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

Question

A 1400 kg sedan goes through a wide intersection traveling from north to south when it is hit by a 2000 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 5.51 m west and 6.43 m south of the impact point.

Part A

How fast was sedan traveling just before the collision?

Express your answer in meters per second to three significant figures.

Part B

How fast was SUV traveling just before the collision?

Express your answer in meters per second to three significant figures.

Explanation / Answer

The acceleration after impact is found from
uR = ug(1400 + 2000) = (1400 + 2000)a
a = ug = 0.75 x 9.8 = 7.35
The distance slid is
SQRT(5.51^2 + 6.43^2) = 8.46 m
Speed after impact from
v^2 - u^2 = 2as
v = 0
-u^2 = -2 x 7.35 x 8.46
u = 11.15 m/s and angle = tan^-1(6.43/5.51) = 49.4 deg south of west.
Resolving into south and west components and equating momentum before and after
1400v[1] = (1400 + 2000) x 11.15 x sin(49.4)
2000v[2] = (1400 + 2000) x 11.15 x cos(49.4)
v[1] = 20.55 m/s ===========A
v[2] = 12.33 m/s==========B

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