9 of 9 A fireworks rocket is fired vertically upward. At its maximum height of 7
ID: 1581645 • Letter: 9
Question
9 of 9 A fireworks rocket is fired vertically upward. At its maximum height of 70.0 m , it explodes and breaks into two pieces, one with mass mA = 1.20 kg and the other with mass mB = 0.290 kg . In the explosion, 820 J of chemical energy is converted to kinetic energy of the two fragments. Part A What is the speed of each fragment just after the explosion? Enter your answers numerically separated by a comma. UA, UB = m/s Submit Request Answer Part 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
Begin by conserving momentum:
initial p = 0 = final p
0 = 1.20kg * vA + 0.290kg * vB
vB = -vA * 1.20 / 0.290 = -4.14*vA
Chemical energy is converted to KE, so
820 J = ½ * 1.20kg * vA² + ½ * 0.290kg * vB²
multiply by 2 and substitute for vB
1640 J = 1.20kg * vA² + 0.290kg * (-4.14*vA)² = 2.4kg * vA²
vA = 26.14 m/s
vB = -4.14*26.14m/s = - 108.2m/s meaning "in the other direction"
Their relative velocity is
v = vA - vB = 134.36 m/s
time to fall t = (2h / g) = (2 * 70.0m / 9.8m/s²) = 3.78 s
(b) so when they land the distance is
d = v*t = 134.36m/s * 3.78s = 507.89 m
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