Question

In the diagram particle A moves along the line y = 25$\sqrt{3}$ m with a constant velocity $\overrightarrow{v}$ of magnitude 5 m/s and parallel to the x axis. At the instant particle A passes the y axis, particle B leaves the origin with a zero initial speed and a constant acceleration $\overrightarrow{a}$ of magnitude 4 m/$s^2$. What angle $\theta$ between $\overrightarrow{v}$ and the positive direction of the y axis would result in a collision?

• A. ${30}^{\circ }$
• B. ${45}^{\circ }$
• C. ${60}^{\circ }$
• D. ${37}^{\circ }$

Answer 1

The correct option is A

${30}^{\circ }$

Resolving the velocity and acceleration of Ball B in x and y direction.

Ball B: $v_x$ = 0

$a_x$ = 4 sin $\theta$

$s_y$ = $\frac{1}{2}\left(4cos\theta \right)t^2$

Where $a_y$ = 4 cos $\theta$

Now ball A's position after time t

Velocity is constant and in x direction

$\Rightarrow$ s = 5t

Coordinates will be ( 5t, 25 $\sqrt{3}$ )

$\Rightarrow$ Ball B should move $25\sqrt{3}$ on y axis to catch up with A

$\Rightarrow$ $s_y$ = 2 cos $\theta t^2$ = $25\sqrt{3}$

t = $\sqrt{\frac{25\sqrt{3}}{2cos\theta }}$ ............(1)

And also Ball B should move 5t on x axis to catch A 2 sin $\theta t^2$ = 5t

Substituting the value of t from equation (1)

$t=\frac{5}{2sin\theta }$ = $\sqrt{\frac{25\sqrt{3}}{2cos\theta }}$

Squaring both sides $\frac{25}{4si{n}^2\theta }$ = $\frac{25\sqrt{3}}{2cos\theta }$

$2si{n}^2\theta$ = $\frac{cos\theta }{\sqrt{3}}$

2 ( 1 - $co{s}^2\theta$) = $\frac{cos\theta }{\sqrt{3}}$

$\Rightarrow$ $2co{s}^2\theta$ + $\frac{cos\theta }{\sqrt{3}}$ - 2 = 0

$cos\theta =\frac{-\frac{1}{\sqrt{3}}\pm \sqrt{\frac{1}{3}+16}}{4}$

$cos\theta =\frac{\frac{6}{\sqrt{3}}}{4}=\frac{\sqrt{3}}{2}$

$\Rightarrow$ $\theta ={30}^{\circ }$