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 $45 °$ with velocities $2 {m s}^{- 1}$ and $14 {m s}^{- 1}$ , respectively.

Question Image

$(a)$ Find the minimum separation between them during motion.
$(b)$ At what time is the separation between them minimum?

Accepted Answer

It is given that two particles, particle $1$ and particle $2$ , are projected from two points $A$ and $B$ as shown in the diagram in the question. We need to find the shortest distance between them during the motion and the time at which it occurs.

If we fix our coordinate system on particle $1$ , the relative motion of particle $2$ can be observed.

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. Hence, particle $2$ will appear to move in a straight line.

The relative motion of the particle $2$ with respect to particle $1$ will appear as shown in the figure below.

$A N$ , the perpendicular drawn from $A$ to the trajectory of the relative motion, is the minimum separation between the particles and $B N$ is the relative displacement of particle $2$ when the distance between $1$ and $2$ is shortest.

We will denote the relative velocity as $v_{12}$ '

From the figure, we can write

$v_{12} \cos (θ) = 14 \cos (45 °) + 2 \cos (45 °) = \frac{16}{\sqrt{2}}$ ... $(i)$
$v_{12} \sin (θ) = 14 \sin (45 °) - 2 \sin (45 °) = \frac{12}{\sqrt{2}}$ .... $(ii)$

From $(i)$ and $(ii)$ , upon taking ratio,

$\tan (θ) = \frac{12}{16} = \frac{3}{4}$
$\Rightarrow θ = 37 ° \Rightarrow \cos (θ) = \frac{4}{5}, \sin (θ) = \frac{3}{5}$

$\Rightarrow v_{12} = \frac{16}{\sqrt{2}} \times \frac{5}{4} = 10 \sqrt{2} m s^{- 1}$
$(a)$ Hence, minimum separation between the particles
$= A N = A B \sin (θ) = 60 \times \frac{3}{5} = 36 m$
$(b)$ The time when the separation between the particles is minimum,
$t = \frac{B N}{|{\vec{v}}_{12}|} \Rightarrow t = \frac{60 \cos (37 °)}{10 \sqrt{2}} = \frac{12 \sqrt{2}}{5} s$

B M Sharma Solutions for Chapter: Kinematics II, Exercise 69: ILLUSTRATION 5.65

B M Sharma Physics Solutions for Exercise - B M Sharma Solutions for Chapter: Kinematics II, Exercise 69: ILLUSTRATION 5.65

Questions from B M Sharma Solutions for Chapter: Kinematics II, Exercise 69: ILLUSTRATION 5.65 with Hints & Solutions

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