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You were driving a car with velocity <19, 0, 23> m/s. You quickly turned and braked, and your velocity became <14, 0, 26> m/s. The mass of the car was 1300 kg. (a) What was the (vector) change in momentum during this maneuver? Pay attention to signs. < -6500 , 0 , 3900 > kg·m/s (b) What was the (vector) impulse applied to the car?

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Answer:

a.<-6500,0,3900>kgm/s

b.<-6500,0,3900>kgm/s

Explanation:

We are given that

Initial velocity of car,u=[tex]<19,0,23>[/tex]m/s

Final velocity of car=[tex]v=<14,0,26>m/s[/tex]

Mass of the car=m=1300 kg

a.We have to find the change in momentum during this manuver.

Change in momentum=[tex]\Delta P=m(v-u)[/tex]

Using the formula

[tex]\Delta P=1300(<14,0,26>-<19,0,23>)=1300(<-5,0,3>)=<-6500,0,3900>kgm/s[/tex]

Hence, the change in momentum during this maneuver=<-6500,0,3900>kgm/s

b.Impulse =Change in momentum of car

Impulse applied to the car=<-6500,0,3900>kgm/s

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