What if check: Am I calculating tension wrong?

[Callista]

Homework Statement

Two objects are connected by a string that passes over a frictionless pulley, where m1<m2 and a1 and a2 are the magnitudes of the respective accelerations. Which mathematical statement is true regarding the magnitude of the acceleration a2 of the mass m2?
(a) a2<g
(b) a2>g
(c) a2=g
(d) a2<a1
(e) a2>a1

Homework Equations

F(net)=ma

The Attempt at a Solution

for m1:
F(net)=m1a
Fx=0
Fy=T-m1g=m1a1
$a1= \frac{T}{m1}-g$

for m2:
F(net)=m2a
Fx=0
Fy=-T+m2g=m2a2
$a2=g- \frac{T}{m2}<g$

Answer= (a)

While the book confirms my answer is right, I wanted to try applying my derived formulas to the situation where m1=5kg, and m2=10kg to see what the result would be. However, when I do this I run into a problem.

$a2=g- \frac{T}{m2}$
->$Fy=-T+m2g=m2a2$
->$T=m2g-m2a2$
->$a2=g- \frac{m2g-m2a2}{m2}$
->$a2=g- g+a2$
->$a2=a2$

A similar thing happens when I try to use the equation I derived for a1. What am I missing?

 

[haruspex]

Fy=T-m1g=0

Why =0?

 

[Callista]

Why =0?

Sorry, looks like I mistyped that. I meant to say
->Fy=T-m1g=m1a1
->Fy=-T+m2g=m2a2
I've update my post to reflect it.

Upon reflection, I realize I was trying to use the same formula twice to solve for a new variable... Thanks for all the help.

 

[Delta2]

You have two equations each one from applying Newton's 2nd law to each body
$T-m_1g=m_1a_1$
$m_2g-T=m_2a_2$

You have to solve for tension from the one equation and replace the tension into the other equation. If you solve for tension from one equation and replace it into the same equation it is a simple mathematical consequence that you will end up with conclusions of the form $a_i=a_i$.