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 90.0m , it explodes and breaks into two pieces, one with mass mA = 1.50kg and the other with mass mB = 0.270kg . In the explosion, 910J 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)Let speeds are Va & Vb.

momentum conservation jst after

maVa = mbVb (1)

0.5maVa^2 + 0.5mbVb^2 = 910 (2)

putting values and solving

Va=13.6 m/s.

Vb=75.581 m/s.

b)time taken to fall = sqrt(2h/g) = 4.2857 s.

So, distance = (Va+Vb)*t = 382.2 m.

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