A body of mass `m/2` moving with velocity `v_0` collides elastically with another mass of `m/3` . Find % change in KE of first body

Options

A. 32
B. 96
C. 34
D. 80

Answer

To solve the problem, we need to find the percentage change in kinetic energy of the first body after an elastic collision with another body. Let's break down the solution step by step.

### Step 1: Understand the initial conditions
- Mass of the first body, \( m_1 = \frac{m}{2} \)
- Initial velocity of the first body, \( v_0 \)
- Mass of the second body, \( m_2 = \frac{m}{3} \)
- Initial velocity of the second body, \( v_2 = 0 \) (since it is at rest)

### Step 2: Calculate the initial kinetic energy of the first body
The initial kinetic energy (KE) of the first body can be calculated using the formula:
\[ KE_{initial} = \frac{1}{2} m_1 v_0^2 = \frac{1}{2} \left(\frac{m}{2}\right) v_0^2 = \frac{m v_0^2}{4} \]

### Step 3: Apply conservation of momentum
Using the conservation of momentum before and after the collision:
\[ m_1 v_0 + m_2 \cdot 0 = m_1 v_1 + m_2 v_2 \]
Substituting the masses:
\[ \frac{m}{2} v_0 = \frac{m}{2} v_1 + \frac{m}{3} v_2 \]
Dividing through by \( m \) (assuming \( m \neq 0 \)):
\[ \frac{1}{2} v_0 = \frac{1}{2} v_1 + \frac{1}{3} v_2 \]
Multiplying through by 6 to eliminate fractions:
\[ 3 v_0 = 3 v_1 + 2 v_2 \quad \text{(Equation 1)} \]

### Step 4: Apply the elastic collision condition
For elastic collisions, the relative velocity of separation equals the relative velocity of approach:
\[ v_2 - v_1 = v_0 \quad \text{(Equation 2)} \]

### Step 5: Solve the equations simultaneously
From Equation 2, we can express \( v_2 \):
\[ v_2 = v_1 + v_0 \]
Substituting \( v_2 \) in Equation 1:
\[ 3 v_0 = 3 v_1 + 2(v_1 + v_0) \]
Expanding and simplifying:
\[ 3 v_0 = 3 v_1 + 2 v_1 + 2 v_0 \]
\[ 3 v_0 - 2 v_0 = 5 v_1 \]
\[ v_0 = 5 v_1 \quad \Rightarrow \quad v_1 = \frac{v_0}{5} \]

### Step 6: Calculate the final kinetic energy of the first body
Now we can find the final kinetic energy of the first body:
\[ KE_{final} = \frac{1}{2} m_1 v_1^2 = \frac{1}{2} \left(\frac{m}{2}\right) \left(\frac{v_0}{5}\right)^2 = \frac{1}{2} \left(\frac{m}{2}\right) \left(\frac{v_0^2}{25}\right) = \frac{m v_0^2}{100} \]

### Step 7: Calculate the percentage change in kinetic energy
Now, we can find the percentage change in kinetic energy:
\[ \text{Percentage Change} = \frac{KE_{initial} - KE_{final}}{KE_{initial}} \times 100 \]
Substituting the values:
\[ \text{Percentage Change} = \frac{\frac{m v_0^2}{4} - \frac{m v_0^2}{100}}{\frac{m v_0^2}{4}} \times 100 \]
Calculating the fraction:
\[ = \frac{\frac{25 m v_0^2}{100} - \frac{m v_0^2}{100}}{\frac{m v_0^2}{4}} \times 100 = \frac{\frac{24 m v_0^2}{100}}{\frac{m v_0^2}{4}} \times 100 \]
\[ = \frac{24}{100} \times 4 \times 100 = 96\% \]

### Final Answer:
The percentage change in kinetic energy of the first body is **96%**.