*****PLEASE INDICATE ANSWER CLEARLY****** A 1600 kg sedan goes through a wide in

Question

*****PLEASE INDICATE ANSWER CLEARLY******

A 1600 kg sedan goes through a wide intersection traveling from north to south when it is hit by a 2200 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 5.45 m west and 6.37 m 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

halt position of the vehicles 5.45 m west and 6.37 m south

so the distance from the impact point = sqrt(5.45^2 + 6.37^2)

so the distance from the impact point = sqrt(5.45^2 + 6.37^2)

so the distance from the impact point = 8.383 m

friction force = coefficient of friction * weight

also force = mass * acceleration

so,

mass * acceleration = coefficient of friction * mass * g

acceleration = coefficient of friction * g

acceleration = 0.75 * 9.8

by third equation of motion

v^2 = u^2 + was

0 = u^2 - 2 * 0.75 * 9.8 * 8.383

u = 11.1 m/s

angle = tan^-1(6.37 / 5.45)

angle = 49.45 degree

by conservation of momentum

initial horizontal momentum = final horizontal momentum

2200 * v1 = (1600 + 2200) * 11.1 * cos(49.45)

v1 = 12.464 m/s

speed of SUV before collision = 12.464 m/s

initial vertical momentum = final vertical momentum

1600 * v2 = (1600 + 2200) * 11.1 * sin(49.45)

v2 = 20.03 m/s

speed of sedan before collision = 20.03 m/s

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