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3 years ago

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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 m

oving 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? vf =

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

1 answer:

Tatiana [17]3 years ago

7 0

Answer:

Explanation:

Given

mass of each railroad is M

first railroad is moving with velocity $v_1$ and other two's with $v_2$

conserving Momentum

$Mv_1+2Mv_2=3M v_f$

where v_f is the final velocity of combined system

$v_f=\frac{v_1+2v_2}{3}$

(b)Initial Kinetic Energy

$=\frac{1}{2}Mv_1^2+\frac{1}{2}2Mv_2^2$

$=\frac{1}{2}Mv_1^2+Mv_2^2$

Final Kinetic Energy

$=\frac{1}{2}3Mv_f^2=\frac{1}{6}M(v_1^2+4v_2^{2}+4v_{1}v_2)$

Loss in Kinetic Energy

$=\frac{1}{2}Mv_1^2+Mv_2^2-\frac{1}{6}M(v_1^2+4v_2^{2}+4v_{1}v_2)$

$=M(v_1-v_2^2)^2$

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