An instrument-carrying projectile accidentally explodes at the top of its trajec

Question

An instrument-carrying projectile accidentally explodes at the top of its trajectory. The horizontal distance between the launch point and the point of explosion is L. The projectile breaks into two pieces which fly apart horizontally. One piece has three times the mass of the other. To the surprise of the scientist in charge, the small piece returns to earth precisely at the launching station. How far away does the larger piece land? Neglect air resistance and effects due to the earth's curvature.

Explanation / Answer

let v be the projectile velocity and theta be the angle

maximum height reached

H = (v*sinthet)^2/(2g)

time taken to reach top = ttop = v*sintheta/g

L = v*costheta*t = v^2*costheta*sinthet/g = v^2*sin2theta /2g

at the top of the trajectory

initial momentum Pi = m*v*costheta

after explosion

for small mass m/4

along vertical

initial velocity voy =0

acceleration ay = -g

distance y = -H

from equation of motion

y = voy*t + 0.5*ay*t^2

t = sqrt(2H/g) = v*sintheta*sqrt(2/g)

along

horizantal

L = v1*t

v1 = L/t

after collision final momentum

Pf = -(m/4)*v1 + (3m/4)*v2

from momentum conservation

Pf = Pi

-(m/4)*v1 + (3m/4)*v2 = m*v*costheta

-L/t + 3*v2 = 4*m*v*costheta

v2 = 4*m*v*costheta + L/t

for larger piece 3m/4

X = v2*t

X = ( 4*v*costheta + L/t ) *t

X = 4*v*costheta*( v*sintheta*sqrt(2/g)) + L

X = 2*v^2*2*sintheta2*costheta*sqrt(2/g) + L

X = 2*v^2*sin2theta *sqrt(2/g) + L

X = 2*2*g*L*sqrt(2/g) + L

X = 4L*sqrt(2g) + L

X = L*( 4+sqrt(2g) ) ............answer

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