Two cars A and B are moving in the same direction along a straight line with speeds 100 km/h and 80 km/h, respectively such that car A is moving ahead of car B. A person in car B throws a stone with a speed $v$ so that it hits the car A with a speed of 5 m/s. The value of $v$ is ______ km/h.
Step 1: Understanding the Concept:
This problem operates on the principle of relative velocity. The stone is thrown from the reference frame of car B, so its absolute velocity depends on both throw speed and car B's speed. We then compute the stone's velocity relative to car A to match the impact speed.
Step 2: Key Formula or Approach:
Absolute velocity of stone: $\vec{v}_s = \vec{v}_{throw} + \vec{v}_B$
Relative velocity of stone with respect to A: $\vec{v}_{sA} = \vec{v}_s - \vec{v}_A$
Convert 5 m/s to km/h by multiplying by $\frac{18}{5}$.
Step 3: Detailed Explanation:
Let the direction of motion of the cars be the positive x-direction.
Velocity of car A, $v_A = 100$ km/h.
Velocity of car B, $v_B = 80$ km/h.
The stone is thrown forward from B with speed $v$ relative to B.
Velocity of the stone relative to the ground is $v_s = v_B + v = 80 + v$.
The stone hits car A. The impact speed is the relative speed of the stone with respect to car A.
Velocity of stone relative to A is $v_{sA} = v_s - v_A = (80 + v) - 100 = v - 20$.
The impact speed is given as 5 m/s. Convert this to km/h:
$5 \text{ m/s} = 5 \times \frac{18}{5} \text{ km/h} = 18 \text{ km/h}$.
The magnitude of the relative impact velocity is 18 km/h.
$|v - 20| = 18$
This gives two possibilities:
$v - 20 = 18 \implies v = 38$ km/h.
$v - 20 = -18 \implies v = 2$ km/h.
Since car A is ahead of car B and moving faster, a stone thrown with $v = 2$ km/h relative to B would have a ground speed of 82 km/h. It would never catch up to car A (which is at 100 km/h). Thus, the stone must be thrown fast enough to exceed A's speed.
Therefore, $v$ must be 38 km/h.
