Question

Two particles A and B, move with constant velocities $\vec{v_1}$ and $\vec{v_2}$. At the initial moment their position vectors are $\vec{r_1}$, and $\vec{r_2}$ respectively. The condition for particles A and B for their collision is

• A. $\vec{r_1} \times \vec{v_1}=\vec{r_2} \times \vec{v_2}$
• B. $\vec{r_1} -\vec{r_2}=\vec{v_1} \times \vec{v_2}$
• C. \(\frac{\vec{r_1}-\vec{r_2}}{|\vec{r_1}-\vec{r_2}|}=\frac{\vec{v_2}-\vec{v_1}}{|\vec{v_2}-\vec{v_1}|}\)
• D. $\vec{r_1} \cdot \vec{v_1}=\vec{r_2} \cdot \vec{v_2}$

Answer 1

The Correct Option is C

To determine the condition for the collision of two particles A and B moving with constant velocities, we need to analyze their positions and velocities vectorially.

1. Given:
   • Particle A has initial position vector $\vec{r_1}$ and velocity $\vec{v_1}$.
   • Particle B has initial position vector $\vec{r_2}$ and velocity $\vec{v_2}$.
2. The position vectors of the particles at time $t$ are:
   • For Particle A: $\vec{r_1}(t) = \vec{r_1} + \vec{v_1} t$
   • For Particle B: $\vec{r_2}(t) = \vec{r_2} + \vec{v_2} t$
3. For the particles to collide at some time $t$, their position vectors must be equal:
   $\vec{r_1} + \vec{v_1} t = \vec{r_2} + \vec{v_2} t$
4. Rearranging the above equation gives:
   $(\vec{r_1} - \vec{r_2}) = (\vec{v_2} - \vec{v_1}) t$
5. For a valid solution of $t$:
   $\frac{\vec{r_1} - \vec{r_2}}{|\vec{r_1} - \vec{r_2}|} = \frac{\vec{v_2} - \vec{v_1}}{|\vec{v_2} - \vec{v_1}|}$

   This indicates that the direction of the relative position vector from A to B is the same as the relative velocity direction, ensuring that A can catch up with or intersect B.

6. Hence, the condition for collision is satisfied by option

   $\frac{\vec{r_1}-\vec{r_2}}{|\vec{r_1}-\vec{r_2}|}=\frac{\vec{v_2}-\vec{v_1}}{|\vec{v_2}-\vec{v_1}|}$

   .