Question

A 1480 - kg car moving at 5.30 m/s is initially traveling north along the positive direction of a y axis. After completing a 90° right-hand turn to the positive x direction in 5.40 s, the inattentive operator drives into a tree, which stops the car in 470 ms. In unit-vector notation, what is the impulse on the car during the turn?

Answer 1

Answer:

$\vec{I}=7844\frac{kg*m}{s}(\hat{i}-\hat{j})$

Explanation:

The impulse of the force exerted on the car is equal to the change of the momentum of it:

$\vec{I}=\Delta \vec{p}\\\vec{I}=\vec{p_f}-\vec{p_i}(1)\\\vec{p_i}=m\vec{v_i}\\\vec{p_i}=1480kg(5.3\frac{m}{s}\hat{j})\\\vec{p_i}=7844\frac{kg*m}{s}\hat{j}(2)\\\vec{p_f}=m\vec{v_f}\\\vec{p_i}=1480kg(5.3\frac{m}{s}\hat{i})\\\vec{p_f}=7844\frac{kg*m}{s}\hat{i}(3)$

Replacing (2) and (3) in (1):

$\vec{I}=7844\frac{kg*m}{s}\hat{i}-7844\frac{kg*m}{s}\hat{j}\\\vec{I}=7844\frac{kg*m}{s}(\hat{i}-\hat{j})$