An impulse `vec(I)` changes the velocity of a particle from `vec(v)_(1)` to `vec(v)_(2)`. Kinetic energy gained by the particle is :-

Options

A. `(1)/(2)vec(I) .(vec(v)_(1) + vec(v)_(2))`
B. `(1)/(2)vec(I).(vec(v)_(1) - vec(v)_(2))`
C. `vec(I).(vec(v)_(1) - vec(v)_(2))`
D. `vec(I).(vec(v)_(1) + vec(v)_(2))`

Explanation

To solve the problem of finding the kinetic energy gained by a particle when its velocity changes from \(\vec{v}_1\) to \(\vec{v}_2\) due to an impulse \(\vec{I}\), we can follow these steps:

### Step 1: Understand the formula for kinetic energy
The kinetic energy (KE) of a particle is given by the formula:
\[ KE = \frac{1}{2} mv^2 \]
where \(m\) is the mass of the particle and \(v\) is its velocity.

### Step 2: Write the expression for initial and final kinetic energy
The initial kinetic energy (\(KE_1\)) when the particle has velocity \(\vec{v}_1\) is:
\[ KE_1 = \frac{1}{2} m v_1^2 \]
The final kinetic energy (\(KE_2\)) when the particle has velocity \(\vec{v}_2\) is:
\[ KE_2 = \frac{1}{2} m v_2^2 \]

### Step 3: Calculate the change in kinetic energy
The change in kinetic energy (\(\Delta KE\)) is given by:
\[ \Delta KE = KE_2 - KE_1 = \frac{1}{2} m v_2^2 - \frac{1}{2} m v_1^2 \]
This can be factored as:
\[ \Delta KE = \frac{1}{2} m (v_2^2 - v_1^2) \]

### Step 4: Use the identity for the difference of squares
We can use the identity \(a^2 - b^2 = (a - b)(a + b)\) to rewrite the expression:
\[ v_2^2 - v_1^2 = (v_2 - v_1)(v_2 + v_1) \]
Thus, we can express the change in kinetic energy as:
\[ \Delta KE = \frac{1}{2} m (v_2 - v_1)(v_2 + v_1) \]

### Step 5: Relate impulse to change in momentum
Impulse \(\vec{I}\) is defined as the change in momentum:
\[ \vec{I} = m \vec{v}_2 - m \vec{v}_1 = m(\vec{v}_2 - \vec{v}_1) \]
From this, we can express the change in velocity in terms of impulse:
\[ \vec{v}_2 - \vec{v}_1 = \frac{\vec{I}}{m} \]

### Step 6: Substitute impulse into the kinetic energy equation
Substituting \(\vec{v}_2 - \vec{v}_1\) into the change in kinetic energy equation gives:
\[ \Delta KE = \frac{1}{2} m \left(\frac{\vec{I}}{m}\right)(\vec{v}_2 + \vec{v}_1) \]
This simplifies to:
\[ \Delta KE = \frac{1}{2} \vec{I} (\vec{v}_2 + \vec{v}_1) \]

### Conclusion
Thus, the kinetic energy gained by the particle is:
\[ \Delta KE = \frac{1}{2} \vec{I} (\vec{v}_2 + \vec{v}_1) \]