- #1

NoPhysicsGenius

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## Homework Statement

A 1200-kg car traveling initially with a speed of 25 m/s in an easterly direction crashes into the rear end of a 9000-kg truck moving in the same direction at 20 m/s. The velocity of the car right after the collision is 18 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? How do you account for this loss in energy?

## Homework Equations

[tex]m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}[/tex]

[tex]K = \frac{1}{2}mv^2[/tex]

## The Attempt at a Solution

I was able to solve part (a) as follows ...

[tex]m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}[/tex]

[tex]\Rightarrow v_{2f} = \frac{m_1v_{1i} + m_2v_{2i} - m_1v_{1f}}{m_2}[/tex]

[tex]\Rightarrow v_{2f} = \frac{(1200 kg)(25 m/s) + (9000 kg)(20 m/s) - (1200 kg)(18 m/s)}{9000 kg} = 20.9 m/s east[/tex]

For part (b), however, I didn't get very far ...

I know the energy loss is due to heat (thermal energy), but I could only calculate the loss in

*kinetic*energy, not

*mechanical*energy:

[tex]K_f - K_i = \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[/tex]

[tex]\Rightarrow K_f - K_i = \frac{1}{2}(1200 kg)(18 m/s)^2 + \frac{1}{2}(9000 kg)(20.9 m/s)^2 - \frac{1}{2}(1200 kg)(25 m/s)^2 - \frac{1}{2}(9000 kg)(20 m/s)^2 = -14,955 Joules = -15.0 kJ [/tex]

The problem is that the answer in the back of the book is

**8.74 kJ**.