Final answer:
The conservation of momentum for a collision between billiard balls is expressed as m_A × v_A = m_A × v_A1 × cos(θ_A) + m_B × v_B1 × cos(θ_B) for the components in the x direction.
Step-by-step explanation:
To solve the problem given in the question, we apply the conservation of momentum principle which states that the total momentum of a closed system remains constant provided no external forces act on it. The equation for the conservation of momentum in the x direction for the collision described by the student can be formulated as follows:
Before the collision, only ball A has momentum in the x direction since ball B is at rest. The momentum of ball A can be written as mA × vA. After the collision, ball A's momentum in the x direction is mA × vA1 × cos(θA) and ball B's momentum in the x direction is mB × vB1 × cos(θB).
The equation for the conservation of momentum in the x direction is:
mA × vA = mA × vA1 × cos(θA) + mB × vB1 × cos(θB)
To express the conservation of momentum for the components in the x direction, we need to consider the momentum before and after the collision. Before the collision, ball A has momentum in the x direction given by mA * vA. Ball B is initially at rest, so it has zero momentum in the x direction. After the collision, ball A is deflected off at an angle of 30.0° with a speed vA1, and ball B is moving with a speed vB1.
Using the conservation of momentum in the x direction, the momentum before the collision is equal to the momentum after the collision. This can be expressed as:
mA * vA = mA * vA1 * cos(θA) + mB * vB1 * cos(θB)