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A projectile is fired with velocity $u$ at an angle $\theta $ with the horizontal. At the highest point of its trajectory, it splits into three segments of masses, $m$, $m$, $2m$. The first part falls with zero initial velocity vertically downward and second part returns back via the same path to the point of projection. Then just after explosion, velocity of third part of mass $2m$ will be-
(A). $u\cos \theta $
(B). $\dfrac{3}{2}u\cos \theta $
(C). $2u\cos \theta $
(D). $\dfrac{5}{2}u\cos \theta $

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Last updated date: 20th Jun 2024
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Answer
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Hint: A projectile stops at its highest point, so its vertical velocity becomes zero. It breaks into three individual systems. Since the system is isolated, the momentum and energy is conserved. Calculating velocity of each particle from the conditions and substituting corresponding values in the equation of law of conservation of momentum, we can calculate the missing velocity.

Formula used:
$mu={{m}_{1}}{{v}_{1}}+{{m}_{2}}{{v}_{2}}+{{m}_{3}}{{v}_{3}}$

Complete step-by-step solution:
The projectile breaks into three particles of masses- $m,\,m,\,2m$
Total mass of the projectile- $m+m+2m=4m$

At the highest point the vertical velocity becomes zero; therefore, the velocity of the projectile will be-
$v=u\cos \theta $, i.e. velocity in the horizontal direction.

As one particle traces back its path to the starting point, so its velocity will be-
$v'=-u\cos \theta $, negative sign indicates velocity is in the opposite direction.

One of the particles falls downwards vertically. So it will have pure vertical velocity and its horizontal velocity is 0.

As there are no external forces acting on the projectile, the system is conserved. This means that the initial momentum of the system is equal to the final momentum of the system. Therefore,
$\begin{align}
  & mu={{m}_{1}}{{v}_{1}}+{{m}_{2}}{{v}_{2}}+{{m}_{3}}{{v}_{3}} \\
 & \Rightarrow 4mu\cos \theta =-u\cos \theta +0+2mv'' \\
 & \Rightarrow 5mu\cos \theta =2mv'' \\
 & \Rightarrow 2v''=5mu\cos \theta \\
 & \therefore v''=\dfrac{5}{2}u\cos \theta \\
\end{align}$
Therefore, the final velocity of the third particle of mass $2m$ is $\dfrac{5}{2}u\cos \theta $.

Hence, the correct option is (D).

Note:
When a system is isolated, it either travels with a constant velocity or remains at rest until an external force acts on it. In a projectile, a body has horizontal as well as vertical motion, so its velocity can be resolved in the horizontal component as well as the vertical component. In a projectile motion, a body follows a parabolic path.