Question

Two cars \(A\) and \(B\) are moving on a road with speeds \(100\,\text{km/h}\) and \(80\,\text{km/h}\) respectively. A stone is thrown from car \(B\) with speed \(V\) km/h relative to it. The stone hits car \(A\) with speed \(5\,\text{m/s}\) relative to car \(A\) (ignore gravity). Find \(V\).

• A. \(18\)
• B. \(38\)
• C. \(48\)
• D. \(20\)

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
This problem involves determining the speed of a projectile relative to its source when its relative speed to a target moving at a different velocity is known. We must work with consistent units.
Step 2: Key Formula or Approach:
Relative velocity formula: \(\vec{v}_{S/A} = \vec{v}_{S} - \vec{v}_{A}\).
Velocity of stone in ground frame: \(\vec{v}_{S} = \vec{v}_{B} + \vec{v}_{S/B}\).
Step 3: Detailed Explanation:
First, convert the relative speed of impact from m/s to km/h:
\[ 5 \text{ m/s} = 5 \times \frac{18}{5} = 18 \text{ km/h} \] Assuming both cars move in the same direction (e.g., along the \(+\hat{i}\) direction):
\(v_{A} = 100 \text{ km/h}\) and \(v_{B} = 80 \text{ km/h}\).
Let the velocity of the stone relative to Car \(B\) be \(V \text{ km/h}\) (forward). The velocity of the stone in the ground frame is:
\[ v_{S} = v_{B} + V = 80 + V \] The velocity of the stone relative to Car \(A\) is:
\[ v_{S/A} = v_{S} - v_{A} = (80 + V) - 100 = V - 20 \] Given that the magnitude of this relative speed is \(18 \text{ km/h}\):
\[ |V - 20| = 18 \] This gives two possible values for \(V\):
1) \(V - 20 = 18 \Rightarrow V = 38 \text{ km/h}\)
2) \(V - 20 = -18 \Rightarrow V = 2 \text{ km/h}\)
Comparing these with the provided options, \(V = 38\) is the matching value.
Step 4: Final Answer:
The speed \(V\) relative to Car \(B\) is 38 km/h, which is option (B).