physics question

Question

A car with a mass of 1443 kg is headed north at a speed of 24.1 m/s. Later, the car is headed east at a speed of 20.1 m/s. Assuming traditional sign conventions (e.g., north is +y and east is +x), what are the magnitude and direction with respect to the +x-axis of the car's change in momentum? (Enter units of momentum as kg*m/s and units for direction as deg.)

Answer 1

Change of momentum for constant mass = m(V2-V1), where V are vectors

For an object of constant mass change of momentum

V ⃗north=0i ̂+24.1j ̂ V ⃗east= 20.1i ̂+ 0j ̂

∆V ⃗=20.1i ̂-24.1j ̂ |∆V ⃗|=√(〖20.1〗^2+〖24.1〗^2 ) =31.4 m/s

∆p=(1443 kg)(31.4 m/s)=45284 kg·m/s

θ=〖tan〗^(-1) ((-24.1)/20.1)=-0.83 deg