Answer
given,
mass of the automobile = 1300 Kg
initial speed of the automobile in north direction= 7.20 j m/s
final speed of the automobile in west direction = -6.50 i m/s
now,
change in momentum
[tex]\Delta P = m (v_f - v_i)[/tex]
[tex]\Delta P = 1300\times (-6.50 \hat{i} - 7.20 \hat{j})[/tex]
[tex]\Delta P = 1300\times \sqrt{(-6.50)^2 + (-7.20)^2}[/tex]
[tex]\Delta P = 1300\times 9.7[/tex]
[tex]\Delta P = 12610\ kg.m/s[/tex]
now, direction calculation
[tex]\theta = tan^{-1}(\dfrac{7.2}{6.5})[/tex]
[tex]\theta = tan^{-1}(1.1076)[/tex]
[tex]\theta = 47.92^0[/tex]
direction of the car is 48° in south west direction.
