Answer:
Solution given:
North car
mass[m1]=1460kg
velocity[u1]=27 m/s
mass[m2]=2165kg
velocity [u2]=18m/s
let v be velocity after collision
we have
From the principle of conservation of linear momentum
m1u1+m2u2=(m1+m2)v
1460*27+2165*18=(1460+2165)v
v=
![(78390)/(3625)](https://img.qammunity.org/2022/formulas/physics/college/bsn9g7kiaqccw2gsge6gjggguh781ou4tq.png)
v=21.6m/s
the speed after the collision in 21.6 m/s.
For angle.
Tan angle =
![(m1u1)/(m2u2)](https://img.qammunity.org/2022/formulas/physics/college/u8y1aa5o5z0wxuwt2ncp5dq480ibx7tpj0.png)
Tan angle =
![(1460*27)/(2165*18)](https://img.qammunity.org/2022/formulas/physics/college/ilsga2rfvl8raabxjkjxhzzxj2p1f2v7k5.png)
Tan angle=327.74
angle=Tan-¹(327.74)=89.82=90°
in degrees north of east is 90°