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# A bomb is at the summit of a cliff and breaks into two equal fragments. One of the fragments attains a horizontal velocity$20\sqrt 3 \dfrac{m}{s}$. The horizontal distance between the two fragments, when their displacement vector is inclined at $60^\circ$ relative to each other is$g = 10\dfrac{m}{{{s^2}}}$.A) $40\sqrt 3$B) $80\sqrt 3$C) $120\sqrt 3$D) $480\sqrt 3$

Last updated date: 04th Aug 2024
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Hint:The linear conservation of momentum is always conserved. The horizontal component of the velocity in the projectile motion has zero acceleration. Simple formula of speed can be used here for the horizontal component of the projectile motion.

Formula used: The formula for calculation of displacement of motion is given by $s = ut + \dfrac{1}{2}a{t^2}$ where u is the initial velocity t is the time taken and a is the acceleration of the body.

Step by step solution:
As the bombshell breaks into two equal parts in which one fragment attains a horizontal velocity of $20\sqrt 3 \dfrac{m}{s}$ and we need to find the value of distance between the two particles when their displacement vector is inclined at$60^\circ$ relative to each other.
The linear momentum will be always conserved,
$\Rightarrow {m_1}{v_1} + {m_2}{v_2} = 0$
Where ${m_1}$ and ${m_2}$ are the masses of two particles with velocity ${v_1}$ and ${v_2}$ respectively.
$\Rightarrow m{v_2} + m\left( {20\sqrt 3 } \right) = 0$
The horizontal component of the velocity is,
$\Rightarrow {v_2} = - 20\sqrt 3 \dfrac{m}{s}\hat i$
The vertical component of the velocity is given by,
$\Rightarrow s = ut + \dfrac{1}{2}a{t^2}\hat j$
$\Rightarrow s = - \dfrac{1}{2}\left( {10} \right){t^2}\hat j$
$\Rightarrow s = - 5{t^2}\hat j$
The displacement for particle 1 will is ${s_1} = 20\sqrt 3 \hat i - 5{t^2}\hat j$ and displacement for second particle is ${s_2} = - 20\sqrt 3 \hat i - 5{t^2}\hat j$.
It is given that the displacement vector of the two particles have $60^\circ$ between them and therefore taking dot product of the two displacement vectors.
$\Rightarrow {s_1} \cdot {s_2}$
Since dot product is given by,
$\vec a \cdot \vec b = \left| {\vec a} \right| \cdot \left| {\vec b} \right|\cos \theta$
Applying dot product
$\Rightarrow \left( {20\sqrt 3 \hat i - 5{t^2}\hat j} \right) \cdot \left( { - 20\sqrt 3 \hat i - 5{t^2}\hat j} \right) = \left( {25{t^4}} \right) \cdot \left( {1200{t^2}} \right) \cdot \cos 60^\circ$
$\Rightarrow \cos 60^\circ = \dfrac{{\left( {25{t^4}} \right) \cdot \left( {1200{t^2}} \right)}}{{\left( {20\sqrt 3 \hat i - 5{t^2}\hat j} \right) \cdot \left( { - 20\sqrt 3 \hat i - 5{t^2}\hat j} \right)}}$
$\Rightarrow \dfrac{1}{2} = \dfrac{{\left( {25{t^4}} \right) + \left( {1200{t^2}} \right)}}{{\left( {20\sqrt 3 \hat i - 5{t^2}\hat j} \right) \cdot \left( { - 20\sqrt 3 \hat i - 5{t^2}\hat j} \right)}}$
$\Rightarrow \dfrac{1}{2} = \dfrac{{\left( {25{t^4}} \right) + \left( {1200{t^2}} \right)}}{{ - 1200{t^4} + 1200{t^2}}}$
$\Rightarrow \dfrac{1}{2} = \dfrac{{\left( {25{t^4}} \right) + \left( {1200{t^2}} \right)}}{{ - 1200{t^4} + 1200{t^2}}}$
$\Rightarrow 25{t^2} + 1200 = 50{t^2} - 2400$
$\Rightarrow 25{t^2} = 3600$
$\Rightarrow t = 12s$
The horizontal distance is equal to,
Since, ${\text{distance}} = {\text{speed}} \times {\text{time}}$
$\Rightarrow d = 40\sqrt 3 \times t$
Replace the value of time$t = 12s$.
$\Rightarrow d = 40\sqrt 3 \times t$
$\Rightarrow d = 40\sqrt 3 \times 12$
$\Rightarrow d = 480\sqrt 3 m$.
So the horizontal distance between the two parts is$d = 480\sqrt 3 m$.

The correct answer for this problem is option D.

Note:There is a negative sign in the vertical component of the displacement because the particles are moving towards the ground which is the negative direction of the y-axis. The two particles move in the exact opposite direction in horizontal direction as their magnitude is the same but the sign is opposite.