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 conservation of kinetic energy. Let's denote the masses and their initial velocities:

- Mass \( m_1 = 10 \, \text{kg} \) with initial velocity \( u_1 = 2 \, \text{m/s} \) (to the right)
- Mass \( m_2 = 2 \, \text{kg} \) with initial velocity \( u_2 = -4 \, \text{m/s} \) (to the left, hence negative)

### Step 1: Write the conservation of momentum equation

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:

\[ 10 \times 2 + 2 \times (-4) = 10 v_1 + 2 v_2 \]

Calculating the left side:

\[ 20 - 8 = 10 v_1 + 2 v_2 \]

This simplifies to:

\[ 12 = 10 v_1 + 2 v_2 \quad \text{(Equation 1)} \]

### Step 2: Write the conservation of kinetic energy equation

Since the collision is elastic, kinetic energy is also conserved:

\[ \frac{1}{2} m_1 u_1^2 + \frac{1}{2} m_2 u_2^2 = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 \]

Substituting the known values:

\[ \frac{1}{2} \times 10 \times (2)^2 + \frac{1}{2} \times 2 \times (-4)^2 = \frac{1}{2} \times 10 v_1^2 + \frac{1}{2} \times 2 v_2^2 \]

Calculating the left side:

\[ \frac{1}{2} \times 10 \times 4 + \frac{1}{2} \times 2 \times 16 = 20 + 16 = 36 \]

This simplifies to:

\[ 36 = 5 v_1^2 + v_2^2 \quad \text{(Equation 2)} \]

### Step 3: Solve the equations

Now we have two equations:

1. \( 10 v_1 + 2 v_2 = 12 \) (Equation 1)
2. \( 5 v_1^2 + v_2^2 = 36 \) (Equation 2)

From Equation 1, we can express \( v_2 \) in terms of \( v_1 \):

\[ 2 v_2 = 12 - 10 v_1 \implies v_2 = 6 - 5 v_1 \]

### Step 4: Substitute \( v_2 \) into Equation 2

Substituting \( v_2 \) into Equation 2:

\[ 5 v_1^2 + (6 - 5 v_1)^2 = 36 \]

Expanding the square:

\[ 5 v_1^2 + (36 - 60 v_1 + 25 v_1^2) = 36 \]

Combining like terms:

\[ 30 v_1^2 - 60 v_1 + 36 - 36 = 0 \]

This simplifies to:

\[ 30 v_1^2 - 60 v_1 = 0 \]

Factoring out \( 30 v_1 \):

\[ 30 v_1 (v_1 - 2) = 0 \]

Setting each factor to zero gives:

\[ v_1 = 0 \quad \text{or} \quad v_1 = 2 \]

### Step 5: Find \( v_2 \)

If \( v_1 = 0 \):

\[ v_2 = 6 - 5(0) = 6 \]

If \( v_1 = 2 \):

\[ v_2 = 6 - 5(2) = -4 \]

### Conclusion

The final velocities after the collision are:

- For the 10 kg mass: \( v_1 = 0 \, \text{m/s} \)
- For the 2 kg mass: \( v_2 = 6 \, \text{m/s} \)

### Final Answer

The final velocities are:
- \( v_1 = 0 \, \text{m/s} \) (for the 10 kg mass)
- \( v_2 = 6 \, \text{m/s} \) (for the 2 kg mass)

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