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Physics Question on Elastic and inelastic collisions

Two particles A and B, move with constant velocities v1⃗\vec{v_1} and v2⃗\vec{v_2}. At the initial moment their position vectors are r1⃗\vec{r_1}, and r2⃗\vec{r_2} respectively. The condition for particles A and B for their collision is

A

r1⃗×v1⃗=r2⃗×v2⃗\vec{r_1} \times \vec{v_1}=\vec{r_2} \times \vec{v_2}

B

r1⃗−r2⃗=v1⃗×v2⃗\vec{r_1} -\vec{r_2}=\vec{v_1} \times \vec{v_2}

C

r1⃗−r2⃗∣r1⃗−r2⃗∣=v2⃗−v1⃗∣v2⃗−v1⃗∣\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

r1⃗⋅v1⃗=r2⃗⋅v2⃗\vec{r_1} \cdot \vec{v_1}=\vec{r_2} \cdot \vec{v_2}

Answer

r1⃗−r2⃗∣r1⃗−r2⃗∣=v2⃗−v1⃗∣v2⃗−v1⃗∣\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}|}

Explanation

Solution

The correct answer is C:r1⃗−r2⃗∣r1⃗−r2⃗∣=(v2⃗−v1⃗)∣v2⃗−v1⃗∣\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}|}
Let the particles A and B collide at time t. For their collision, the position vectors of both particles should be same at time t, i.e.
r1⃗+v1⃗t=r2⃗+v2⃗t\vec{r_1} + \vec{v_1}t=\vec{r_2} + \vec{v_2}t
r1⃗−r2⃗=v2⃗t−v1⃗t\vec{r_1} - \vec{r_2}=\vec{v_2}t - \vec{v_1}t
=(v2⃗−v1⃗)t...(i)=(\vec{v_2}-\vec{v_1})t ... (i)
Also, ∣r1⃗−r2⃗∣=∣v1⃗−v1⃗∣t|\vec{r_1}-\vec{r_2}|=|\vec{v_1}-\vec{v_1}|t or t=∣r1⃗−r2⃗∣∣v2⃗−v1⃗∣t=\frac{|\vec{r_1}-\vec{r_2}|}{|\vec{v_2}-\vec{v_1}|}
Substituting this value of t in eqn. (i), we get
r1⃗−r2⃗=(v2⃗−v1⃗)∣r1⃗−r2⃗∣∣v2⃗−v1⃗∣\vec{r_1}-\vec{r_2}=(\vec{v_2}-\vec{v_1})\frac{|\vec{r_1}-\vec{r_2}|}{|\vec{v_2}-\vec{v_1}|}
or r1⃗−r2⃗∣r1⃗−r2⃗∣=(v2⃗−v1⃗)∣v2⃗−v1⃗∣\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}|}