Question

In the figure, block A (mass 1.6 kg) slides into block B (mass 2.4 kg), along a frictionless surface. The directions of velocities before and after the collision are indicated; the corresponding speeds are v_Ai = 5.4 m/s, v_Bi = 2.0 m/s, and v_Bf = 4.5 m/s. What is velocity v_Af (including sign, where positive denotes motion to the right)

In the figure, block A (mass 1.6 kg) slides into block B (mass 2.4 kg), along a frictionless surface. The directions of velocities before and after the collision are indicated; the corresponding speeds are vAi = 5.4 m/s, vBi = 2.0 m/s, and vBf = 4.5 m/s. What is velocity vAf (including sign, where positive denotes motion to the right) ?

Answer 1

Momentum before = Momentum after
(1.1)(5.6) + (2.9)(2.2) = (1.1)(vAf) + (2.9)(4.9)
I found vAf.
But when I calculated the kinetic energy before and after the collision, I got an increase in kinetic energy, as in : kinetic energy before collision < kinetic energy after collision.

Kinetic energy before collision:
(1/2)(1.1)(5.6)^2 + (1/2)(2.9)(2.2)^2 = 24.3 J.

Kinetic energy after collision, using vAf = -1.52, found when conserving momentum:
(1/2)(1.1)(-1.52)^2 + (1/2)(2.9)(4.9)^2 = 36.1 J

Answer 2