Question

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

Answer 1

Hint: In order to solve the question we need to understand the definition of projectile motion which states that the path traced by a launched object under only force of gravitation and it continues its motion by virtue of its own inertia. Also relative motion is that motion than in which frame of reference has been fixed to one body and other body motion is analyzed.

Complete step by step answer:
[a] For minimum separation we need to analyze the motion of body $A$ in respect of body $B$.According to the given question, the velocity of body $A$ is ${v_1} = 2\,m{s^{ - 1}}$ and ${v_2} = 14\,m{s^{ - 1}}$. So velocity of body $B$ with respect to body $A$ is \[{\vec v_{BA}} = {\vec v_2} - {\vec v_1}\]. Adding according rule of vector addition we get \[{v_{BA}} = \sqrt {\left( {{{14}^2}} \right) + \left( {{2^2}} \right)} \] as angle between them is $90^\circ $

The projectile diagram is shown as

So ${v_{BA}} = 14.142m{s^{ - 1}}$ and angle that it makes with \[{\vec v_B}\] is given by
$\tan \left( {\theta + 45} \right) = \dfrac{2}{{14}}$
$\Rightarrow \theta + 45 = {\tan ^{ - 1}}(\dfrac{2}{{14}}) = 8.130$
$\Rightarrow \theta = 8.130 - 45 = - 36.86^\circ $
Here negative sign implies that ${\vec v_{BA}}$ is below the horizontal by $36.86^\circ $
So we considered body A to be at rest then minimum separation between then is distance d
So $d = 60\sin (36.86)$
$\therefore d = 35.99\,m$

So the minimum separation between them is $35.99\,m$.

[b] For time we use that distance $BC = {d_1} = 60\cos (36.86) = 48.0062\,m$
So the time taken by particle to travel distance BC with speed ${\vec v_{BA}} = 14.142\,m{s^{ - 1}}$ and with $0$ relative acceleration ( ${\vec a_{BA}} = {\vec a_B} - {\vec a_A} = g - g = 0$ ) is $t = \dfrac{{{d_1}}}{{{{\vec v}_{BA}}}}$.
$t = \dfrac{{48.0062\,m}}{{14.142\,m\,{{\sec }^{ - 1}}}} \\
\therefore t= 3.394\,s$

So the minimum time required is $3.394\,s$.

Note: It should be remembered that relative projectile motion would only be applied if one particle is considered at rest and only other moves. Here body $A$ is considered to be at rest while $B$ is moving but if reverse analysis done means $B$ is at rest and $A$ is moving the result would be the same. Also in relative projectile motion particles travel in a straight line.