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 = 1.25 kg and the other with mass mB = 0.220 kg . In the explosion, 940 J of chemical energy is converted to kinetic energy of the two fragments. What is the speed of each fragment just after the explosion? 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

use from conservatio principle

Initail momentum = final moemtnum

m1u1 = m2v2

V1 = 0.22V2/1.25

V1 = 0.176 V2

ENergy E = 940 = 0.5 m1u1^2 + 0.5 m2v2^2

0.5 * 1.25 * (0.176 V)^2 + 0.5 * 0.22* v^2 = 940

0.01936V^2 + 0.11v^2 = 940

V2 = sqrt(940/(0.11+ 0.01936)

V2 = 85.24 m./s

so v1 = 0.176 * 85.24

V1 = 15 m/s

--------------------------

Distance travelled is Y = 0.6 gt^2

here t^2 = 2H/g

t =sqrt(2 * 85/9.8)

t = 4.164 secs

so distance D2 = V1 t = 85.24 * 4.164 = 355m

D1 = 15* 4.164 = 62.46 m

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