At a classic auto show,a 840kg 1955 Nash Metropolitan motors by 9.0m/s , followe

Question

At a classic auto show,a 840kg 1955 Nash Metropolitan motors by 9.0m/s , followed by 1620kg 1957 Packard Clipper purring past at 5.0m/s. a) Which car as the greater kinetic energy? What is the ratio of the kinetic energy of the Nash on that of the Packard? b) Which car as the greater magnitude of momentum? What is the ratio of magnitude of momentum of the Nash to that of the Packard? c) Let FN be the net force required to stop the Nash in time t, and let Fp be the net force required to stop the Packard in the same time. Which is large FN or Fp? What is the ratio FN/Fp of these two forces? d) Now let be FN be the net force required to stop the Nash in a distance d, and let Fp be the net force required to stop the Packard in the same distance. Which is large : FN or Fp? What is the ratio FN/Fp?

Explanation / Answer

A).

K.E1 = 0.5 x 840 x 81 = 34020 J

K.E2 = 0.5 x 1620 x 25 = 20250 J

Motor 1 has more kinetic energy.

B).

P1 = 840 x 9 = 7560

P2 = 1620 x 5 = 8100

Motor 2 has more momentum

C).

Since, v1 > v2

to stop both motor in same time motor 1 required more reatrdation mean more force.

hence Fn > Fp

Fn = 840 x (9/t)

Fp = 1620 x (5/t)

Fn/Fp = 14/15=0.9333

D).

Fn = 840 x (9*9/2d) = 34020 / d

Fp = 1620 x (5*5/2d) = 20250 / d

Fn > Fp

Fn/Fp = 42/25 = 1.68

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