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

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* (-6.50 \hat{i} - 7.20 \hat{j})`

`\Delta P = 1300* √((-6.50)^2 + (-7.20)^2)`

`\Delta P = 1300* 9.7`

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

now, direction calculation

`\theta = tan^(-1)((7.2)/(6.5))`

`\theta = tan^(-1)(1.1076)`

`\theta = 47.92^0`

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