Question

A light body of momentum \( P_L \) and a heavy body of momentum \( P_H \), both have the same kinetic energy, then:

• A. \( P_L > P_H \)
• B. \( P_L < P_H \)
• C. \( P_L = P_H \)
• D. \( P_H = 2 P_L \)

Hint:

When two bodies have the same kinetic energy, the body with the larger mass will have a smaller momentum, since momentum is proportional to the square root of mass when kinetic energy is fixed.

Answer 1

The Correct Option is B

Given that both bodies have the same kinetic energy, we know that: \[ KE = \frac{P^2}{2m} \] Since the momentum of the two bodies is \( P_L \) and \( P_H \), and both have the same kinetic energy: \[ \frac{P_L^2}{2m_L} = \frac{P_H^2}{2m_H} \] The above equation implies that the body with the larger mass will have a smaller momentum because the kinetic energy is constant for both. Therefore, the light body must have less momentum than the heavy body. Thus, \( P_L < P_H \).

Answer 2

Given:
A light body with momentum \( P_L \) and a heavy body with momentum \( P_H \), both having the same kinetic energy.

Step 1: Recall the kinetic energy formula:
\[ K = \frac{p^2}{2m} \] where \( p \) is momentum and \( m \) is mass.

Step 2: For the light and heavy bodies, since their kinetic energies are equal:
\[ \frac{P_L^2}{2 m_L} = \frac{P_H^2}{2 m_H} \] \[ \Rightarrow \frac{P_L^2}{m_L} = \frac{P_H^2}{m_H} \] \[ \Rightarrow \frac{P_L^2}{P_H^2} = \frac{m_L}{m_H} \] \[ \Rightarrow \frac{P_L}{P_H} = \sqrt{\frac{m_L}{m_H}} \]

Step 3: Since the light body has less mass (\( m_L < m_H \)), it follows that:
\[ \frac{P_L}{P_H} < 1 \implies P_L < P_H \]

Therefore, the momentum of the light body is less than the momentum of the heavy body:
\[ \boxed{P_L < P_H} \]