A block of mass `0.50kg` is moving with a speed of `2.00m//s` on a smooth surface. It strikes another mass of `1 kg` at rest and they move as a single body. The energy loss during the collision is

Options

A. 0.16 J
B. 1.00 J
C. 0.67 J
D. 0.34 J

Answer

To find the energy loss during the collision between the two blocks, we can follow these steps:

### Step 1: Identify the masses and initial speeds
- Mass of the first block, \( m_1 = 0.50 \, \text{kg} \)
- Initial speed of the first block, \( v_1 = 2.00 \, \text{m/s} \)
- Mass of the second block, \( m_2 = 1.00 \, \text{kg} \)
- Initial speed of the second block, \( v_2 = 0 \, \text{m/s} \) (since it is at rest)

### Step 2: Calculate the initial momentum
The total initial momentum \( P_{\text{initial}} \) of the system is given by:
\[ P_{\text{initial}} = m_1 v_1 + m_2 v_2 = (0.50 \, \text{kg} \times 2.00 \, \text{m/s}) + (1.00 \, \text{kg} \times 0) = 1.00 \, \text{kg m/s} \]

### Step 3: Calculate the final mass and use conservation of momentum
After the collision, the two blocks stick together and move as a single body. The combined mass \( m_f \) is:
\[ m_f = m_1 + m_2 = 0.50 \, \text{kg} + 1.00 \, \text{kg} = 1.50 \, \text{kg} \]
Using conservation of momentum:
\[ P_{\text{final}} = m_f v_f \]
Setting \( P_{\text{initial}} = P_{\text{final}} \):
\[ 1.00 \, \text{kg m/s} = 1.50 \, \text{kg} \times v_f \]
Solving for \( v_f \):
\[ v_f = \frac{1.00 \, \text{kg m/s}}{1.50 \, \text{kg}} = \frac{2}{3} \, \text{m/s} \]

### Step 4: Calculate the initial kinetic energy
The initial kinetic energy \( KE_{\text{initial}} \) of the system is:
\[ KE_{\text{initial}} = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 \]
Substituting the values:
\[ KE_{\text{initial}} = \frac{1}{2} (0.50 \, \text{kg}) (2.00 \, \text{m/s})^2 + \frac{1}{2} (1.00 \, \text{kg}) (0)^2 \]
Calculating:
\[ KE_{\text{initial}} = \frac{1}{2} \times 0.50 \times 4 = 1.00 \, \text{J} \]

### Step 5: Calculate the final kinetic energy
The final kinetic energy \( KE_{\text{final}} \) after the collision is:
\[ KE_{\text{final}} = \frac{1}{2} m_f v_f^2 = \frac{1}{2} (1.50 \, \text{kg}) \left(\frac{2}{3} \, \text{m/s}\right)^2 \]
Calculating:
\[ KE_{\text{final}} = \frac{1}{2} \times 1.50 \times \frac{4}{9} = \frac{1.50 \times 4}{18} = \frac{6}{18} = \frac{1}{3} \, \text{J} \approx 0.33 \, \text{J} \]

### Step 6: Calculate the energy loss
The energy loss during the collision is given by:
\[ \text{Energy loss} = KE_{\text{initial}} - KE_{\text{final}} = 1.00 \, \text{J} - 0.33 \, \text{J} \approx 0.67 \, \text{J} \]

### Final Answer
The energy loss during the collision is approximately \( 0.67 \, \text{J} \).

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