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Question: A shell is fired from a cannon with a velocity \[v\] at an angle \[\theta \] with the horizontal dir...

A shell is fired from a cannon with a velocity vv at an angle θ\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. 3vcosθ3v\cos \theta
B. 2vcosθ2v\cos \theta
C. 32vcosθ\dfrac{3}{2}v\cos \theta
D. 32vcosθ\dfrac{{\sqrt 3 }}{2}v\cos \theta

Explanation

Solution

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
m1u1+m2u2=m1v1+m2v2{m_1}{u_1} + {m_2}{u_2} = {m_1}{v_1} + {m_2}{v_2} …… (1)
Here, m1{m_1} and m2{m_2} are the masses of the two objects, u1{u_1} and u2{u_2} are the initial velocities of the two objects before collision and v1{v_1} and v2{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 vv at angle of θ\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 vcosθv\cos \theta which is constant throughout the projectile motion of the shell.
Let the shell fired from the cannon have mass 2m2m and after explosion it divides into two equal pieces each of mass mm.
One piece retraces the original path of the shell with the same horizontal velocity vcosθv\cos \theta as before but in the opposite direction.
Let the velocity of the other piece is VV.
Calculate the velocity VV of the second piece of the shell using law of conservation of linear momentum in the horizontal direction.
2mvcosθ=mvcosθ+mV2mv\cos \theta = - mv\cos \theta + mV
The negative sign of mvcosθmv\cos \theta represents that the first piece retraces the same path.
Rearrange the above equation for VV.
V=3mvcosθmV = \dfrac{{3mv\cos \theta }}{m}
V=3vcosθ\Rightarrow V = 3v\cos \theta
Therefore, the speed of the other piece immediately after explosion is 3vcosθ3v\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.