A light and a heavy body have equal momentum. Which one has greater K.E. ?

Options

A. the light body
B. both have equal K.E.
C. the heavy body
D. data given in incomplete

Answer

To solve the problem, we need to analyze the kinetic energy of two bodies (a light body and a heavy body) that have equal momentum. Let's denote:

- Mass of the lighter body: \( m_1 \)
- Mass of the heavier body: \( m_2 \) (where \( m_2 > m_1 \))
- Velocity of the lighter body: \( v_1 \)
- Velocity of the heavier body: \( v_2 \)

### Step 1: Write the expression for momentum
The momentum \( p \) of an object is given by the formula:
\[ p = mv \]
For the lighter body:
\[ p_1 = m_1 v_1 \]
For the heavier body:
\[ p_2 = m_2 v_2 \]
Since both bodies have equal momentum, we can set their momenta equal to each other:
\[ m_1 v_1 = m_2 v_2 \]

### Step 2: Express kinetic energy
The kinetic energy (K.E.) of an object is given by the formula:
\[ K.E. = \frac{1}{2} mv^2 \]
For the lighter body, the kinetic energy \( K.E. \) is:
\[ K.E._1 = \frac{1}{2} m_1 v_1^2 \]
For the heavier body, the kinetic energy is:
\[ K.E._2 = \frac{1}{2} m_2 v_2^2 \]

### Step 3: Substitute velocity from momentum equation
From the momentum equation \( m_1 v_1 = m_2 v_2 \), we can express \( v_1 \) in terms of \( v_2 \):
\[ v_1 = \frac{m_2}{m_1} v_2 \]

### Step 4: Substitute \( v_1 \) in kinetic energy of lighter body
Now, substitute \( v_1 \) in the kinetic energy of the lighter body:
\[ K.E._1 = \frac{1}{2} m_1 \left(\frac{m_2}{m_1} v_2\right)^2 \]
This simplifies to:
\[ K.E._1 = \frac{1}{2} m_1 \frac{m_2^2}{m_1^2} v_2^2 = \frac{1}{2} \frac{m_2^2}{m_1} v_2^2 \]

### Step 5: Write kinetic energy of heavier body
The kinetic energy of the heavier body remains:
\[ K.E._2 = \frac{1}{2} m_2 v_2^2 \]

### Step 6: Compare the two kinetic energies
Now we will compare \( K.E._1 \) and \( K.E._2 \):
\[ \frac{K.E._1}{K.E._2} = \frac{\frac{1}{2} \frac{m_2^2}{m_1} v_2^2}{\frac{1}{2} m_2 v_2^2} = \frac{m_2^2}{m_1 m_2} = \frac{m_2}{m_1} \]
Since \( m_2 > m_1 \), it follows that:
\[ \frac{K.E._1}{K.E._2} > 1 \implies K.E._1 > K.E._2 \]

### Conclusion
Thus, the kinetic energy of the lighter body is greater than that of the heavier body. Therefore, the answer is:

**The lighter body has greater kinetic energy.**