Question

A fireworks rocket is moving at a speed of v = 45.7 m/s. The rocket suddenly breaks into two pieces of equal mass, which fly off with velocities v1 at an angle of theta1 = 30.7° and v2 at an angle of theta2 = 59.3° as shown in the drawing below.

Answer 1

give data

Initial mass of rocket M = M
Initial speed v = 45.7 m / s

Initial momentum of the rocket P = M v = M(45.7) kg.m/s

Mass of rocket of each piece m = M / 2

when inyial direction of rocket

the momentum of the rockets

       P' = ( M/2) v₁ cos 30 .7+ ( M/2) v₂ cos 59.3

            = M [ 0.42 v₁ + 0.25 v₂ ]

from law of conservation of momentum ,

                      P = P'

          M(45.7) =   M [ 0.42 v₁ + 0.25 v₂ ]

45.7 = [ 0.42 v₁+0.25 v₂]--------1   

perpendicular to the initial direction of rocket is

  0 = ( M/2 ) v₁ sin30.7 + (M/2)(-v₂ )sin59.3

0 = M [ 0.25 v ₁ - 0.429 v₂ ]-------2

from 1 and 2

v_{2=47 m/s}

_{and substitute this value in 1 or 2}

v₁=80.82 m/s