Answer:
[tex]p_t_o_t_a_l=250kg\frac{m}{s}[/tex]
Explanation:
The total momentum of a system is defined by:
[tex](mv)_t_o_t=m_1v_1+m_2v_2+...[/tex]
Where,
[tex](mv)_t_o_t[/tex] is the total momentum or it could be expressed also as [tex]p_t_o_t_a_l[/tex].
[tex]m_1[/tex] and [tex]m_2[/tex] represents the masses of the objects interacting in the system.
[tex]v_1[/tex] and [tex]v_2[/tex] are the velocities of the objects of the system.
Remember: The momentum is a fundamental physical magnitude of vector type.
We have:
[tex]m_1=725 kg[/tex]
[tex]v_1=15\frac{m}{s}\\m_2=625 kg[/tex]
We are going to take the east side as positive, and the west side as negative. Then the velocity of the car B, has to be negative. It goes in a different direction from car A.
[tex]v_2=-17\frac{m}{s}[/tex]
Then the total momentum of the system is:
[tex]p_t_o_t_a_l=m_1v_1+m_2v_2\\p_t_o_t_a_l=(725kg)(15\frac{m}{s})+(625kg)(-17\frac{m}{s})\\p_t_o_t_a_l=10875kg\frac{m}{s}-10625kg\frac{m}{s}\\p_t_o_t_a_l=250kg\frac{m}{s}[/tex]
