A steel ball of radius `2 cm` is at rest on a frictionless surface. Another ball of radius `4 cm` moving at a velocity of `81 cm//sec` collides elastically with first ball. After collision the smaller ball moves with speed of

Options

A. `81 cm//sec`
B. `63 cm//sec`
C. `144 cm//sec`
D. None of these

Explanation

To solve the problem, we need to find the speed of the smaller steel ball (radius 2 cm) after it collides elastically with the larger steel ball (radius 4 cm) that is moving at a speed of 81 cm/s.

### Step-by-step Solution:

1. **Identify the Variables:**
- Let the mass of the smaller ball (radius 2 cm) be \( m_1 \).
- Let the mass of the larger ball (radius 4 cm) be \( m_2 \).
- The initial velocity of the smaller ball, \( u_1 = 0 \) cm/s (at rest).
- The initial velocity of the larger ball, \( u_2 = 81 \) cm/s.

2. **Calculate the Masses:**
- The mass of a sphere is given by the formula \( m = \rho \cdot V \), where \( V = \frac{4}{3} \pi r^3 \).
- Since both balls are made of steel, they have the same density \( \rho \).
- For the smaller ball:
\[ m_1 = \rho \cdot \frac{4}{3} \pi (2)^3 = \rho \cdot \frac{4}{3} \pi \cdot 8 = \frac{32}{3} \pi \rho \]
- For the larger ball:
\[ m_2 = \rho \cdot \frac{4}{3} \pi (4)^3 = \rho \cdot \frac{4}{3} \pi \cdot 64 = \frac{256}{3} \pi \rho \]

3. **Use Conservation of Momentum:**
- The conservation of momentum before and after the collision gives us:
\[ m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2 \]
- Substituting the known values:
\[ 0 + \left(\frac{256}{3} \pi \rho\right) (81) = \left(\frac{32}{3} \pi \rho\right) v_1 + \left(\frac{256}{3} \pi \rho\right) v_2 \]
- Simplifying (we can cancel \( \frac{1}{3} \pi \rho \)):
\[ 256 \cdot 81 = 32 v_1 + 256 v_2 \quad \text{(Equation 1)} \]

4. **Use the Coefficient of Restitution:**
- For elastic collisions, the coefficient of restitution \( e = 1 \):
\[ e = \frac{v_2 - v_1}{u_1 - u_2} \]
- Substituting the known values:
\[ 1 = \frac{v_2 - v_1}{0 - 81} \]
- Rearranging gives:
\[ v_2 - v_1 = -81 \quad \text{or} \quad v_1 - v_2 = 81 \quad \text{(Equation 2)} \]

5. **Solve the System of Equations:**
- From Equation 2, we have:
\[ v_1 = v_2 + 81 \]
- Substitute \( v_1 \) in Equation 1:
\[ 256 \cdot 81 = 32(v_2 + 81) + 256 v_2 \]
- Expanding gives:
\[ 256 \cdot 81 = 32 v_2 + 2592 + 256 v_2 \]
- Combine like terms:
\[ 256 \cdot 81 - 2592 = 288 v_2 \]
- Calculate \( 256 \cdot 81 = 20736 \):
\[ 20736 - 2592 = 288 v_2 \]
- Thus:
\[ 18144 = 288 v_2 \]
- Solving for \( v_2 \):
\[ v_2 = \frac{18144}{288} = 63 \text{ cm/s} \]

6. **Find \( v_1 \):**
- Substitute \( v_2 \) back into Equation 2:
\[ v_1 = 63 + 81 = 144 \text{ cm/s} \]

### Final Answer:
The speed of the smaller ball after the collision is **144 cm/s**.