A billiard ball moving with a speed of \(8 \mathrm{~m} / \mathrm{s}\) collides with an identical ball originally at rest. If the first ball stops after collision, then the second ball will move forward with a speed of .... (elastic collision) (A) \(8 \mathrm{~m} / \mathrm{s}\) (B) \(4 \mathrm{~m} / \mathrm{s}\) (C) \(16 \mathrm{~m} / \mathrm{s}\) (D) \(1.0 \mathrm{~m} / \mathrm{s}\)

In an elastic collision, the linear momentum is conserved. We can write the conservation of linear momentum as: \(m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}\), where \(m_1\) is the mass of the first ball, \(m_2\) is the mass of the second ball, \(v_{1i}\) is the initial velocity of the first ball, \(v_{2i}\) is the initial velocity of the second ball, \(v_{1f}\) is the final velocity of the first ball, and \(v_{2f}\) is the final velocity of the second ball. Since the two balls are identical, their masses are equal, so \(m_1 = m_2 = m\). Also, we know the initial velocities of both balls: \(v_{1i} = 8 \mathrm{~m} / \mathrm{s}\) and \(v_{2i} = 0\). The final velocity of the first ball is also given: \(v_{1f} = 0\). Now our equation becomes: \(mv_{1i} + 0 = 0 + mv_{2f}\)

Now we can solve the equation for the final velocity of the second ball, \(v_{2f}\): \(mv_{1i} = mv_{2f}\) Since both masses are identical, we can divide both sides of the equation by m: \(v_{1i} = v_{2f}\) Substituting the given value of \(v_{1i}\): \(8 \mathrm{~m} / \mathrm{s} = v_{2f}\) So, the final velocity of the second ball is 8 m/s. Therefore, the correct answer is: (A) \(8 \mathrm{~m} / \mathrm{s}\)