How do you calculate this if not given a mass?

In an elastic collision, both momentum and kinetic energy are conserved AND the relative velocities are reversed. Since the masses are equal, Ball A rebounds with B's speed and vice versa.

i.e. vA = -3 m/s and vB = +2 m/s <= ANS

(so total momentum p = 2m - 3m = -1m, and relative velocity vA - vB = -5 m/s)

I can tell you right off the bat that because the balls have the same mass, they will exchange velocities so the final velocity of A is -3 m/s and the final velocity of B is 2 m/s. The conservation of momentum and the conservation of kinetic energy are both needed to solve this mathematically. The equation for the former is as follows:

m_A*v_Ai + m_B*v_Bi = m_A*v_Af + m_A*v_Bf

where:

m_A = mass of ball A

m_B = mass of ball B

v_Ai = initial velocity of ball A = 2 m/s

v_Bi = initial velocity of ball B = -3 m/s

v_Af = final velocity of ball A

v_Bf = final velocity of ball B

So we want to solve for v_Af and v_Bf.

Since the masses fo the balls are the same, m_A = m_B so we will call them both 'm'. This simplifies the equation to:

m*v_Ai + m*v_Bi = m*v_Af + m*v_Bf

m*(v_Ai + v_Bi) = m*(v_Af + v_Bf)

v_Ai + v_Bi = v_Af + v_Bf

We see that because the masses are the same, they cancel out in this equation. We will use what we have derived to find a term for v_Af in terms of v_Bf:

v_Ai + v_Bi = v_Af + v_Bf

v_Ai + v_Bi - v_Bf = v_Af

Plugging in the values we know already:

2 + (-3) - v_Bf = v_Af

v_Af = -1 - v_Bf

We will substitute v_Af for the right hand side of this equation once we derive what we need from the kinetic energy equation.

The conservation of kinetic energy equation is (remember kinetic energy is (1/2)*mv^2):

.5*m_A*(v_Ai)^2 + .5*m_B*(v_Bi)^2 = .5*m_A*(v_Af)^2 + .5*m_B*(v_Bf)^2

Again, we simplify since m_A = m_B = m:

.5*m*(v_Ai)^2 + .5*m*(v_Bi)^2 = .5*m*(v_Af)^2 + .5*m*(v_Bf)^2

.5*m*( (v_Ai)^2 + (v_Bi)^2) = .5*m*( (v_Af)^2 + (v_Bf)^2)

(v_Ai)^2 + (v_Bi)^2 = (v_Af)^2 + (v_Bf)^2

Here we make our substitutions of known values v_Ai = 2, v_Bi = -3, v_Af = -1 - v_Bf:

(2)^2 + (-3)^2 = (-1 - v_Bf)^2 + (v_Bf)^2

4 + 9 = 1 + v_Bf + v_Bf + (v_Bf)^2 + (v_Bf)^2

13 = 1 + 2*v_Bf + 2*(v_Bf)^2

2*(v_Bf)^2 + 2*v_Bf - 12 = 0 <----- quadratic equation

x = [-b +/- sqrt(b^2 - 4*a*c)]/(2*a), where 'a' is the coefficient of v_Bf^2, b is the coefficient of v_Bf, and c = -12.

Solving this gives 2 or -3. We know that ball B cannot have the same velocity after the collision as it had before it (think of Newton's third law of motion), so the throw-away solution is -3. Therefore, v_Bf = 2 m/s.

Now that we know the value of v_Bf, we can plug it into the relationship we derived from the conservation of momentum equation before:

v_Af = -1 - v_Bf

v_Af = -1 - 2 m/s

v_Af = -3 m/s

*****ANSWER*****

velocity of ball A after collision = -3 m/s

velocity of ball B after collision = 2 m/s