Loss of Mechanical Energy after Collision

[lizzyb]

Homework Statement

A 1200-kg car traveling initially with a speed of 25.0 m/s in an easterly direction crashes into the rear end of a 9000-kg truck moving in the same direction at 20.0 m/s. The velocity of the car right after the collision is 18.0 m/s to the east. (a) What is the velocity of the truck right after the collision? (b) How much mechanical energy is lost in the collision? Account for this loss of energy.

It's sub-question (b) that's causing trouble.

Homework Equations

$$m_1 v_{1i} + m_2 v_{2i} = m_1 v_{1f} + m_2 v_{2f}$$
$$\frac{1}{2}m_1 v_{1f}^2 + \frac{1}{2} m_2 v_{2f}^2 - (\frac{1}{2} m_1 v_{1i}^2 + \frac{1}{2} m_2 v_{2i}^2) = - \Delta KE$$

The Attempt at a Solution

I did part (a) fine and substituted those values into the second equation above but the answer was not the same as in the book. What could possibly be wrong?

 

[cristo]

What answer did you get, and what does the book say is the correct answer? We'll be able to point out which is correct!

 

[lizzyb]

ok, for solving (a) where vehicle 2 is the truck, 1 is the car:

$$v_{2f} = \frac{m_1(v_{1i}-v_{1f})}{m2} + v_{2i} = -20.933 m/s$$

So we have:

$$\frac{1}{2}m_1 v_{1f}^2 + \frac{1}{2} m_2 v_{2f}^2 - (\frac{1}{2} m_1 v_{1i}^2 + \frac{1}{2} m_2 v_{2i}^2) = \frac{1}{2} 1200 (-25)^2 + \frac{1}{2} 9000 (-20.93)^2 - (\frac{1}{2} 1200 (-25)^2 + \frac{1}{2} 9000 (-20)^2) = -9307.95$$

and the book says 8.68 kJ.

 

[cristo]

I get the same answer as you for the loss in kinetic energy (9307J). Although, the speed of the truck after the collision is 20.933m/s (not -20.933).

 

[lizzyb]

i put the negative direction as going to the east.

 

[cristo]

Oh, ok, I can see that in the equation now! Well, both your solutions are correct then.

 

[lizzyb]

ok thanks - normally the book is correct. i'll keep plugging away then! :-)