Answer: The velocity of the second ball after the collision is -2 m/s due east.
Step-by-step explanation:
We can use the law of conservation of momentum to solve this problem. According to this law, the total momentum of a system before a collision is equal to the total momentum of the system after the collision, provided that no external forces act on the system.
The momentum p of an object is given by the product of its mass m and velocity v: p = mv.
Before the collision, the momentum of the system is:
p_initial = m_1 * v_1 + m_2 * v_2
where m_1 and v_1 are the mass and velocity of the first ball, m_2 and v_2 are the mass and velocity of the second ball.
Plugging in the given values, we get:
p_initial = (0.50 kg)(6.0 m/s) + (1.00 kg)(-12.0 m/s) = -6.0 kg*m/s
After the collision, the momentum of the system is:
p_final = m_1 * v'_1 + m_2 * v'_2
where v'_1 and v'_2 are the velocities of the first and second ball after the collision.
We are given that the first ball moves away at -14 m/s after the collision, so:
v'_1 = -14 m/s
We can now use the conservation of momentum to solve for v'_2:
p_initial = p_final
m_1 * v_1 + m_2 * v_2 = m_1 * v'_1 + m_2 * v'_2
Plugging in the given values, we get:
(0.50 kg)(6.0 m/s) + (1.00 kg)(-12.0 m/s) = (0.50 kg)(-14 m/s) + (1.00 kg)(v'_2)
Solving for v'_2, we get:
v'_2 = -2 m/s
Therefore, the velocity of the second ball after the collision is -2 m/s due east.