According to a FIR in police station, car A was at rest waiting for a red light at a crossing when it was hit by similar car B from rear side. Both cars had then hard brekes on, and from their skid marks it is surmiesed that they skidded together about 8m in the original direction of travel before coming to rest. Assuming a stopping force of about 0.4 times the combined weights of the cars. the speed of car B just before the collision is approximately `[g=10 m//s^(2)]`

Options

A. 8m/s
B. 4m/s
C. 16 m/s
D. 32 m/s

Explanation

To solve the problem, we need to determine the speed of car B just before the collision using the information provided.

### Step-by-Step Solution:

1. **Understanding the Scenario**:
- Car A is at rest (initial velocity \( u_A = 0 \)).
- Car B is moving and collides with car A.
- After the collision, both cars skid together for a distance of 8 m before coming to rest.

2. **Using the Conservation of Momentum**:
- Let the mass of each car be \( m \).
- Let the initial velocity of car B be \( v_0 \).
- After the collision, both cars move together with a common velocity \( v \).
- By the conservation of momentum:
\[ mv_0 = 2mv \]
- Simplifying gives:
\[ v = \frac{v_0}{2} \]

3. **Calculating the Deceleration**:
- The stopping force is given as \( 0.4 \) times the combined weight of the cars.
- The combined weight of the two cars is \( 2mg \).
- Therefore, the stopping force \( F \) is:
\[ F = 0.4 \times 2mg = 0.8mg \]
- Using Newton's second law, the deceleration \( a \) can be calculated as:
\[ a = \frac{F}{\text{combined mass}} = \frac{0.8mg}{2m} = 0.4g \]

4. **Applying the Third Equation of Motion**:
- We know the initial velocity \( v \), the final velocity \( 0 \) (since they come to rest), the distance \( s = 8 \, \text{m} \), and the deceleration \( a = 0.4g \).
- Using the equation:
\[ v^2 = u^2 + 2as \]
- Substituting \( u = v \) and \( a = -0.4g \):
\[ 0 = v^2 - 2(0.4g)(8) \]
- Rearranging gives:
\[ v^2 = 2 \times 0.4g \times 8 \]
- Simplifying:
\[ v^2 = 6.4g \]

5. **Substituting \( g = 10 \, \text{m/s}^2 \)**:
- Now substituting \( g \):
\[ v^2 = 6.4 \times 10 = 64 \]
- Taking the square root:
\[ v = 8 \, \text{m/s} \]

6. **Finding the Initial Speed of Car B**:
- Recall that \( v = \frac{v_0}{2} \):
\[ 8 = \frac{v_0}{2} \]
- Therefore:
\[ v_0 = 16 \, \text{m/s} \]

### Final Answer:
The speed of car B just before the collision is approximately **16 m/s**.