A railroad car of mass M moving at a speed v1 collides and couples with two coup

Question

A railroad car of mass M moving at a speed v1 collides and couples with two coupled railroad cars, each of the same mass M and moving in the same direction at a speed v2.

(a) What is the speed vf of the three coupled cars after the collision in terms of v1 and v2? (Use M, v_1 for v1, and v_2 for v2 as appropriate.)


(b) How much kinetic energy is lost in the collision? Answer in terms of M, v1, and v2. (Use M, v_1 for v1, and v_2 for v2 as appropriate.)

besides the correct answer i need to know how to enter this equations answer into the computer web assign the right way I thought i had the right answer but It doesnt seem to like the way im entering it i have one try left please help!!!!!

Explanation / Answer

Given Mass of the railroad car is M Speed of the car is v1 Mass of two coupled cars is 2M Speed of the two cars is v2 a) By law of conservation of momentum    M v1 + 2M v2 = (3M) vf       M(v1+2 v2) =  (3M) vf        vf = ( v1 + 2 v2 ) / 3  ----1 is the final speed of the three coupled cars b) Kinetic energy before collision is ,     KEi = 1/2 Mv1^2 +1/2(2M) v2^2 = 1/2 M (v1^2 +2 v2^2)     Kinetic energy after collision is ,     KEf = 1/2 (3M ) vf^2 = 3/2 M ((v1 + 2 v2 ) / 3)^2   [from eq1]             = (M / 6) (v1^2 + 4 v2^2 + 4 v1 v2) Loss of KE is = KEi - KEf                       =   1/2 M (v1^2 +2 v2^2) - (M / 6) (v1^2 + 4 v2^2 + 4 v1 v2)                       = 1/3 Mv1^2 + 1/3 Mv2^2  - 2/3 Mv1 v2          

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