1. Tardigrade
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  4. A projectile is fired with velocity u at an angle θ with horizontal. At the highest point of its trajectory it splits up into three segments of masses m, m and 2 m. First part falls vertically downward with zero initial velocity and second part returns via same path to the point of projection. The velocity of third part of mass 2 m just after explosion will be

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Q. A projectile is fired with velocity $u$ at an angle $\theta$ with horizontal. At the highest point of its trajectory it splits up into three segments of masses $m, m$ and $2\, m$. First part falls vertically downward with zero initial velocity and second part returns via same path to the point of projection. The velocity of third part of mass $2\, m$ just after explosion will be

Laws of Motion

Solution:

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along $x$-axis no internal force exists, hence momentum will be conserved along $x$-axis
$\left.\left.P_{i}\right)_{x}=P_{f}\right)_{x}$
$(m+m+2 m) \cos \theta=-m u \cos \theta+0+2 m V$
$\Rightarrow 2 m v=5 m u \cos \theta$
$v=\frac{5}{2} u \cos \theta$
and along $y$ direction $\left.P_{i}\right)_{y}=0$
So $\left.P_{f}\right)_{y}=0$