Question

Two particles $A$ and $B$ are moving with uniform velocity as shown in figure given below at $t = 0$ .

(a) Will the two particles collide?
(b) Find out the shortest distance between two particles.

Accepted Answer

Let the relative velocity of $A$ with respect to $B$ be ${\vec{v}}_{A B}$
${\vec{v}}_{A B} = {\vec{v}}_{A} - {\vec{v}}_{B} = {\vec{v}}_{A} + (- {\vec{v}}_{B})$

The angle $θ$ that ${\vec{v}}_{A B}$ makes with the vertical axis, satisfies
$\tan (θ) = \frac{10}{20} = \frac{1}{2}$

$\Rightarrow \cos (θ) = \frac{1}{\sqrt{1 + \tan^{2} (θ)}} = \frac{1}{\sqrt{1 + {(\frac{1}{2})}^{2}}} = \frac{2}{\sqrt{5}}$

Also, it can be seen in the diagram below that
$\tan (θ) = \frac{A D}{C D} = \frac{A D}{40}$

Hence,

$\frac{A D}{40} = \frac{1}{2}$
$\Rightarrow A D = 20 m$

$\Rightarrow D O = A O - A D = 30 - 20 = 10 m$
$\Rightarrow B C = 10 m$

Hence, the shortest distance between the particles
$d = B C \times \cos (θ) = 10 \times \cos (θ) = \frac{10 \times 2}{\sqrt{5}} = 4 \sqrt{5} m$

Since the closest distance is non-zero, therefore, the particles will not collide.

B M Sharma Solutions for Chapter: Kinematics II, Exercise 72: ILLUSTRATION 5.68

B M Sharma Physics Solutions for Exercise - B M Sharma Solutions for Chapter: Kinematics II, Exercise 72: ILLUSTRATION 5.68

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