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Momentum

A 24kg proj...

Question

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 far (in m) from the point of firing does the second fragment strike the ground? (Assume the ground is level)

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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 total distance covered by the second fragment along x component is
$x^{'} = 60 t + x = 60 (4) + 120 = 360 m$

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