A cannon elevated at some angle (unknown) fires a shell with some initial speed

Question

A cannon elevated at some angle   (unknown) fires a shell with some initial speed v0 (unknown) that has a range R when fired over level ground (see Figure (a)). During a test fire, a shell explodes at the top of the trajectory into two unequal mass fragments. The shell explodes in such a manner that neither fragment experiences a change in momentum in the y direction. However, 10% of the shell (m2) continues onward and the other fragment (m1) falls straight down. Determine the distance D(in terms of R) that the forward moving fragment lands from the cannon (see Figure (b)).
D = _____

Explanation / Answer

for fig (a)

before exploding

Range R = vo*sin(2theta) / g

maximum heigt h = vo^2*(sintheta)^2/2g

at the maximum height

speed of the cannon before exploding = vo*costheta

initial momentum Pix = M*vo*costheta

after exploding

speed of m1 = v1 = 0

speed of m2 = v2 = ?

final momentum Pf = m1*v1 + m2*v2

given m1 = 0.9 M

m2 = 0.1 m

from momentum conservation

Pf = Pi

0.1*v2 = vo*costheta

v2 = 10*vo*costheta

distance travelled by m2 in horizantl direction after exploding

x = v2*sqrt(2h/g)

x = 10*vo*costheta*sqrt(vo^2*(sintheta)^2/g^2 )

x = 10*vo^2*sintheta*costheta/g

x = 5*vo^2*sin(2theta)/g = 5R

D = R/2 + X = R/2 + 5R = 11R/2 = 5.5R

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