Question

You were driving a car with velocity ‹ 28, 0, 14 › m/s. You quickly turned and braked, and your velocity became ‹ 21, 0, 18 › m/s. The mass of the car was 1100 kg.

(a) What was the (vector) change in momentum ? during this maneuver? Pay attention to signs.
? = kg · m/s
(b) What was the (vector) impulse applied to the car by the ground?
(vector) impulse: N · s
(c) If the maneuver took 7 seconds, what was the the average (vector) force exerted on the car? (The net force is due to the ground and the Earth; the y components of these forces cancel.)

Answer 1

Answer:

a) <-7700, 0, 4400> kgm/s

b) <-7700, 0, 4400> kgm/s

c) <-1100, 0, 628.57> N

Explanation:

We can calculate the change in velocity first

$\Delta v = v_2 - v_1 = <21, 0, 18> - <28, 0, 14> = <-7, 0 , 4>m/s$

Then the change in momentum is the product of velocity change and the car mass

$\Delta p = m\Delta v = 1100 < -7, 0, 4> = <-7700, 0, 4400> kgm/s$

This change in momentum would also equal to the impulse applied to the car by the ground, which is <-7700, 0, 4400> kgm/s

The average force would be the impulse divided by time duration t

$F = \Delta p / \Delta t=<-7700, 0, 4400> / 7 = <-1100, 0, 628.57> N$

Answer 2

Momentum is the be defined as a product of mass and velocity. The change in momentum of the given car is -7700 kgm/s.

Change in Momentum:

Change in momentum can be defined as the product of mass and change in velocity.

$\Delta p = m \times \Delta v$

$\Delta p$ - change ion momentum

$\Delta v$- chnge in velocity = 21 -28 = -7 m/s.

$m$ - mass of car  = 1100 kg

Put the values in the formula,

$\Delta p = 1100 \times -7 \\\\ \Delta p = -7700$

Therefore, the change in momentum of the given car is -7700 kgm/s.

Learn more about change in momentum:

https://brainly.com/question/2284371