Student A has half of the kinetic energy of student B. Student A speeds up by 1.0 m/s?

Question

Student A has half of the kinetic energy of student B. Student A speeds up by 1.0 m/s, at which point she has the same kinetic energy as student B. If student A's mass is twice the mass of student B, what was the original speed of student A?

Answer 1

4MbVa^2 = MbVb^2
2Mb(Va + 1)^2 = MbVb^2
The Mb's cancel, leaving
4Va^2 = Vb^2
2(Va + 1)^2 = Vb^2
2(Va + 1)^2 = 4Va^2

Va^2 + 2Va + 1 = 2Va^2
Va^2 - 2Va - 1 = 0
Va = (2 ± √(4 + 4))/2
Va = (1 ± √2)
Rejecting the negative root,
Va = 2.4142 m/s

(Va + 1)^2 = 2Va^2
11.6569 = 11.6569