Question

Two vehicles approach an intersection, a 2500kg pickup travels from E to W at 14.0m/s and a 1500kg car from S to N at 23.0m/s. Find P net of this system (direction and magnitude)

Answer 1

The magnitude of the momentum is 49145.19 kg.m/s

The direction of the two vehicles is 44.6° North West

Step-by-step 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)

`P_x = mv_x\\\\P_x = 2500kg(-14 \ m/s)\\\\P_x = -35000 \ kg.m/s`

momentum of the second vehicle in y-direction (S to N is in positive y-direction)

`P_y = m_2v_y\\\\P_y = 1500kg(23 \ m/s)\\\\P_y = 34500 \ kg.m/s`

Magnitude of the momentum of the system;

`P= √(P_x^2 + P_y^2) \\\\P = √((-35000)^2+(34500)^2) \\\\P = 49145.19 \ kg.m/s`

Direction of the two vehicles;

`tan \ \theta = (P_y)/(|P_x|) \\\\tan \ \theta = (34500)/(35000) \\\\tan \ \theta = 0.9857\\\\\theta = tan^(-1) (0.9857)\\\\\theta = 44.6^0`North West