So close to solving - force/friction

[ff4930]

A mass m1 on a horizontal shelf is attached by a thin string that passes over a frictionless pulley to a 3.0 kg mass (m2) that hangs over the side of the shelf 1.3 m above the ground. The system is released from rest at t = 0 and the 3.0 kg mass strikes the ground at t = 0.78 s. The system is now placed in its initial position and a 1.2 kg mass is placed on top of the block of mass m1. Released from rest, the 3.0 kg mass now strikes the ground 1.3 seconds later.
(a) Determine the mass m1.
kg

(b) Determine the coefficient of kinetic friction between m1 and the shelf.


What I done:
d = ½at²
In this case:
1.3m = ½(a_before)(0.78s)²
and:
1.3m = ½(a_after)(1.3s)²

Ab = 4.3m/s^2
Aa = 1.5m/s^2

In the "before" case:
1)F_net1_before = T_before - (μk)(m1)(g) = (m1)(a_before)
and:
2)F_net2_before = (m2)(g) - T_before = (m2)(a_before)

In the "after" case:
3)F_net1_after = T_after - (μk)(m1+1.2kg)(g) = (m1+1.2kg)(a_after)
and
4)F_net2_after = (m2)(g) - T_after = (m2)(a_after)

I combined equations 1 and 2 to eliminate Tb
I combined equations 3 and 4 to eliminate Ta

I plugged in the numbers and got up to
16.53 - (Uk)(M1)(9.81) = (M1)(4.3)
and
20.43 -(Uk)(M1+1.2)(9.81) = (M1+1.2)(1.5)

First equation, solved for Uk:
Uk = (16.53 - (M1)(4.3)) / ((M1)(9.81))

Second equation, solved for Uk:
Uk = (20.43 - (M1+1.2)(1.5)) / (M1+1.2)(9.81)

(16.53 - (M1)(4.3)) / ((M1)(9.81)) = (20.43 - (M1+1.2)(1.5)) / (M1+1.2)(9.81)

I did that and ended up with quadratic equations and I plug that in the formula and ended up with a value for M1. When I plug in M1 to the 2 Uk equations, they did not match.

What I did wrong?
Please help, I spent hours on this, please help. Thanks!
I believe is more of algebraic mistake but I've been going through this so much, if anyone can point out where I went wrong.

 

[Doc Al]

What I done:
d = ½at²
In this case:
1.3m = ½(a_before)(0.78s)²
and:
1.3m = ½(a_after)(1.3s)²

Ab = 4.3m/s^2
Aa = 1.5m/s^2

OK, but don't round off your calculations until the end.

In the "before" case:
1)F_net1_before = T_before - (μk)(m1)(g) = (m1)(a_before)
and:
2)F_net2_before = (m2)(g) - T_before = (m2)(a_before)

In the "after" case:
3)F_net1_after = T_after - (μk)(m1+1.2kg)(g) = (m1+1.2kg)(a_after)
and
4)F_net2_after = (m2)(g) - T_after = (m2)(a_after)

I combined equations 1 and 2 to eliminate Tb
I combined equations 3 and 4 to eliminate Ta

Good.

I plugged in the numbers and got up to
16.53 - (Uk)(M1)(9.81) = (M1)(4.3)
and
20.43 -(Uk)(M1+1.2)(9.81) = (M1+1.2)(1.5)

Double check that 20.43 number. (Best to redo all numbers, to avoid roundoff errors.)

First equation, solved for Uk:
Uk = (16.53 - (M1)(4.3)) / ((M1)(9.81))

Second equation, solved for Uk:
Uk = (20.43 - (M1+1.2)(1.5)) / (M1+1.2)(9.81)

(16.53 - (M1)(4.3)) / ((M1)(9.81)) = (20.43 - (M1+1.2)(1.5)) / (M1+1.2)(9.81)

This looks OK.

I did that and ended up with quadratic equations and I plug that in the formula and ended up with a value for M1. When I plug in M1 to the 2 Uk equations, they did not match.

When you put the above into standard form for a quadratic you might have made an error. I would have solved the first for Uk, as you did, but then plugged that into the second equation to solve for M1.

 

[ff4930]

I finally got it! After so much math crunching and leaving all the decimal places as they are have paid off. Thank you Doc for the advice of redoing the numbers.
It is always sweeter when you solve a problem all by your self(with exception of your help teehee). Thanks!