Two steel balls A and B of mass 10 kg and 10 g rolls towards each other with 5m/s and 1 m/s respectively on a smooth floor. After collision, with what speed B moves [perfectly elastic collision]?

Options

A. 8 m/s
B. 10 m/s
C. 11 m/s
D. Zero

Answer

To solve the problem of two steel balls A and B colliding elastically, we will follow these steps:

### Step 1: Identify the masses and initial velocities
- Mass of ball A, \( m_A = 10 \, \text{kg} = 10000 \, \text{g} \)
- Mass of ball B, \( m_B = 10 \, \text{g} \)
- Initial velocity of ball A, \( u_A = 5 \, \text{m/s} \) (towards the right)
- Initial velocity of ball B, \( u_B = -1 \, \text{m/s} \) (towards the left)

### Step 2: Apply the conservation of momentum
The total momentum before the collision must equal the total momentum after the collision. The equation for momentum conservation is:

\[ m_A u_A + m_B u_B = m_A v_A + m_B v_B \]

Where \( v_A \) and \( v_B \) are the final velocities of balls A and B respectively after the collision.

Substituting the known values:

\[ (10000 \, \text{g})(5 \, \text{m/s}) + (10 \, \text{g})(-1 \, \text{m/s}) = (10000 \, \text{g}) v_A + (10 \, \text{g}) v_B \]

Calculating the left side:

\[ 50000 \, \text{g m/s} - 10 \, \text{g m/s} = 49990 \, \text{g m/s} \]

So we have:

\[ 49990 = 10000 v_A + 10 v_B \quad \text{(Equation 1)} \]

### Step 3: Apply the conservation of kinetic energy
For a perfectly elastic collision, kinetic energy is also conserved. The equation for kinetic energy conservation is:

\[ \frac{1}{2} m_A u_A^2 + \frac{1}{2} m_B u_B^2 = \frac{1}{2} m_A v_A^2 + \frac{1}{2} m_B v_B^2 \]

Substituting the known values:

\[ \frac{1}{2} (10000)(5^2) + \frac{1}{2} (10)(-1)^2 = \frac{1}{2} (10000) v_A^2 + \frac{1}{2} (10) v_B^2 \]

Calculating the left side:

\[ \frac{1}{2} (10000)(25) + \frac{1}{2} (10)(1) = 125000 + 5 = 125005 \]

So we have:

\[ 125005 = 5000 v_A^2 + 5 v_B^2 \quad \text{(Equation 2)} \]

### Step 4: Solve the equations
From Equation 1, we can express \( v_A \) in terms of \( v_B \):

\[ v_A = \frac{49990 - 10 v_B}{10000} \]

Substituting this expression for \( v_A \) into Equation 2:

\[ 125005 = 5000 \left(\frac{49990 - 10 v_B}{10000}\right)^2 + 5 v_B^2 \]

This will lead to a quadratic equation in terms of \( v_B \). However, given that the mass of A is much larger than B, we can simplify our calculations.

### Step 5: Approximate solution
Since \( m_A \) is much larger than \( m_B \), we can use the formula for the final velocity of B in a perfectly elastic collision:

\[ v_B = \frac{(m_A - m_B) u_A + 2 m_B u_B}{m_A + m_B} \]

Substituting the values:

\[ v_B = \frac{(10000 - 10)(5) + 2(10)(-1)}{10000 + 10} \]

Calculating:

\[ v_B = \frac{9990 \cdot 5 - 20}{10010} = \frac{49950 - 20}{10010} = \frac{49930}{10010} \approx 4.97 \, \text{m/s} \]

### Conclusion
The speed of ball B after the collision is approximately \( 4.97 \, \text{m/s} \).