A 1300 kg car moving at 4.9 m/s is initially traveling north along the positive

Question

A 1300 kg car moving at 4.9 m/s is initially traveling north along the positive direction of a y axis. After completing a 90° right-hand turn to the positive x direction in 4.8 s, the inattentive operator drives into a tree, which stops the car in 400 ms.

In unit-vector notation, what is the impulse on the car during the collision? x-component?

y-component?

What is the magnitude of the average force that acts on the car during the turn?

What is the magnitude of the average force that acts on the car during the collision?

What is the direction of the average force during the turn?

Explanation / Answer

initial velocity v1 = 4.9 j

final velocity v2 = 4.9 i

time interval t = 4.8 s

Implse = change in momentm

J = m*V2 - m*v1

J = (1300*4.9)i - (1300*4.9j)

J = 6370i - 630 j

======================

x-component = 6370

y-component = -6370

average force = J/t = ( 6370i - 630 j )/(4.8)

Favg = 1327 i - 1327 j

magnitde = sqrt(1327^2+1327^2) = 1876.6 N <<<====ANSWER

========================

during collision

initial velocity vi = 4.9 i

final velocity vf = 0

time taken = 400 ms = 400*10^-3 s

Favg = m*(Vf-vi)/t

Favg = 1300*(0-4.9)i/(400*10^-3) = -15925 N

magnitude = 15925 N <<<<======answer

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direction = tan^-1(-1327/1327) = -45 = 315 degrees conterclockwise with +x axis

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