Let the speeds of drivers e and n prior to the collision be denoted by ve and vn, respectively. find v2, the square of the speed of the two-car system the instant after the collision.

Thank you for posting your question here at . I hope the answer will help you. Below is the solution:

Make a momentum balance: 
m_e · v_e + m_n · v_n = (m_e + m_n) · v 
where v is the speed of the two cars instant after the collision. 
Because the cars have the same mass, this simplifies to 
v_e + v_n = 2v 
The square of the speed instant after collision is: 
v² = 1/4·(v_e + v_n)² 

Ibelieve the correct answer would be c.