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Momentum. Kinetic energy lost in the collision.

  • Thread starter qwerty159
  • Start date
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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 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.)

a) (M*v_1+M*v_2+M*v_2)/(3M)

(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.)

b) I don't get this I had this for the answer. 1/2*M*[-2/3*v_1^2 +4/3*v_1*v_2-2/3v_2^2]
but it is wrong...

thanks
 

Answers and Replies

gneill
Mentor
20,536
2,626
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.)

a) (M*v_1+M*v_2+M*v_2)/(3M)
While the expression is correct, you can simplify it some more.
(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.)

b) I don't get this I had this for the answer. 1/2*M*[-2/3*v_1^2 +4/3*v_1*v_2-2/3v_2^2]
but it is wrong...
Can you show how you arrived at your result?
 
ehild
Homework Helper
15,361
1,778
b) I don't get this I had this for the answer. 1/2*M*[-2/3*v_1^2 +4/3*v_1*v_2-2/3v_2^2]
but it is wrong...

thanks
Note the sign: It is the change of KE, not the lost KE.
And simplify the expression.

ehild
 

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