1. Identify the type of collision

Since the two masses move together as a single body after collision, this is a perfectly inelastic collision.

So, momentum is conserved, but kinetic energy is not.

2. Given data

• First block: $m_1 = 0.50\,\text{kg}, \quad u_1 = 2.00\,\text{m s}^{-1}$
• Second block: $m_2 = 1.0\,\text{kg}, \quad u_2 = 0$

After collision, common velocity $v$ is produced.

3. Apply conservation of momentum

Initial momentum: $p_i = m_1u_1 + m_2u_2 = (0.50)(2.00) + (1.0)(0) = 1.0\,\text{kg m s}^{-1}$

Final momentum: $p_f = (m_1+m_2)v = (0.50+1.0)v = 1.5v$

Equating: $1.5v = 1.0$ $v = \frac{1.0}{1.5} = \frac{2}{3}\,\text{m s}^{-1}$

4. Calculate initial kinetic energy

$K_i = \frac{1}{2}m_1u_1^2 + \frac{1}{2}m_2u_2^2$ $K_i = \frac{1}{2}(0.50)(2.00)^2 + 0$ $K_i = 0.25 \times 4 = 1.00\,\text{J}$

5. Calculate final kinetic energy

$K_f = \frac{1}{2}(m_1+m_2)v^2$ $K_f = \frac{1}{2}(1.5)\left(\frac{2}{3}\right)^2$ $K_f = 0.75 \times \frac{4}{9} = \frac{3}{4}\times\frac{4}{9} = \frac{1}{3}\,\text{J}$ $K_f \approx 0.33\,\text{J}$

6. Energy loss during collision

$\Delta K = K_i - K_f$ $\Delta K = 1.00 - \frac{1}{3} = \frac{2}{3}\,\text{J}$ $\Delta K \approx 0.67\,\text{J}$

7. Match with options

$0.67\,\text{J}$ corresponds to Option C.

8. Comparison with stored answer

Stored correct answer: C

My derived answer: C

So, they agree.