Blocks A and B are moving towards each other along the x axis. A has mass of 2.0 kg and a velocity of 10 m/s (in the positive x direction), B has a mass of 3.0 kg and a velocity of –5 m/s (in the negative x direction). They suffer an elastic collision and move off along the x axis. After the collision, the velocities of A and B, respectively, are:

Options

A. `–10 and +0.5 m/s`
B. `–8.0 and +7.0 m/s`
C. `–9.0 and +6.0 m/s`
D. `–5.0 and +10 m/s`

Explanation

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

### Step 1: Define the given values
- Mass of block A (m1) = 2.0 kg
- Velocity of block A (u1) = 10 m/s (positive x direction)
- Mass of block B (m2) = 3.0 kg
- Velocity of block B (u2) = -5 m/s (negative x direction)

### Step 2: Write the conservation of momentum equation
In an elastic collision, the total momentum before the collision is equal to the total momentum after the collision. Therefore, we can write:

\[ m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2 \]

Substituting the known values:

\[ (2.0 \, \text{kg} \times 10 \, \text{m/s}) + (3.0 \, \text{kg} \times -5 \, \text{m/s}) = 2.0 \, \text{kg} \times v_1 + 3.0 \, \text{kg} \times v_2 \]

Calculating the left side:

\[ 20 - 15 = 2v_1 + 3v_2 \]

This simplifies to:

\[ 5 = 2v_1 + 3v_2 \quad \text{(Equation 1)} \]

### Step 3: Write the equation for elastic collision
For elastic collisions, the relative velocity of separation is equal to the relative velocity of approach. This can be expressed as:

\[ v_2 - v_1 = -(u_2 - u_1) \]

Substituting the known values:

\[ v_2 - v_1 = -(-5 - 10) \]

This simplifies to:

\[ v_2 - v_1 = 15 \quad \text{(Equation 2)} \]

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

1. \( 2v_1 + 3v_2 = 5 \)
2. \( v_2 - v_1 = 15 \)

From Equation 2, we can express \( v_2 \) in terms of \( v_1 \):

\[ v_2 = v_1 + 15 \]

Now substitute \( v_2 \) in Equation 1:

\[ 2v_1 + 3(v_1 + 15) = 5 \]

Expanding this gives:

\[ 2v_1 + 3v_1 + 45 = 5 \]

Combining like terms:

\[ 5v_1 + 45 = 5 \]

Subtracting 45 from both sides:

\[ 5v_1 = -40 \]

Dividing by 5:

\[ v_1 = -8 \, \text{m/s} \]

### Step 5: Find \( v_2 \)
Now substitute \( v_1 \) back into the equation for \( v_2 \):

\[ v_2 = -8 + 15 = 7 \, \text{m/s} \]

### Final Answer
The velocities after the collision are:
- Velocity of block A (v1) = -8 m/s
- Velocity of block B (v2) = 7 m/s

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