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A 24kg proj...

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A $24 k g$ projectile is fired at an angle of $53^{\circ}$ above the horizontal with an initial speed of $50 m / s$ . At the highest point in its trajectory, the projectile explodes into two fragments of equal mass, the first of which fall vertically with zero initial speed. How much energy in J was released during the explosion?

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Solution

Here, $m = 24 k g, u = 50 m / s$
$v_{x} = \frac{d x}{d t} = u cos 53 \Rightarrow x = u cos 53. t . . . (1)$ and
$v_{y} = \frac{d y}{d t} = - g t + u sin 53 \Rightarrow y = - \frac{1}{2} g t^{2} + u sin 53. t . . . (2)$
At highest point, $v_{y} = 0$ , so $- g t + u sin 53 = 0 \Rightarrow t = \frac{50 sin 53}{10} = 4 s$ and from (1) $x = 50 cos 50. 4 = 120 m$
Momentum is conserved in exploration . If $v_{x}^{'}$ be the velocity of second particle , then $m v_{x} = (m / 2) v_{x}^{'} \Rightarrow v_{x}^{'} = 2 v_{x} = 2 (50 cos 53) = 60 m / s$
The energy in J released during the explosion $= (m / 2) ((v_{x}^{' 2}) / 2 - v_{x}^{2}) = 12 ((60^{2}) / 2 - 30^{2}) = 10800$

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