Question

An automobile of mass 1300 kg has an initial velocity of 7.20 m/s toward the north and a final velocity of 6.50 m/s toward the west. The magnitude and direction of the change in momentum of the car are?

Answer 1

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

$\Delta P = m (v_f - v_i)$

$\Delta P = 1300\times (-6.50 \hat{i} - 7.20 \hat{j})$

$\Delta P = 1300\times \sqrt{(-6.50)^2 + (-7.20)^2}$

$\Delta P = 1300\times 9.7$

$\Delta P = 12610\ kg.m/s$

now, direction calculation

 $\theta = tan^{-1}(\dfrac{7.2}{6.5})$

 $\theta = tan^{-1}(1.1076)$

 $\theta = 47.92^0$

direction of the car is 48° in south west direction.