Question

Particle $A$ moves along the line $y=4\sqrt{3}\,m$ with constant velocity v of magnitude $2.0 \,m/s$ and directed parallel to the positive $x$ -axis (see figure). Particle $B$ starts at the origin with zero speed and constant accelerationa (of magnitude $4.0\,m/s^{2}$ ) 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)

$30^{\circ}$

Answer 2

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