What is the velocity of cart B after the boy lands on it?

[x86]

Homework Statement

The 75kg boy leaps off the cart A with a horizontal velocity of v' = 3 m/s to the left measured relative to the cart. Determine the velocity of cart A just after the jump. If he then lands on cart B with the same velocity that he left cart A, determine the velocity of cart B just after he lands on it. Carts A and B have the same mass of 50kg and are originally at rest.

Homework Equations

Relative motion: Va = Va/b + Vb
Momentum: m1V1 = m2V2

The Attempt at a Solution

the subscript j will represent the boy, while a/b represent the respective carts. the coordinate system: left represents negative, right positive

Relative motion
Vj = Vj/a + Va
(1) Vj = -3 + Va

Momentum
0 + 0 = ma*Va - mj*Vj
mj*Vj = ma*Va
(2) Vj = (ma)/(mj)Va = (50/75)Va

Substituting (1) and (2)
(50/75)Va = -3 + Va
Va(50/75-1)=-3
Va = 9 m/s (to the right)
Vj = -3 + 9m/s = 6m/s (to the left)

Now I find Vb (assume direction right)
mj*Vj + mb*Vb = (mj+mb)V
75*(-6) + 50*0 = (50+75)V

V = (75*6)/(50+75) = 3.6 (left)

But this is wrong,the answer is 0.720 m/s (left)

 

[haruspex]

0 + 0 = ma*Va - mj*Vj

Need to be consistent about signs. Are you measuring all velocities as positive right, or the cart's as positive right and the boy's as positive left?

Va = 9 m/s (to the right)
Vj = -3 + 9m/s = 6m/s (to the left)

What relative velocity would that be?