A 1400 kgcar moving at 5.3 m/sis initially traveling north along the positive direction of a yaxis. After completing a$\mathbf{90}\mathbf{°}$right-hand turn in 4.6 s, the inattentive operator drives into a tree, which stops the car in 350 ms. In unit-vector notation, what is the impulse on the car (a) due to the turn and (b) due to the collision? What is the magnitude of the average force that acts on the car (c) during the turn and (d) during the collision? (e) What is the direction of the average force during the turn?

Mass of the car is, m = 1400 kg .

The initial velocity of the car is,$v_i=5.3\frac{m}{s}$ .

Time taken by the car to take$90°$a right-hand turn is, t = 4.6 s .

Time taken by the car to stop is, t = 350 ms = 0.35 s .

We can find the impulse on the car due to the turn and collision using the formula for it. Using the relation between impulse and average force, we can find the average force acting on the car during the turn and collision.

The initial velocity of the car is,and it is moving in the y direction.

Then,

$\overrightarrow{v_i}=5.3\hat{j}\frac{m}{s}$.

After aright-hand turn, the velocity of the car is,

role="math" localid="1661318124691" $\overrightarrow{v_f}=5.3\hat{i}\frac{m}{s}$.

Impulse on the car due to the turn can be written as,

$$\begin{gathered} \overrightarrow{J}=∆\overrightarrow{P} \\ \overrightarrow{J}=m\left(\overrightarrow{v_f}-\overrightarrow{v_i}\right) \end{gathered}$$

Substitute the values in the above expressions, and we get,

$$\begin{gathered} \overrightarrow{J}=\left(1400\right)\left(0-\left(5.3\hat{i}\right)\right).\left(1kg\times 1m/s\times \frac{1N}{1kg.m/s^2}\right) \\ =\left(-7.4\times {10}^3\hat{i}\right)N.s \end{gathered}$$

Therefore, the impulse on the car due to the turn is$\left(-7.4\times {10}^3\hat{i}\right)N.s$ .

The average force on the car during the turn is,

${\overrightarrow{F}}_{avg}=\frac{\overrightarrow{J}}{t}$

Substitute the values in the above expressions, and we get,

$$\begin{gathered} {\overrightarrow{F}}_{avg}=\frac{7.4\times {10}^3\left(\hat{i}-\hat{j}\right)N.s}{4.6s} \\ =1609\left(\hat{i}-\hat{j}\right)N \\ ~1600\left(\hat{i}-\hat{j}\right)N \end{gathered}$$.

The magnitude of the average force can be calculated as,

$$\begin{gathered} F_{avg}=1600\sqrt{{\left(1\right)}^2+{\left(-1\right)}^2} \\ =1600\sqrt{2} \\ =2.3\times {10}^{3}N \end{gathered}$$

Therefore, the magnitude of the average force on the car during the turn is$2.3\times {10}^{3}N$ .

${\overrightarrow{F}}_{avg}=\frac{\overrightarrow{J}}{t}$

Substitute the values in the above expressions, and we get,

$$\begin{gathered} {\overrightarrow{F}}_{avg}=\frac{-7.4\times {10}^3\hat{i}N.s}{0.35s} \\ =\left(-2.1\times {10}^4\hat{i}\right)N \\ =2.1\times {10}^{4}N \end{gathered}$$

Therefore, the magnitude of the average force on the car during the collision is$2.1\times {10}^{4}N$ .

The average force on the car during the turn is,${\overrightarrow{F}}_{avg}=1600\left(\hat{i}-\hat{j}\right)N$ .

The y component of it is,${\overrightarrow{F}}_{avg.y}=1600\left(-\hat{j}\right)N$ .

The x component of it is,${\overrightarrow{F}}_{avg,x}=1600\left(\hat{i}\right)N$ .

The direction of${\overrightarrow{F}}_{avg}$can be calculated as,

$\tan \theta =\frac{{\overrightarrow{F}}_{avg,y}}{{\overrightarrow{F}}_{avg,x}}$

Substitute the values in the above expressions, and we get,

$$\begin{gathered} \tan \theta =-\frac{1600}{1600} \\ \tan \theta =\left(-1\right) \\ \theta ={\tan }^{-1}\left(-1\right) \\ =-45° \end{gathered}$$

Therefore, the direction of the average force on the car during the turn is$\theta =-45°$ .