Question

In the given figure, a particle $A$ moves along the line $y=30\text{m}$ with a constant velocity of magnitude $3\text{m/s}$ and parallel to the $x-$ axis. At the instant particle $A$ passes the $y-$ axis, the particle $B$ leaves the origin with a zero initial speed and a constant acceleration of magnitude $0.40{\text{m/s}}^2$. What angle $\theta$ between${\overrightarrow{a}}_B$ and the positive direction of the $y-$axis would result in a collision?

• A. ${15}^{\circ }$
• B. ${30}^{\circ }$
• C. ${45}^{\circ }$
• D. ${60}^{\circ }$

Answer 1

The correct option is D ${60}^{\circ }$

Given that:
Vertical distance between $A$ and $B$, $\left(y\right)=30\text{m}$
Velocity of $A$, $\left(u_A\right)=3\text{m/s}$
Initial velocity of $B$, $\left(u_B\right)=0\text{m/s}$
Acceleration of $B$, $\left(a_B\right)=0.40{\text{m/s}}^2$


Let the particles collide at point $P$.
In vertical direction for $B$:
$y=\frac{1}{2}{a}_y{t}^2$
$\Rightarrow 30=\frac{1}{2}\left(0.40\right)\cos \theta t^2$
$\Rightarrow 150=\cos \theta t^2$ ...........$\left(i\right)$
In horizontal direction for $A$ and $B$:
$u_{A}t=\frac{1}{2}{a}_x{t}^2$
$\Rightarrow t=\frac{2u_A}{a_x}$
$\Rightarrow t=\frac{2\times 3}{\left(0.40\right)\sin \theta }$
$\Rightarrow t=\frac{15}{\sin \theta }$
Substituting $t$ in equation $\left(i\right)$,
$150=\cos \theta \times {\left(\frac{15}{\sin \theta }\right)}^2$
$\Rightarrow 150=\cos \theta \times \frac{225}{1-{\cos }^2\theta }$
$\Rightarrow 1-{\cos }^2\theta =\frac{3\cos \theta }{2}$
$\Rightarrow 2-2{\cos }^2\theta =3\cos \theta$
$\Rightarrow 2{\cos }^2\theta +3\cos \theta -2=0$

Solving this, we get
$\cos \theta =\frac{1}{2}$ or $\cos \theta =-2$ $[$neglect$]$
$\Rightarrow \cos \theta =\frac{1}{2}$
$\Rightarrow \theta ={60}^{\circ }$
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