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

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?

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