Relative Velocity of Stone Thrown Between Two Moving Cars

To find the speed $v$ with which the person in car $B$ throws the stone relative to car $B$, we analyze the relative motion of the two cars and the stone along the straight line.

1. Given Data:

• Velocity of car $A$ relative to the ground: $v_A = 100 \text{ km/h}$
• Velocity of car $B$ relative to the ground: $v_B = 80 \text{ km/h}$
• Speed of the stone relative to car $A$ when it hits car $A$: $v_{\text{stone}/A} = 5 \text{ m/s}$

2. Unit Conversion: Convert the relative hitting speed from $\text{m/s}$ to $\text{km/h}$: $v_{\text{stone}/A} = 5 \text{ m/s} = 5 \times \frac{18}{5} \text{ km/h} = 18 \text{ km/h}$

3. Formulation of Relative Velocities: Let the stone be thrown forward with a speed $v$ relative to car $B$. The velocity of the stone relative to the ground is: $v_{\text{stone}} = v_B + v = 80 + v$

The velocity of the stone relative to car $A$ is given by: $v_{\text{stone}/A} = v_{\text{stone}} - v_A$

Substitute $v_{\text{stone}}$ and $v_A$ into the relative velocity equation: $v_{\text{stone}/A} = (80 + v) - 100 = v - 20$

4. Solving for $v$: Since the stone hits car $A$ with a relative speed of $18 \text{ km/h}$: $v - 20 = 18$ $v = 38 \text{ km/h}$

Thus, the value of $v$ is $38 \text{ km/h}$.

Correct Option: C