A fireworks rocket is fired vertically upward. At its maximum height of 85.0 m ,

Question

A fireworks rocket is fired vertically upward. At its maximum height of 85.0 m , it explodes and breaks into two pieces, one with mass mA = 1.35 kg and the other with mass mB = 0.250 kg . In the explosion, 950 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

A) At the maximum height, velocity of the rocket is zero.

Conservation of momentum:

mAvA + mBvB = 0

=> vB = -(mA/mB)vA = -(1.35/0.25)vA = -5.4vA

Total kinetic energy of the two masses = 950 J

=> mAvA2/2 + mBvB2/2 = 950

=> 1.35vA2 + 0.25(-5.4vA)2 = 1900

=> vA = 14.83 m/s

Magnitude of vB = 5.4vA = 5.4 * 14.83 = 80.1 m/s

B) Since the fragments hit the ground at the same time, we can deduce that their initial velocity in vertical direction is zero since that's the only way their initial momentum can be conserved in vertical direction.

Time taken to come down, t = (2s/a)1/2 = (2 * 85 / 9.81)1/2 = 4.16 s

Distance between the particles, d = (80.1 + 14. 83) * 4.16 = 394.9 m

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