Answer :
Answer:
The magnitude of the momentum is 49145.19 kg.m/s
The direction of the two vehicles is 44.6° North West
Explanation:
Given;
speed of first vehicle, v₁ = 14 m/s (East to west)
mass of first vehicle, m₁ = 1500 kg
speed of second vehicle, v₂ = 23 m/s (South to North)
momentum of the first vehicle in x-direction (E to W is in negative x-direction)
[tex]P_x = mv_x\\\\P_x = 2500kg(-14 \ m/s)\\\\P_x = -35000 \ kg.m/s[/tex]
momentum of the second vehicle in y-direction (S to N is in positive y-direction)
[tex]P_y = m_2v_y\\\\P_y = 1500kg(23 \ m/s)\\\\P_y = 34500 \ kg.m/s[/tex]
Magnitude of the momentum of the system;
[tex]P= \sqrt{P_x^2 + P_y^2} \\\\P = \sqrt{(-35000)^2+(34500)^2} \\\\P = 49145.19 \ kg.m/s[/tex]
Direction of the two vehicles;
[tex]tan \ \theta = \frac{P_y}{|P_x|} \\\\tan \ \theta = \frac{34500}{35000} \\\\tan \ \theta = 0.9857\\\\\theta = tan^{-1} (0.9857)\\\\\theta = 44.6^0[/tex]North West