Question

A 1600 kg sedan goes through a wide intersection traveling from north to south when it is hit by a 2300 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 5.54 m west and 6.19 m south of the impact point. How fast was sedan traveling just before the collision? How fast was SUV traveling just before the collision?

Answer 1

Step-by-step explanation:

momentum of sedan of 1600 kg = 1600x v , where v is its velocity

momentum of suv of 2300 kg = 2300 x u where u is its velocity .

force of friction = ( 1600 + 2300 ) x 9.8 x .75 ( fiction = μ mg )

= 28665 N

distance by which friction acted = √ (5.54² + 6.19²)

= 8.3 m

work done by friction

= 28665 x 8.3

= 237919.5 J

Total kinetic energy of cars = work done by friction

1/2 x 1600 x v² + 1/2 x 2300 u² = 237919.5

16 v² + 23 u² = 4758.4

1600 x v / 2300 u = 6.19 / 5.54

v / u = 1.6

v = 1.6 u

putting this equation in fist equation

40.96 u² + 23 u² = 4758.4

= 63.96 u² = 4758.4

u² = 74.4

u = 8.62 m /s

v = 13.8 m /s