A ball of mass '2m' moving with velocity `u hat(i)` collides with another ball of mass 'm' moving with velocity `- 2 u hat(i)`. After the collision, mass 2m moves with velocity `(u)/(2) ( hat(i) - hat(j))`. Then the change in kinetic energy of the system ( both balls ) is

Options

A. zero
B. `(3)/(2) m u^(2)`
C. `(3)/(2) m u^(2) ( 1- ( 1)/( 2 sqrt(2)))`
D. `(3)/(2) m u^(2) ( 1+ ( 1)/( 2 sqrt(2)))`

Answer

To solve the problem, we need to find the change in kinetic energy of the system after the collision of two balls. Let's break down the solution step by step.

### Step 1: Identify the initial conditions
- Mass of the first ball (m1) = 2m, initial velocity (u1) = u î
- Mass of the second ball (m2) = m, initial velocity (u2) = -2u î

### Step 2: Calculate the initial kinetic energy (KE_initial)
The kinetic energy (KE) of an object is given by the formula:
\[ KE = \frac{1}{2} m v^2 \]

For the first ball (mass = 2m):
\[ KE_{1} = \frac{1}{2} (2m) (u^2) = mu^2 \]

For the second ball (mass = m):
\[ KE_{2} = \frac{1}{2} (m) ((-2u)^2) = \frac{1}{2} (m) (4u^2) = 2mu^2 \]

Thus, the total initial kinetic energy is:
\[ KE_{initial} = KE_{1} + KE_{2} = mu^2 + 2mu^2 = 3mu^2 \]

### Step 3: Identify the final conditions
After the collision, the first ball (mass = 2m) moves with a velocity of:
\[ v_{1} = \frac{u}{2} (î - ĵ) \]
The second ball (mass = m) has an unknown final velocity (let's denote it as \(v_{2}\)).

### Step 4: Calculate the final kinetic energy (KE_final)
First, we need to find the final velocity of the second ball using the conservation of momentum.

#### Step 4.1: Apply conservation of momentum
Initial momentum:
\[ p_{initial} = m_{1} u_{1} + m_{2} u_{2} = (2m)(u) + (m)(-2u) = 2mu - 2mu = 0 \]

Final momentum:
\[ p_{final} = m_{1} v_{1} + m_{2} v_{2} = (2m) \left(\frac{u}{2} (î - ĵ)\right) + m v_{2} \]

Setting initial momentum equal to final momentum:
\[ 0 = (2m) \left(\frac{u}{2} (î - ĵ)\right) + m v_{2} \]

This simplifies to:
\[ 0 = mu (î - ĵ) + m v_{2} \]

Dividing through by m:
\[ 0 = u (î - ĵ) + v_{2} \]

Thus:
\[ v_{2} = -u (î - ĵ) = -u î + u ĵ \]

#### Step 4.2: Calculate the final kinetic energy
Now we can calculate the final kinetic energy for both balls.

For the first ball (mass = 2m):
\[ KE_{1, final} = \frac{1}{2} (2m) \left(\frac{u}{2}\right)^2 + \left(-\frac{u}{2}\right)^2 = \frac{1}{2} (2m) \left(\frac{u^2}{4} + \frac{u^2}{4}\right) = \frac{1}{2} (2m) \left(\frac{u^2}{2}\right) = mu^2 \]

For the second ball (mass = m):
\[ KE_{2, final} = \frac{1}{2} m \left(\sqrt{(-u)^2 + u^2}\right)^2 = \frac{1}{2} m (2u^2) = mu^2 \]

Thus, the total final kinetic energy is:
\[ KE_{final} = KE_{1, final} + KE_{2, final} = mu^2 + mu^2 = 2mu^2 \]

### Step 5: Calculate the change in kinetic energy
The change in kinetic energy is given by:
\[ \Delta KE = KE_{final} - KE_{initial} = 2mu^2 - 3mu^2 = -mu^2 \]

### Final Answer
The change in kinetic energy of the system is:
\[ \Delta KE = -mu^2 \]