Two cars collide at right angles. car 1 (mass 1,100 kg) is going north with a speed 31 m/s. car 2 (mass 1,200 kg) is moving east with a speed 19 m/s. the cars stick together after the collision. what is the magnitude of the velocity of the cars after the collisions? what angle does the velocity vector make with the x axis?

17.84 m/s

56.31 deg

Explanation:

Consider east direction along positive x-direction and north direction along positive y-direction.

m₁ = mass of car 1 = 1100 kg

v₁ = initial velocity of car 1 = 0 i + 31 j

m₂ = mass of car 2 = 1200 kg

v₂ = initial velocity of car 2 = 19 i + 0 j

v = final velocity of the combination of cars after collision

using conservation of momentum

m₁ v₁ + m₂ v₂ = (m₁ + m₂ ) v

(1100) (0 i + 31 j ) + (1200) (19 i + 0 j) = (1100 + 1200) v

34100 j + 22800 i = 2300 v

v = 9.91 i + 14.83 j

magnitude of the velocity is given as

|v| = sqrt(9.91² + 14.83²)

|v| = 17.84 m/s

Angle is given as

θ = tan⁻¹(14.83/9.91)

θ = 56.31 deg

resonant frequency

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west