A fireworks rocket is fired vertically upward. At its maximum height of 90.0 m ,

Question

A fireworks rocket is fired vertically upward. At its maximum height of 90.0 m , it explodes and breaks into two pieces, one with mass mA = 1.35 kg and the other with mass mB = 0.290 kg . In the explosion, 800 J of chemical energy is converted to kinetic energy of the two fragments.

a)What is the speed of each fragment just after the explosion?

b)It is observed that the two fragments hit the ground at the same time. What is the distance between the points on the ground where they land? Assume that the ground is level and air resistance can be ignored.

Explanation / Answer

Hmax = 90 m

ma = 1.35 kg

mb = 0.29 kg

E = 800 J

a. speed of ma after explosion = ua

speed of mb after explosion = ub

hence frpm conservation of meomtnum

maua + mbub = 0

1.35ua = -0.29ub

ub = -4.65517ua

also

E = 0.5maua^2 + 0.5mbub^2 = 0.5maua^2 + 0.5mb*(4.65517^2)ua^2 = 800

ua = 14.4767 m/s

ub = -67.3916225 m/s

b. the speeds ub and ua are horizontal hence

relative speed = ua - ub = 81.8683225 m/s

time of fall = t

Hmax = 0.5gt^2

t = 4.2835 s

hence

distance = (ua - ub)t = 350.6853638 m

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