Answer:
Car B will be travelling after the crash at a constant speed of 20 mph in the same direction as Car A was travelling before the crash
Explanation:
Let [tex]m_A, m_B[/tex] be the mass of car A resp. car B, [tex]v_{A0} = 20 \ \text{mph}[/tex] the velocity of car A before the crash, [tex]v_{B1}[/tex] the velocity of car B after the crash. For car B being at rest before the crash and car A at rest after, it holds [tex]v_{B0} = v_{A1} = 0\ \text{mph}[/tex].
With conservation of momentum, it holds
[tex]m_A v_{A0} + m_B v_{B0} = m_A v_{A1} + m_B v_{B1}[/tex]
Zero terms eliminated: [tex]m_A v_{A0} = m_B v_{B1} \iff m_B = m_A\frac{v_{A0}}{v_{B1}}[/tex]
With conservation of energy and assuming that no conversion of mechanical energy into other forms of energy occurs (through friction etc.), the mechanical energy before and after the crash is simply equated:
[tex]\frac{1}{2} m_A v_{A0}^2 = \frac{1}{2} m_B v_{B1}^2[/tex]
Inserting the [tex]m_B[/tex] from above: [tex]\frac{1}{2}m_A v_{A0}^2 = \frac{1}{2}m_A\frac{v_{A0}}{v_{B1}} v_{B1}^2 \iff v_{A0} = v_{B1} = 20\ \text{mph}[/tex]
Bonus: [tex]m_B = m_A\frac{v_{A0}}{v_{B1}} = m_A[/tex] (both cars have the same weight)
