Question

PLEASE SHOW CALCULATIONS

A mass M has initial velocity v=5.2뛰+ 8.9 -8. which have values M1-M, M2 = M3 = After the breakup, the masses have velocities: i-1.491-5.55+ 3.5%k k. It subsequently splits into three smaller masses Mi,M2, and M3 What angle (in degrees) does the vector v3 make with the positive z-axis (i.e. with k)? Answer: 30.2 The correct answer is: 129.03

Answer 1

momentum before collision = momentum after collision

M*v = M1*v1 + M2*v2 + M3*v3

M*(5.2i + 8.9 j - 8.1 k) = (M/2)*(1.4i - 5.5j + 3.5 k) + (M/4)*(2.8 + 5j - 3.5 k) + (M/4)*v3

(5.2i + 8.9 j - 8.1 k) = (1/2)*(1.4i - 5.5j + 3.5 k) + (1/4)*(2.8 + 5j - 3.5 k) + (1/4)*v3

(5.2i + 8.9 j - 8.1 k) = 0.7i - 2.75 j + 1.75 k + 0.7 i + 1.25 j - 0.875 k + v3/4

v3 = 4*(5.2 - 0.7 - 0.7)i + 4*(8.9 + 2.75 - 1.25)j -4*(8.1 + 1.75 - 0.875 )k

v3 = 15.2 i + 41.6 j - 35.9 k

costheta = v3z/v3

costheta = -35.9/(sqrt(15.2^2+41.6^2+35.9^2)

costheta = -0.6297

theta = cos^-1(-0.6297) = 129.03 <<<<_-------------ANSWER