# Loss of Mechanical Energy after Collision

#### lizzyb

1. The problem statement, all variables and given/known data

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.

2. Relevant equations

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

3. 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

Staff Emeritus
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.

Last edited:

#### cristo

Staff Emeritus
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

Staff Emeritus
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! :-)

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