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 2100kg 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 5.57m west and 6.49m 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 + 2100) = (1400 + 2100)a
a = ug = 0.75 x 9.8 = 7.35
The distance slid is
SQRT(5.57^2 + 6.49^2) = 8.55 m
Speed after impact from
v^2 - u^2 = 2as
v = 0
-u^2 = -2 x 7.35 x 8.55
u = 11.21 m/s and angle = tan^-1(6.49/5.57) = 47.19 deg south of west.
Resolving into south and west components and equating momentum before and after
1400v[1] = (1400 + 2100) x 11.21 x sin(47.19)
2100v[2] = (1400 + 2100) x 11.21 x cos(47.19)
v[1] = 20.56 m/s ...............(a)
v[2] = 12.70 m/s.................(b)

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.