G 1470-kilogram truck moving with a speed of 24.0 m/s runs into the rear end of a 1110-kilogram stationary car. if the collision is completely inelastic, how much kinetic energy is lost in the collision?

Kinetic energy = Potential energy

(1/2) mV^2 = mgh

where

m = mass of the car

V = velocity of the car = 15 m/s (given)

g = acceleration due to gravity = 9.8 m/sec^2 (constant)

h = vertical distance travelled by car = d (sin 19)

d = distance travelled by the car along the incline

Substituting values,

(1/2) (m) (529) = m (9.8) h

and since "m" appears on both sides of the equation, it will simply cancel out, hence the above becomes

(1/2) (529) = 9.8h

h = 225/19.6 = 26,98 m

Let

d = distance that the car will coast up the hill before coming down

h/d = sin 19

and therefore,

d = h/sin 19

and substituting values,

d = 26,98/sin 19

d = 82,870 m - - - after coasting up the hill for this distance, the car will start to roll down.