Question
A 1000kg car moving at 15m/s to the right collided head-on with another 1500 kg car who was coming in the opposite direction at a speed of 20m/s. After collision the two cars interlocked to each other, and the wreckage dragged with a new speed. What is the lost of kinetic energy during the collision ?

A 1000 kg moving at 15 m/s to the right collided head-on with another 1500 kg car who was coming in the opposite direction at
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Answer #1

Using the momentum conservation equation,

$$\begin{array}{l}{m_1}{v_1} + {m_2}{v_2} = \left( {{m_1} + {m_2}} \right)v\\\left( {1000kg} \right)\left( {15m/s} \right) + \left( {1500kg} \right)\left( { - 20m/s} \right) = \left( {1000kg + 1500kg} \right)v\\\left| v \right| = 6m/s\end{array}$$

The lost energy is:

$$\begin{array}{l}\Delta E = {E_i} - {E_f}\\\Delta E = \left( {\frac{1}{2}{m_1}v_1^2 + \frac{1}{2}{m_2}v_2^2} \right) - \frac{1}{2}\left( {{m_1} + {m_2}} \right){v^2}\\\Delta E = \left( {\frac{1}{2}\left( {1000kg} \right){{\left( {15m/s} \right)}^2} + \frac{1}{2}\left( {1500kg} \right){{\left( {20m/s} \right)}^2}} \right) - \frac{1}{2}\left( {1000kg + 1500kg} \right){\left( {6m/s} \right)^2}\\\Delta E = 367500J\end{array}$$

Therefore, option B is correct.

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