A 10 kg mass travelling 2m /s meets and collides elastically with a 2 kg mass travelling 4 / m s in the opposite direction. Find the final velocities of both objects .

Options

A. `V_(Af)=2m//s,V_(Bf)=3m//s`
B. `V_(Af)=0m//s,V_(Bf)=6m//s`
C. `V_(Af)=5m//s,V_(Bf)=8m//s`
D. `V_(Af)=4m//s,V_(Bf)=2m//s`

Answer

To solve the problem of two masses colliding elastically, we will use the principles of conservation of momentum and the coefficient of restitution. Let's break down the solution step by step.

### Step 1: Define the Variables
Let:
- Mass of object A (mA) = 10 kg
- Initial velocity of object A (uA) = 2 m/s (to the right)
- Mass of object B (mB) = 2 kg
- Initial velocity of object B (uB) = -4 m/s (to the left, hence negative)

### Step 2: Calculate Initial Momentum
The initial momentum (P_initial) of the system can be calculated using the formula:
\[ P_{\text{initial}} = m_A \cdot u_A + m_B \cdot u_B \]
Substituting the values:
\[ P_{\text{initial}} = (10 \, \text{kg} \cdot 2 \, \text{m/s}) + (2 \, \text{kg} \cdot -4 \, \text{m/s}) \]
\[ P_{\text{initial}} = 20 \, \text{kg m/s} - 8 \, \text{kg m/s} = 12 \, \text{kg m/s} \]

### Step 3: Set Up the Conservation of Momentum Equation
According to the conservation of momentum:
\[ P_{\text{initial}} = P_{\text{final}} \]
Let the final velocities of objects A and B be \( v_A \) and \( v_B \) respectively. Then:
\[ P_{\text{final}} = m_A \cdot v_A + m_B \cdot v_B \]
So we have:
\[ 12 = 10v_A + 2v_B \]
This is our first equation (Equation 1).

### Step 4: Set Up the Coefficient of Restitution Equation
For an elastic collision, the coefficient of restitution (e) is defined as:
\[ e = \frac{\text{Velocity of separation}}{\text{Velocity of approach}} \]
Since e = 1 for elastic collisions, we can write:
\[ 1 = \frac{v_B - v_A}{u_A - u_B} \]
Substituting the initial velocities:
\[ 1 = \frac{v_B - v_A}{2 - (-4)} \]
\[ 1 = \frac{v_B - v_A}{6} \]
Thus, we have:
\[ v_B - v_A = 6 \]
This is our second equation (Equation 2).

### Step 5: Solve the Equations Simultaneously
Now we have two equations:
1. \( 10v_A + 2v_B = 12 \) (Equation 1)
2. \( v_B - v_A = 6 \) (Equation 2)

From Equation 2, we can express \( v_B \) in terms of \( v_A \):
\[ v_B = v_A + 6 \]

Substituting this into Equation 1:
\[ 10v_A + 2(v_A + 6) = 12 \]
\[ 10v_A + 2v_A + 12 = 12 \]
\[ 12v_A + 12 = 12 \]
\[ 12v_A = 0 \]
\[ v_A = 0 \]

Now substituting \( v_A = 0 \) back into Equation 2 to find \( v_B \):
\[ v_B = 0 + 6 = 6 \]

### Final Result
The final velocities are:
- \( v_A = 0 \, \text{m/s} \) (the 10 kg mass comes to rest)
- \( v_B = 6 \, \text{m/s} \) (the 2 kg mass moves to the right)