Question

A car of mass 1200kg traveling northward at 17m/s collides in an intersection with a 1600kg car traveling westward at 12m/s. (a) Assuming that the two cars remain attached to each-other, find the velocity of the wreck immediately after the collision. (b) Calculate the speed of the wreck and the angle at which it is moving with respect to due North. (c) Calculate the difference between the final kinetic energy and the total initial kinetic energy.

Answer #1

Here ,

let the east is the positive x direction

north is positive y direction

a) Using conservation of momentum

1200 * (17 j ) - 1600 * 12i = (1200 + 1600) * v

v = 7.3 i - 6.9 j m/s

v = sqrt(7.3^2 + 6.9^2) atan(6.9/7.3) degree

v = 10 m/s at 43.9 degree north of west

just after the collision

the velocity of wreck is 10 m/s at 43.9 degree north of west

b)

speed of wreck is 10 m/s

the angle with the north is (90 - 43.9) = 46.6 degree

c)

difference in final kinetic energy and initial kinetic energy

loss in KE = 0.50 * 1200 * 17^2 + 0.50 * 1600 * 12^2 - 0.50 * 2800 * 10^2

loss in KE = 148600 J

the difference in KE is 148600 J

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