Final answer:
In an inelastic collision between two billiard balls with equal masses, the balls stick together and move as one mass after the collision. Using the principle of conservation of momentum, we can find the velocity of the combined mass of the two balls after the collision. In this case, the velocity of the combined mass is 1 m/s.
Step-by-step explanation:
The question states that two smooth billiard balls A and B have equal masses of 200 g each. Ball A strikes ball B with a velocity of 2 m/s. Since the collision is not specified as elastic or inelastic, we can assume it is inelastic. In an inelastic collision, the two objects stick together after the collision and move as one mass.
Using the principle of conservation of momentum, we can find the velocity of the combined mass of ball A and ball B after the collision. Momentum is defined as the product of mass and velocity.
Let the final velocity of the combined mass be vf. The initial momentum of ball A is given by mvA1, where m is the mass and vA1 is the initial velocity of ball A. The initial momentum of ball B is zero since it is at rest. The final momentum of the combined mass is (ma + mb)vf, where ma and mb are the masses of ball A and ball B respectively.
According to the conservation of momentum, the initial momentum should equal the final momentum:
mvA1 = (ma + mb)vf
Substituting the given values:
(0.2 kg)(2 m/s) = (0.2 kg + 0.2 kg)vf
0.4 kg m/s = 0.4 kg vf
vf = 1 m/s
Therefore, the velocity of the combined mass (ball A + ball B) after the collision is 1 m/s.