Conservation of momentum railroad car collision

[mandy9008]

Homework Statement

A railroad car of mass 27000 kg moving at 2.00 m/s collides and couples with two coupled railroad cars, each of the same mass as the single car and moving in the same direction at 1.20 m/s. (a) What is the speed of the three coupled cars after the collision?

Homework Equations

mv + mv = mv + mv

The Attempt at a Solution

Mrc = 27000 kg
Vrc = 2.00 m/s
M2rc = 54000 kg
V2rc = 1.20 m/s

(Mrc)(Vrc)i + (M2rc)(V2rc)i = (Mrc)(Vrc)f + (M2rc)(V2rc)f
(27000 kg)(2.00 m/s) + (54000 kg)(1.20 m/s) = (27000 kg)(0 m/s) + (54000 kg)(V2rc)f
V2rcf=2.20 m/s

 

[rock.freak667]

(Mrc)(Vrc)i + (M2rc)(V2rc)i = (Mrc)(Vrc)f + (M2rc)(V2rc)f
(27000 kg)(2.00 m/s) + (54000 kg)(1.20 m/s) = (27000 kg)(0 m/s) + (54000 kg)(V2rc)f
V2rcf=2.20 m/s

The left side of your equation is correct. Remember, the three cars couple together, so your new mass would be the sum of the masses which are all moving at the same final velocity.

 

[mandy9008]

(27000 kg)(2.00 m/s) + (81000 kg)(1.20 m/s) = (27000 kg)(0 m/s) + (81000 kg)(V2rc)f
V2rcf= 1.87 m/s

or

(27000 kg)(2.00 m/s) + (54000 kg)(1.20 m/s) = (27000 kg)(0 m/s) + (81000 kg)(V2rc)f
V2rcf= 1.47m/s

 

[rock.freak667]

Remember that conservation of momentum is basically that momentum before = momentum after

(27000 kg)(2.00 m/s) + (81000 kg)(1.20 m/s) = (27000 kg)(0 m/s) + (81000 kg)(V2rc)f
V2rcf= 1.87 m/s

The terms on the left side didn't happen before.

(
(27000 kg)(2.00 m/s) + (54000 kg)(1.20 m/s) = (27000 kg)(0 m/s) + (81000 kg)(V2rc)f
V2rcf= 1.47m/s

This is the correct one.

 

[mandy9008]

(b) How much kinetic energy is lost in the collision?
KE = 1/2 mv²
KE = 1/2 (81000 kg)(1.47 m/s)²
KE = 87516.45 J

what is the other KE that I need to calculate to determine the difference?

 

[rock.freak667]

(b) How much kinetic energy is lost in the collision?
KE = 1/2 mv²
KE = 1/2 (81000 kg)(1.47 m/s)²
KE = 87516.45 J

what is the other KE that I need to calculate to determine the difference?

You calculated the final KE. Initially, the car is moving at 2 m/s and the couple at 1.20 m/s. So find the total KE then and find the difference.

 

[mandy9008]

KE= 1/2 (27000 kg)(2.0 m/s)^2
KE = 54000 J

KE = KE= 1/2 (54000 kg)(1.2 m/s)^2
KE = 38880 J

KEi = 92880 J
KEf = 87516.45 J

KElost = 5363.55

 

[rock.freak667]

That should be correct.

 

[mandy9008]

it says this:
Your response is within 10% of the correct value. This may be due to roundoff error, or you could have a mistake in your calculation. Carry out all intermediate results to at least four-digit accuracy to minimize roundoff error.
I reworked it and I keep getting the same exact answer, but I only have one more submission left before I will get it wrong.

 

[rock.freak667]

(27000 kg)(2.00 m/s) + (54000 kg)(1.20 m/s) = (27000 kg)(0 m/s) + (81000 kg)(V2rc)f
V2rcf= 1.47m/s

From this equation, use the exact value of V2rcf and see what you get.

 

[mandy9008]

okay that worked because the value was 1.4666666667 and I rounded it to 1.47. Thanks