Question

Two bodies having masses in the ratio 1:3 have equal linear momentum. Their respective kinetic energies are in the ratio:

• A. 3:1
• B. 1:2
• C. 1:3
• D. 4:1
• E. 2:1

Hint:

For equal momentum, the lighter body always has higher kinetic energy.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
This problem explores the relationship between kinetic energy (KE) and linear momentum (p). We need to find the ratio of kinetic energies given the ratio of masses and the fact that their momenta are equal.
Step 2: Key Formula or Approach:
1. Linear momentum: \( p = mv \) 2. Kinetic energy: \( KE = \frac{1}{2}mv^2 \) 3. Relationship between KE and p: We can write KE in terms of p by substituting \( v = p/m \) into the KE formula. \[ KE = \frac{1}{2}m\left(\frac{p}{m}\right)^2 = \frac{1}{2}m\frac{p^2}{m^2} = \frac{p^2}{2m} \] Step 3: Detailed Explanation:
We are given: - Ratio of masses: \( m_1 : m_2 = 1 : 3 \). Let \( m_1 = m \) and \( m_2 = 3m \). - Equal linear momentum: \( p_1 = p_2 = p \). We need to find the ratio of their kinetic energies, \( KE_1 : KE_2 \). Using the formula \( KE = \frac{p^2}{2m} \): For the first body: \[ KE_1 = \frac{p_1^2}{2m_1} = \frac{p^2}{2m} \] For the second body: \[ KE_2 = \frac{p_2^2}{2m_2} = \frac{p^2}{2(3m)} = \frac{p^2}{6m} \] Now, find the ratio \( \frac{KE_1}{KE_2} \): \[ \frac{KE_1}{KE_2} = \frac{\frac{p^2}{2m}}{\frac{p^2}{6m}} = \frac{p^2}{2m} \times \frac{6m}{p^2} \] Cancel the common terms \( p^2 \) and \( m \): \[ \frac{KE_1}{KE_2} = \frac{6}{2} = \frac{3}{1} \] So, the ratio of their kinetic energies is 3:1.
Step 4: Final Answer:
The respective kinetic energies are in the ratio 3:1.