Answer
$J = (7420~\hat{i}-7420~\hat{j})~kg~m/s$
Work Step by Step
We can express the initial momentum in unit-vector notation:
$p_i = m~v_i$
$p_i = (1400~kg)~(5.3~\hat{j})~m/s$
$p_i = (7420~\hat{j})~kg~m/s$
We can express the momentum after the turn in unit-vector notation:
$p_f = m~v_f$
$p_f = (1400~kg)~(5.3~\hat{i})~m/s$
$p_f = (7420~\hat{i})~kg~m/s$
We can find the change in momentum:
$\Delta p = p_f-p_i$
$\Delta p = (7420~\hat{i}-7420~\hat{j})~kg~m/s$
The impulse on the car is equal to the change in momentum.
We can express the impulse on the car due to the turn:
$J = (7420~\hat{i}-7420~\hat{j})~kg~m/s$
