Answer to Question #98306 in Physics for Luis

Question #98306
You are a world-famous physical lawyer who defends a client who has been injured in a traffic accident. His client's car was hit from behind by another vehicle while he was in custody. The judge who is taking the case, argues that if the vehicle of the affected had been in motion, the injuries would not have been so serious. Is the judge correct? Or do you think you can win this case? Investigate and raise your arguments, based on collisions in one dimension.

A) It shows a detailed diagram (design) of the situation, where it shows the conditions before and after.

B)It shows a correct procedure of conservation laws and the mathematical calculations that support their arguments.

C)It gives a final conclusion of its found results.
1
Expert's answer
2019-11-13T09:24:27-0500

Consider different scenarios for the first car while the second (moving from behind) was moving with speed "u".

1) The first car's brakes were activated. In this case we can say that most of the kinetic energy of the car moving from behind was transferred to destruction "D", and some part of it was transferred motion and to overcoming friction of the hit car:


"E_{before}=E_{after},\\\\\nm_2u^2\/2=D+\\mu m_1gx_1+m_1v^2\/2,\\\\\nD=m_2u^2\/2-\\mu m_1gx_1-m_1v^2\/2."

where "x_1"- braking distance (track) of the first car, "v" - its speed after the driver put his leg off the braking pedal (because of safety considerations, can be determined from dash cam). We need more information for this case.

2) The first car' brakes weren't activated and it was resting. After the accident both cars were moving as one. In that case we can try to apply conservation of momentum in addition to the law of conservation of energy:


"p_{before}=p_{after},\\\\\nm_2u=(m_2+m_1)v,\\\\\nm_2u^2\/2=D+(m_1+m_2)v^2\/2,\\\\\nD=\\frac{m_2u^2}{2}\\bigg(1-\\frac{m_2}{m_1+m_2}\\bigg)."

3) the first car's brakes weren't activated and it was in motion. In this case:


"m_2u=m_2u_1+m_1v,\\\\\nm_2u^2\/2=D+m_2u_1^2\/2+m_1v^2\/2."

As a conclusion, we can say that the last case would bring the least serious consequences, but to estimate the situation quantitatively, we need more information about mass of the cars and at least velocity of the cars before the interaction and after the interaction).



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