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A shell is fired from a cannon with a velocity \[v\] at an angle \[\theta \] with the horizontal direction. At its highest point in its path, it explodes into two pieces of equal mass. One of the piece retraces its path. Speed of the other piece immediately after the explosion is:
A. \[3v\cos \theta \]
B. \[2v\cos \theta \]
C. \[\dfrac{3}{2}v\cos \theta \]
D. \[\dfrac{{\sqrt 3 }}{2}v\cos \theta \]

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Last updated date: 13th Jun 2024
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Answer
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Hint: Use law of conservation of linear momentum.
 Use this conservation law only in a horizontal direction.

Formula used:
The conservation of linear momentum of two objects that undergoes collision is
\[{m_1}{u_1} + {m_2}{u_2} = {m_1}{v_1} + {m_2}{v_2}\] …… (1)
Here, \[{m_1}\] and \[{m_2}\] are the masses of the two objects, \[{u_1}\] and \[{u_2}\] are the initial velocities of the two objects before collision and \[{v_1}\] and \[{v_2}\] are the final velocities of the two objects after collision.

Complete step by step answer:
The shell undergoes projectile motion when fired from a cannon with a velocity \[v\] at angle of \[\theta \]. At the maximum height, the shell explodes into two equal masses.
The shell (projectile) has no vertical velocity at the maximum height. Hence, at the maximum height, the shell has only horizontal velocity \[v\cos \theta \] which is constant throughout the projectile motion of the shell.
Let the shell fired from the cannon have mass \[2m\] and after explosion it divides into two equal pieces each of mass \[m\].
One piece retraces the original path of the shell with the same horizontal velocity \[v\cos \theta \] as before but in the opposite direction.
Let the velocity of the other piece is \[V\].
Calculate the velocity \[V\] of the second piece of the shell using law of conservation of linear momentum in the horizontal direction.
\[2mv\cos \theta = - mv\cos \theta + mV\]
The negative sign of \[mv\cos \theta \] represents that the first piece retraces the same path.
Rearrange the above equation for \[V\].
\[V = \dfrac{{3mv\cos \theta }}{m}\]
\[ \Rightarrow V = 3v\cos \theta \]
Therefore, the speed of the other piece immediately after explosion is \[3v\cos \theta \].

So, the correct answer is “Option A”.

Note:
Since the vertical velocity of the cannon at its highest point is zero,
The linear momentum conservation law is applied only in horizontal direction.