Question

A steel solid sphere of diameter \(2\) cm collides head-on elastically with another steel solid sphere of radius \(2\) cm at rest. The fraction of the initial kinetic energy of the smaller sphere transferred to the larger sphere during collision is

• A. \(\dfrac{16}{27}\)
• B. \(\dfrac{16}{9}\)
• C. \(\dfrac{8}{9}\)
• D. \(\dfrac{32}{81}\)

Hint:

For a one-dimensional elastic collision where the second body is initially at rest, \[ \boxed{ \text{Fraction of kinetic energy transferred} = \frac{4m_1m_2}{(m_1+m_2)^2}. } \] For solid spheres of the same material, \[ \boxed{m\propto r^3.} \]

Answer 1

The Correct Option is D

Step 1: Find the mass ratio of the spheres. Since both spheres are made of steel, they have the same density. Hence, \[ m\propto r^3. \] The smaller sphere has radius \[ 1\text{ cm}, \] and the larger sphere has radius \[ 2\text{ cm}. \] Therefore, \[ m_1:m_2 = 1^3:2^3 = 1:8. \]

Step 2: Use the formula for elastic collision. For a head-on elastic collision with the second sphere initially at rest, the fraction of kinetic energy transferred to the second sphere is \[ \frac{K_2}{K_1} = \frac{4m_1m_2}{(m_1+m_2)^2}. \] Substituting \[ m_1=1,\qquad m_2=8, \] we get \[ \frac{K_2}{K_1} = \frac{4(1)(8)}{(1+8)^2} = \frac{32}{81}. \]

Step 3: Write the answer. Hence, the required fraction is \[ \boxed{\frac{32}{81}}. \] Thus, \[ \boxed{(D)} \] is the correct answer.