Question

Two particles A and B are projected simultaneously with velocities $V_A=11\text{m/s}$ and $V_B=15\text{m/s}$ as shown in the figure. Find the time (in seconds) when they collide.

• A. $1\text{s}$
• B. $2\text{s}$
• C. $4\text{s}$
• D. $3\text{s}$

Answer 1

The correct option is B $2\text{s}$

Solving from frame of reference of particle A,
Relative acceleration vertical direction ${a_y}_{BA}=0$

At the time of collision, relative horizontal displacement of particle B with respect to particle A will be
$X_{BA}=40\text{m}$

Now, relative horizontal velocity of B with respect to A will be:-
$\overrightarrow{V_{BA}}=V_B-V_A=15\sin {37}^{\circ }\hat{i}+15\cos {37}^{\circ }\hat{j}-(-11\hat{i})$
$\therefore \overrightarrow{V_{BA}}=9\hat{i}+12\hat{j}+11\hat{i}=20\hat{i}+12\hat{j}$



Applying the condition of collision for particles:
$X_{BA}={V_{BA}}_x\times t$

$\therefore t=\frac{X_{BA}}{{V_{BA}}_x}=\frac{40}{20}=2\text{s}$