A 1400kg sedan goes through a wide intersection traveling from north to south wh

Question

A 1400kg sedan goes through a wide intersection traveling from north to south when it is hit by a 2000kg 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 0.750, and the cars slide to a halt at a point 5.39m west and 6.19m south of the impact point.

How fast was sedan traveling just before the collision?

How fast was SUV traveling just before the collision?

Explanation / Answer

The acceleration after impact is found from
uR = ug(1400 + 2000) = (3400)a
a = ug = 0.75 x 9.8 = 7.35
The distance slid is
SQRT(5.54^2 + 6.25^2) = 8.35 m
Speed after impact from
v^2 - u^2 = 2as
v = 0
-u^2 = -2 x 7.35 x 8.35
u = 11.1 m/s and angle = tan^-1(6.25/5.54) = 48.4 deg south of west.
Resolving into south and west components and equating momentum before and after
1400v[1] = (1400 + 2000) x 11.1 x sin(48.4)
2000v[2] = (1400 + 2000) x 11.1 x cos(48.4)
v[1] = 20.15 m/s
v[2] = 12.52 m/s

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