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 = 1.45 kg and the other with mass mB = 0.230 kg . In the explosion, 920 J of chemical energy is converted to kinetic energy of the two fragments.

What is the speed of each fragment just after the explosion? Enter your answers numerically separated by a comma.

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

here

after explosion

m1 * v1 = m2 * v2

v1 = 0.23 * v2 / 1.45 = 0.158 v2

then

E = 920 = 0.5 * m1 * v1^2 + 0.5 * m2 * v2^2

920 = 0.5 * 1.45 * 0.025 * v2^2 + 0.5 * 0.23 * v2^2

920 = 0.133125 * v2^2

v2 = 83.1 m/s

v1 = 0.158 * 83.1 = 13.13 m/s

b)

by using second equation of motion

y = u*t + 0. 5 * a * t^2

90 = 0 + 0.5 * 9.8 * t^2

t = sqrt( 90 * 2 / 9.8)

t = 4.28 sec

then

x1 = 13.13* 4.28 = 56.2 m

x2 = 83.1 * 4.28 = 355.6 m

D = x1+ x2 = 355.6 + 56.2 = 411.8 m

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