Answer to Question #106028 in Mechanics | Relativity for Valentine Agun

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? (b) How much kinetic energy is lost in the collision? Answer in terms of M, v1, and v2.

Answer 1

(a)

With the use of conservation of momentum law

$Mv_1+2Mv_2=3Mv\to v=\frac{v_1+2v_2}{3}$

(b)

$\Delta KE=KE_f-KE_i$

$KE_f=\frac{3M\cdot v^2}{2}=\frac{M(v_1+2v_2)^2}{6}$

$KE_i=\frac{Mv_1^2}{2}+\frac{2Mv_2^2}{2}$

$\Delta KE=\frac{M(v_1+2v_2)^2}{6}-\frac{Mv_1^2}{2}-\frac{2Mv_2^2}{2}=$

$=\frac{M}{6}(-2v_1^2-2v_2^2+4v_1v_2)=\frac{M}{3}(-v_1^2-v_2^2+2v_1v_2)=$

$=-\frac{M}{3}(v_1^2-2v_1v_2+v_2^2)=-\frac{M}{3}(v_1-v_2)^2$

$KE_{lost}=\lvert \Delta KE\rvert=\frac{M}{3}(v_1-v_2)^2$