Suppose a 120 kg fireworks shell is shot into the air with a

Suppose a 12.0 kg fireworks shell is shot into the air with an initial velocity of 72.0 m/s at an angle of 80.0° above the horizontal. At the highest point of its trajectory, a small explosive charge separates it into two pieces, neither of which ignite (two duds). One 9.00 kg piece falls straight down, having zero velocity just after the explosion. Neglect air resistance (a poor approximation, but do it anyway). (a) At what horizontal distance (in m) from the starting point does the 9.00 kg piece hit the ground? m (b) Calculate the velocity (in m/s) of the 3.00 kg piece just after the separation. m/s (c) At what horizontal distance (in m) from the starting point does the 3.00 kg piece hit the ground? m

Solution

a)

the center of mass of shell from the starting point xcm = vo^2*sin(2theta)/g

xcm = 72^2*sin(2*80)/9.8 = 181 m

horizontal distance of 9 kg mass, x1 = R/2 = 90.5 m

(b)

at the maximum height

from conservation of momentum

M*V0*costheta = m1*v1 + m2*v2

12*72*cos80 = (9*0) + 3*v2

speed of 3 kg mass v2 = 50 m/s

(c)

Xcm = ( m1*x1 + m2*x2)/(m1+m2)

181 = ((9*90.5)+(3*x2))/(9+3)

x2 = 452.5 m <<<-----ANSWER