Question

Two particles are located on a horizontal plane at a distance 60 m. At t = 0 both the particles are simultaneously projected at angle ${45}^o$ with velocities $2m{s}^{-1}$ and $14m{s}^{-1}$, respectively. Find
a. Minimum separation between them during motion
b. At what time is the separation between them minimum?

Answer 1

In relative motion, observer considers himself at rest and observes the motion of object. Graphically, we can draw the direction of motion of particle 2 w.r.t. particle l.

Both the particles are moving in gravitation field with same acceleration g. Hence, relative acceleration of particle 2 as seen from particle 1 will be zero. It means the relative velocity of the particle 2 w.r.t. particle 1 will be constant and will be equal to initial relative velocity. Graphically, we can draw the situation as shown in Fig.

AN is the minimum separation between the particles and BN is the relative separation between the particles when the distance between 1 and 2 is shortest. From figure, we can write

$v_{12}\cos \theta =14\cos {45}^o+2\cos {45}^o$ ...(i)

$v_{12}\sin \theta =14sin{45}^o-2\sin {45}^o$ ...(ii)

From (i) and (ii), $v_{12}=10\sqrt{2}m{s}^{-1}$

$cos\theta =\frac{4}{5}$ and $sin\theta =\frac{3}{5}$, as $\theta ={37}^o$

Hence, minimum separation between the particles

$=AN=AB\sin \theta =60\times \frac{3}{5}=36m$

The time when separation between the particles is minimum,

$t=\frac{BN}{|{\overrightarrow{v}}_{12}|}\Rightarrow t=\frac{60\cos {37}^o}{10\sqrt{2}}=\frac{12\sqrt{2}}{5}s$