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 0.750, and the cars slide to a halt at a point 5.54 m west and 6.43 m south of the impact point.

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

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

After part A I got vo=6.57 for sedan and for part B I got vo=3.96 but they are not correct:/

Answer 1

given,

mass of sedan = 1400 kg

mass of SUV = 2000 kg

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(5.54^2 + 6.43^2) = 8.487 m

Speed after impact from

v^2 - u^2 = 2as

v = 0

-u^2 = -2 x 7.35 x 8.487

u = 11.169 m/s

angle = tan^-1(6.43/5.54)

angle = 49.25 deg south of west.

Resolving into south and west components and equating momentum before and after

1400v[1] = (1400 + 2000) x 11.169 x sin(49.25)

2000v[2] = (1400 + 2000) x 11.169 x cos(49.25)

v[1] = 20.5487 m/s

v[2] = 12.3941 m/s