A smooth sphere of mass M moving with velocity u directly collides elastically with another sphere of mass m at rest. After collision their final velocities are V and v respectively. The value of v is

Options

A. `(2uM)/(m)`
B. `(2um)/(M)`
C. `(2u)(1+(m)/(M))`
D. `(2u)/(1+(m)/(M))`

Answer

To solve the problem of two spheres colliding elastically, we will use the principles of conservation of momentum and conservation of kinetic energy. Let's go through the solution step by step.

### Step 1: Write down the conservation of momentum equation.
Before the collision, the momentum of the system is given by the mass of the moving sphere multiplied by its velocity, while the second sphere is at rest.

\[ \text{Initial Momentum} = M \cdot u + m \cdot 0 = M \cdot u \]

After the collision, the momentum is given by the sum of the momenta of both spheres:

\[ \text{Final Momentum} = M \cdot V + m \cdot v \]

Setting these equal gives us the equation:

\[ M \cdot u = M \cdot V + m \cdot v \quad \text{(1)} \]

### Step 2: Write down the conservation of kinetic energy equation.
The initial kinetic energy before the collision is:

\[ \text{Initial Kinetic Energy} = \frac{1}{2} M u^2 + 0 = \frac{1}{2} M u^2 \]

After the collision, the kinetic energy is:

\[ \text{Final Kinetic Energy} = \frac{1}{2} M V^2 + \frac{1}{2} m v^2 \]

Setting these equal gives us the equation:

\[ \frac{1}{2} M u^2 = \frac{1}{2} M V^2 + \frac{1}{2} m v^2 \quad \text{(2)} \]

### Step 3: Simplify the equations.
From equation (1), we can express \(V\) in terms of \(u\) and \(v\):

\[ V = \frac{M u - m v}{M} \quad \text{(3)} \]

Now substitute equation (3) into equation (2):

\[ M u^2 = M \left(\frac{M u - m v}{M}\right)^2 + m v^2 \]

### Step 4: Expand and simplify.
Expanding the squared term:

\[ M u^2 = \frac{(M u - m v)^2}{M} + m v^2 \]

This expands to:

\[ M u^2 = \frac{M^2 u^2 - 2M m u v + m^2 v^2}{M} + m v^2 \]

Multiplying through by \(M\) to eliminate the fraction:

\[ M^2 u^2 = M^2 u^2 - 2M m u v + m^2 v^2 + M m v^2 \]

### Step 5: Rearranging terms.
Rearranging gives:

\[ 0 = -2M m u v + (m^2 + M m) v^2 \]

Factoring out \(v\):

\[ 0 = v(-2M m u + (m + M)v) \]

This gives us two solutions: \(v = 0\) or:

\[ v = \frac{2M u}{m + M} \]

### Final Answer:
Thus, the final velocity \(v\) of the smaller mass after the collision is:

\[ v = \frac{2M u}{m + M} \]