Question

Two particles A and B are thrown vertically upward with velocity, 5 m/s and 10 m/s respectively ($g=10m/s^2$). Find separation between them after one second. (Use relative motion approach)

• A. 0 m
• B. 5 m
• C. 10 m
• D. 15 m

Answer 1

The correct option is B

5 m

Method-I
Position of A after 1 sec
$S_A=ut-\frac{1}{2}g{t}^2=5t-\frac{1}{2}times10\times t^2=5\times 1-\frac{1}{2}\times 10\times 1^2=5-5=0$
i.e., the particle will return to ground at t=1 sec
The position of B after 1 sec; $S_B=ut-\frac{1}{2}g{t}^2=10\times 1^2=10-5=5m$
Hence separation between A and B, $S_B-S_A=5m$.

Method-II
Relative velocity method:
Accleration of B w.r.t A $\overrightarrow{a_{BA}}=\overrightarrow{a_B}-\overrightarrow{a_A}=(-10)-(-10)=0$
Initial relative velocity Also $\overrightarrow{v_{BA}}=\overrightarrow{v_B}-\overrightarrow{v_A}=10-5=5m/s$

Hence relative separation between particles
$\therefore \overrightarrow{S_{BA}}\left(in1sec\right)=\overrightarrow{v_{BA}}\times t=5\times ‘=5m$
$\therefore$ Distance between A and B after 1 sec = 5 m.