Question

Two towers $AB$ and $CD$ are situated at distance d apart as shown in Fig. $AB$ is $20m$ high and $CD$ is $30m$ high from the ground. An object of mass m is thrown from the top of $AB$ horizontally with a velocity of $10m{s}^{-1}$ towards $CD$. Simultaneously, another object of mass $2m$ is thrown from the lay of $CD$ at an angle ${60}^o$ to the horizontal towards $AB$ with the same magnitude of initial velocity us that of the first object. The two objects move in the same vertical plane, collide in mid-air, and stick to each other.
a. Calculate the distance d between the towers.
b. Find the position where the objects hit the ground.

Answer 1

Method 1:

a. Let $t$ be the time taken for collision. For mass $m$ thrown horizontally from $A$ for horizontal motion,

$PM=10t$ ...(i)

For vortical motion, $u_y=0,s_y=y,a_y=g$

$S_y=u_{y}t+\frac{1}{2}{a}_y{t}^2\Rightarrow Y=\frac{1}{2}g{t}^2$ ....(ii)

$v_y=u_y+a_{y}t=gt$ ....(iii)

For mass $2m$ thrown from $C$, for horizontal motion

$QM=\left[10\cos {60}^o\right]t\Rightarrow QM=5t$ .....(iv)

For vertical motion: $u_y=10\sin {60}^o=5\sqrt{3}$

$\Rightarrow v_y=5\sqrt{3}+gt$ .....(v)

$a_y=g,S_y=y+10,S=ut+\frac{1}{2}a{t}^2$

$\Rightarrow y+10=5\sqrt{3}t+\frac{1}{2}g{t}^2$ .....(vi)

From (i) and (vi),$\frac{1}{2}g{t}^2+10=5\sqrt{3}t+\frac{1}{2}g{t}^2\Rightarrow t=\frac{2}{\sqrt{3}}s$

$BD=PM+MQ$

$=10t+5t$

$=15t=15\times \frac{2}{\sqrt{3}}$

$=10\sqrt{3}=17.32m$

b. Applying the conservation of linear momentum (during collision of the masses at $M$) in the horizontal direction, we have

$m\times 10-2m10cos{60}^o$

$=3m\times v_x$

$\Rightarrow 10m-10m=3m\times v_x$

$\Rightarrow v_x=0$

Since the horizontal momentum comes out to be zero, the combination of masses will drop vertically downwards and fall at $E,BE=PM=10t=10\times \frac{3}{\sqrt{3}}=11.547m$

Method 2:

Acceleration of $A$ and Cboth is $9.8ms$ ${}^{-1}$ downward. Therefore, the relative acceleration between them is zero, i.e., the relative motion between them will be a straight line.

Now assuming A to be at rest, the condition of collision will be that ${\overrightarrow{v}}_{CA}={\overrightarrow{v}}_C-{\overrightarrow{v}}_A$ = relative velocity of $C$ w.r.t. $A$ should be along $CA$.

${\overrightarrow{v}}_A=10\hat{i}$

${\overrightarrow{v}}_B=-5\hat{i}-5\sqrt{3}\hat{j}$

${\overrightarrow{v}}_{BA}=-5\hat{i}-5\sqrt{3}\hat{j}-10\hat{i}$

$\therefore {\overrightarrow{v}}_{BA}=-15\hat{i}-5\sqrt{3}\hat{j}$

$\therefore tan{60}^o=\frac{d}{10}$

$ord=10\sqrt{3}m$