Question

A ball 'A' of mass 1.2 kg moving with a velocity of 8.4 m/s makes a one-dimensional elastic collision with a ball 'B' of mass 3.6 kg at rest. The percentage of kinetic energy transferred by ball 'A' to ball 'B' is:

• A. \( 25\% \)
• B. \( 50\% \)
• C. \( 75\% \)
• D. \( 60\% \)

Hint:

For one-dimensional elastic collisions, use the formula \( \frac{4 m_1 m_2}{(m_1 + m_2)^2} \) to calculate the fraction of kinetic energy transferred from one body to another.

Answer 1

The Correct Option is C

Step 1: Understanding the Energy Transfer Formula In an elastic collision, the fraction of kinetic energy transferred from mass \( m_1 \) to mass \( m_2 \) is given by: \[ \text{Fraction of energy transferred} = \frac{4 m_1 m_2}{(m_1 + m_2)^2} \] Multiplying by 100 gives the percentage transfer. 

Step 2: Substituting Given Values Given: \[ m_1 = 1.2 \text{ kg}, \quad m_2 = 3.6 \text{ kg} \] \[ \text{Percentage of kinetic energy transferred} = \left( \frac{4(1.2)(3.6)}{(1.2 + 3.6)^2} \right) \times 100 \] \[ = \left( \frac{17.28}{23.04} \right) \times 100 \] \[ = 0.75 \times 100 = 75\% \] Thus, the percentage of kinetic energy transferred is: \[ \mathbf{75\%} \]