Question

Particle $A$ moves along the line $y=4\sqrt{3}m$ with constant velocity v of magnitude $2.0m/s$ and directed parallel to the positive $x$ -axis (see figure). Particle $B$ starts at the origin with zero speed and constant accelerationa (of magnitude $4.0m/s^2$ ) at the same instant that the particle A passes the y axis. The angle $\theta$ between a and the positive $y$ axis that would result in a collision between these two particles should have a value equal to

• A. $30^{\circ }$
• B. $45^{\circ }$
• C. $50^{\circ }$
• D. $60^{\circ }$

Answer 1

Answer: (A)

Given that $y=4\sqrt{3}m$
For particle $A$, $v=2m/s$
and For particle $Ba=4m/s^2$
Let particle are collide after $tsec$.
distance covered by $A$ in $tsec.=2t$
and $B,=\frac{1}{2}\times 4\times t^2$
For collision $2t=\frac{1}{2}\times 4\times t^2$
$t=1sec$.
Velocity of $B=4\times 1=4m/s$
Now, from $MBC\sin \theta =\frac{2}{4}=\frac{1}{2}$
$\theta =30^{\circ }$