Question

Plz Solve the question.​

Answer 1

The component of acceleration of particle B along y axis will be $\displaystyle\sf {a\cos\theta}$ and that along x axis will be $\displaystyle\sf {a\sin\theta.}$

The particle A moves along the line $\displaystyle\sf {y=30\ m,}$ which is perpendicularly away from x axis by a distance of 30 m.

This means $\displaystyle\sf {AB=30\ m.}$

Let both the particles collide each other after a time $\displaystyle\sf {t.}$ During this time let the particle A travel a distance $\displaystyle\sf {x}$ along the line $\displaystyle\sf {y=30\ m.}$

So the horizontal displacement of particle B will also be this $\displaystyle\sf {x,}$ since they're about to collide.

Initial speed of particle B is zero.

By second equation of motion, the displacement of particle B along y axis (vertical displacement) is given by,

$\displaystyle\sf{\longrightarrow 30=0\cdot t+\dfrac {1}{2}\cdot a\cos\theta\cdot t^2}$

$\displaystyle\sf{\longrightarrow 30=\dfrac {1}{2}\cdot0.4\cos\theta\cdot t^2}$

$\displaystyle\sf{\longrightarrow 30=0.2\cos\theta\cdot t^2}$

$\displaystyle\sf{\longrightarrow t^2=\dfrac {150}{\cos\theta}}$

$\displaystyle\sf{\longrightarrow t=\sqrt{\dfrac {150}{\cos\theta}}}$

By second equation of motion, the displacement of particle B along x axis (horizontal displacement) is given by,

$\displaystyle\sf{\longrightarrow x=0\cdot t+\dfrac {1}{2}\cdot a\sin\theta\cdot t^2}$

$\displaystyle\sf{\longrightarrow x=\dfrac {1}{2}\cdot 0.4\sin\theta\cdot\dfrac {150}{\cos\theta}}$

$\displaystyle\sf{\longrightarrow x=\dfrac {30\sin\theta}{\cos\theta}\quad\quad\dots(1)}$

The particle A moves with constant velocity $\displaystyle\vec {\sf {v}}$ so it has no acceleration.

By second equation of motion, displacement of particle A,

$\displaystyle\sf{\longrightarrow x=vt+\dfrac {1}{2}\cdot0\cdot t^2}$

$\displaystyle\sf{\longrightarrow x=3\sqrt{\dfrac {150}{\cos\theta}}\quad\quad\dots(2)}$

From (1) and (2),

$\displaystyle\sf{\longrightarrow \dfrac {30\sin\theta}{\cos\theta}=3\sqrt{\dfrac {150}{\cos\theta}}}$

$\displaystyle\sf{\longrightarrow \dfrac {100\sin^2\theta}{\cos^2\theta}=\dfrac {150}{\cos\theta}}$

$\displaystyle\sf{\longrightarrow \dfrac {\sin^2\theta}{\cos\theta}=\dfrac {3}{2}}$

$\displaystyle\sf{\longrightarrow \dfrac {1-\cos^2\theta}{\cos\theta}=\dfrac {3}{2}}$

$\displaystyle\sf{\longrightarrow 2\cos^2\theta+3\cos\theta-2=0}$

$\displaystyle\sf{\longrightarrow 2\cos^2\theta+4\cos\theta-\cos\theta-2=0}$

$\displaystyle\sf{\longrightarrow 2\cos\theta(\cos\theta+2)-(\cos\theta+2)=0}$

$\displaystyle\sf{\longrightarrow(2\cos\theta-1)(\cos\theta+2)=0}$

Since $\displaystyle\sf {-1\leq\cos\theta\leq1,}$

$\displaystyle\sf{\longrightarrow\cos\theta=\dfrac {1}{2}}$

$\displaystyle\sf{\longrightarrow\underline {\underline {\theta=60^o}}}$

Therefore, the particles undergo collision if particle B moves at an angle $\displaystyle\bf {60^o}$ with positive y axis.