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 finding the speed of ball B after a perfectly elastic collision, we can follow these steps:

### Step 1: Understand the given information
We have two steel balls:
- Ball A (mass \( m_1 = 10 \, \text{kg} \)) moving towards ball B with a velocity \( u_1 = 5 \, \text{m/s} \).
- Ball B (mass \( m_2 = 10 \, \text{g} = 0.01 \, \text{kg} \)) moving towards ball A with a velocity \( u_2 = -1 \, \text{m/s} \) (negative because it's in the opposite direction).

### Step 2: Use the coefficient of restitution
For a perfectly elastic collision, the coefficient of restitution \( e \) is equal to 1. The formula for the coefficient of restitution is given by:
\[ e = \frac{v_2 - v_1}{u_1 - u_2} \]
Where:
- \( v_1 \) is the final velocity of ball A,
- \( v_2 \) is the final velocity of ball B.

Since \( e = 1 \):
\[ v_2 - v_1 = u_1 - u_2 \]

Substituting the values:
\[ v_2 - v_1 = 5 - (-1) = 6 \]
So, we can express \( v_1 \) in terms of \( v_2 \):
\[ v_1 = v_2 - 6 \tag{1} \]

### Step 3: Apply conservation of momentum
The total momentum before the collision must equal the total momentum after the collision:
\[ m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2 \]

Substituting the known values:
\[ (10)(5) + (0.01)(-1) = (10)(v_1) + (0.01)(v_2) \]
This simplifies to:
\[ 50 - 0.01 = 10 v_1 + 0.01 v_2 \]
\[ 49.99 = 10 v_1 + 0.01 v_2 \tag{2} \]

### Step 4: Substitute equation (1) into equation (2)
Now, substitute \( v_1 \) from equation (1) into equation (2):
\[ 49.99 = 10(v_2 - 6) + 0.01 v_2 \]
Expanding this gives:
\[ 49.99 = 10 v_2 - 60 + 0.01 v_2 \]
Combining like terms:
\[ 49.99 + 60 = 10 v_2 + 0.01 v_2 \]
\[ 109.99 = 10.01 v_2 \]

### Step 5: Solve for \( v_2 \)
Now, divide both sides by 10.01 to find \( v_2 \):
\[ v_2 = \frac{109.99}{10.01} \approx 10.99 \, \text{m/s} \]

### Conclusion
The speed of ball B after the collision is approximately \( 11 \, \text{m/s} \).