Question

A 900-kg car traveling east at 15.0 m/s collides with a 750-kg car traveling north at 20.0 m/s. The cars stick together. Assume that any other unbalanced forces are negligible.

a. What is the speed of the wreckage just after the collision?

b. In what direction does the wreckage move just after the collision?

c. How much kinetic energy is lost in this collision.

Answer 1

N usina 10 15m AS V2 S Vn Vcosa 900 mz 20ml 2om/s 750kg mi All the given data is shown in figure above No external force is acting on the system So momentum remains conserved Pinitial = P final from momentum conservation in horizontal direction m2v2 = (m, + m2 )vcosa m V2 = Vcosa m+m2 from momentum conservation in vertical direction mi v = (m, + m2 )vsina mv. = vsina m+m2 Now v2sin’a + vcosa = (m>vz+m?v?) (m+m2)

Vmžvž+m7v; V= m,+m2) (900 x 15)2 + ( 750 x 20) (900+ 750) v = 12.231 x m/s vsina mi = 750 x 20 900 x 15 vcosa m V2 A = tan!(1.111) a = 48 Therefore just after the collision the wreckage move at velocity 12.231 x m/s at an angle 48° counter clockwise from East Intial kinetic energy K;=5m, v* + mavž K; = (750) x (20)2+ ten * (900) (15) K; = 251250J final kinetic energy K = }x (mi +m n2 )u2 K = < (750 + 900) (12.23192

Kg = 123417.8J Loss in kinetic energy AK = K;-K AK = 251250 - 123417.8 AK = 127832.2