A 1200 kg car moves due north with a speed of 15m/s. An identical car moves due east with the same speed of 15m/s what are the direction and the magnitude of the system’s total momentum

Respuesta :

a.

The direction of the total momentum is 45°

The momentum of the first car is given by p = mv where m = mass of car = 1200 kg and v = velocity of car = 15 m/sj (since it moves due north).

So, p = mv

= 1200 kg × (15 m/s)j

= (18000 kgm/s)j

Also, the momentum of the identical car, p' = mv' where m = mass of car = 1200 kg and v' = velocity of car = (15 m/s)i (since it moves due east).

So, p' = mv'

= 1200 kg × (15 m/s)i

= (18000 kgm/s)i

So, the total momentum of the system P = p + p'

=  (18000 kgm/s)j +  (18000 kgm/s)i

=  (18000 kgm/s)i +  (18000 kgm/s)j

The direction of the total momentum of the system P is gotten from

tanФ = p'/p

= 18000 kgm/s ÷ 18000 kgm/s

= 1

Ф = tan⁻¹(1)

= 45°

The direction of the total momentum is 45°

b.

The magnitude of the total momentum of the system is 25455.84 kgm/s

The magnitude of the total momentum of the system P = √(p'² + p²)

= √[(18000 kgm/s)² + (18000 kgm/s)²]

= (18000 kgm/s)√(1 + 1)

= (18000 kgm/s)√2

= 25455.84 kgm/s

The magnitude of the total momentum of the system is 25455.84 kgm/s

Learn more about total momentum here:

https://brainly.com/question/25635296

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