9: Calculate Fractional Energy Transfer
The fractional transfer of energy is given by:
\[ \text{Fractional Transfer} = \frac{KE_i - KE_f}{KE_i} \]
Substituting the values:
\[ \text{Fractional Transfer} = \frac{\frac{1}{2} mu^2 - \frac{mu^2}{2(n+1)^2} \left((1-n)^2 + 4n\right)}{\frac{1}{2} mu^2} \]
\[ = 1 - \frac{(1-n)^2 + 4n}{(n+1)^2} \]
\[ = \frac{4n}{(n+1)^2} \]

### Step 10: Maximize the Fractional Transfer
To find the value of \( n \) for maximum transfer, we differentiate:
\[ f(n) = \frac{4n}{(n+1)^2} \]
Setting the derivative to zero and solving gives \( n = 1 \) for maximum energy transfer.

### Final Result
The fractional transfer of energy by the first ball is:
\[ \frac{4n}{(n+1)^2} \]
The value of \( n \) for which the transfer is maximum is:
\[ n = 1 \]