A ball A of mass 3 kg and a ball B of mass 4 kg are moving alogn the same straight line with speeds of 7m/s and 5 m/s respectively. They approach each other and collide. What is the speed of B after collision, If the coefficient of restitution is `3/4`?

Options

A. 4 m/s
B. `-5` m/s
C. `-4` m/s
D. `6m/s

Explanation

To solve the problem, we need to apply the principles of conservation of momentum and the coefficient of restitution. Here’s a step-by-step breakdown of the solution:

### Step 1: Identify the given values
- Mass of ball A, \( m_A = 3 \, \text{kg} \)
- Mass of ball B, \( m_B = 4 \, \text{kg} \)
- Initial speed of ball A, \( u_A = 7 \, \text{m/s} \) (approaching)
- Initial speed of ball B, \( u_B = -5 \, \text{m/s} \) (approaching, hence negative)

### Step 2: Write the conservation of momentum equation
The total momentum before the collision must equal the total momentum after the collision. Let \( v_A \) and \( v_B \) be the final velocities of balls A and B respectively.

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

Substituting the known values:

\[ 3 \cdot 7 + 4 \cdot (-5) = 3 \cdot v_A + 4 \cdot v_B \]

Calculating the left side:

\[ 21 - 20 = 3v_A + 4v_B \]
\[ 1 = 3v_A + 4v_B \quad \text{(Equation 1)} \]

### Step 3: Write the coefficient of restitution equation
The coefficient of restitution (e) relates the relative speeds of separation and approach. Given \( e = \frac{3}{4} \):

\[ e = \frac{v_B - v_A}{u_A - u_B} \]

Substituting the values:

\[ \frac{3}{4} = \frac{v_B - v_A}{7 - (-5)} \]
\[ \frac{3}{4} = \frac{v_B - v_A}{12} \]

Cross-multiplying gives:

\[ 3 \cdot 12 = 4(v_B - v_A) \]
\[ 36 = 4v_B - 4v_A \]
\[ 4v_B - 4v_A = 36 \quad \text{(Equation 2)} \]

### Step 4: Solve the equations
Now we have two equations:

1. \( 3v_A + 4v_B = 1 \)
2. \( 4v_B - 4v_A = 36 \)

From Equation 2, we can express \( v_B \):

\[ 4v_B = 36 + 4v_A \]
\[ v_B = 9 + v_A \quad \text{(Equation 3)} \]

Substituting Equation 3 into Equation 1:

\[ 3v_A + 4(9 + v_A) = 1 \]
\[ 3v_A + 36 + 4v_A = 1 \]
\[ 7v_A + 36 = 1 \]
\[ 7v_A = 1 - 36 \]
\[ 7v_A = -35 \]
\[ v_A = -5 \, \text{m/s} \]

Now substituting \( v_A \) back into Equation 3 to find \( v_B \):

\[ v_B = 9 + (-5) \]
\[ v_B = 4 \, \text{m/s} \]

### Conclusion
The speed of ball B after the collision is \( v_B = 4 \, \text{m/s} \).

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