A1500 kg car travels at 90 km/h east and collides with a 1000 kg car traveling at 60 km/h north. the two cars stick together after the collision. the road surface conditions were wet such that us= 0.5 and uk = 0.4. what is the speed of the cars, in m/s, immediately after the collision? what is the direction of the velocity of the cars immediately after collision? how far did the two vehicles skid, in meters, after the collision?

direction =

velocity =

distance skid = 145.2 meters

Explanation:

Let the mass of the car 1 m1 = 1500 kg, m2 = 1000 kg, v1 = 90 km/h and v2 = 60 km/h.

v1 = 90 i and v2 = 60 j.

applying conservation of linear momentum in i and j direction since there is no external force involved.

along i direction:

          1

along j direction:

         2

where v is the new velocity of combined car and Ф is the angle made with x axis.

equation 2÷1

therefore \tan \Phi = \frac{4}{9}

therefore the direction is

and velocity will be

the car will skid until the total energy of the car is dissipiated by friction

therefore

substituting the values we get x = 145.2 meters

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft