8.73 A 1600 kg sedan goes through a wide intersection traveling from north to so

Question

8.73

A 1600 kg sedan goes through a wide intersection traveling from north to south when it is hit by a 2100 kg SUV traveling from east to west. The two cars become enmeshed due to the impact and slide as one thereafter. On-the-scene measurements show that the coefficient of kinetic friction between the tires of these cars and the pavement is 0.75, and the cars slide to a halt at a point 5.45 m west and 6.40 m south of the impact point.

A. How fast was sedan traveling just before the collision?

ANSWER: V = ________ m/s

B. How fast was SUV traveling just before the collision?

ANSWER: V = ________ m/s

Explanation / Answer

Let v[1] is the velocity of the sedan just before the collision and v[2] is the velocity of the SUV just before the collision.

Now,

The acceleration after impact is found from
uR = ug(1600 + 2100) = (1600 + 2100)a
a = ug = 0.75 x 9.8 = 7.35
The distance slid is
SQRT(5.45^2 + 6.40^2) = 8.41 m
Speed after impact from
v^2 - u^2 = 2as
v = 0
-u^2 = -2 x 7.35 x 8.41
u = 11.12 m/s and angle = tan^-1(6.40/5.45) = 49.58 deg south of west.
Resolving into south and west components and equating momentum before and after
1600v[1] = (1600 + 2100) x 11.12 x sin(49.58)
2100v[2] = (1600 + 2100) x 11.12 x cos(49.58)

Solving the above two equations, we get,
v[1] = 19.58 m/s
v[2] = 12.70 m/s

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