Answer:
The coefficient of friction between the cars and the road is 0.859.
Explanation:
The two cars collide each other inelastically, then we can determine the resulting velocity by the Principle of Momentum Conservation:
[tex]m_{A}\cdot v_{A} + m_{B}\cdot v_{B} = (m_{A} + m_{B})\cdot v[/tex] (1)
Where:
[tex]m_{A}[/tex], [tex]m_{B}[/tex] - Masses of the cars, in kilograms.
[tex]v_{A}[/tex], [tex]v_{B}[/tex] - Initial velocities of the cars, in meters per second.
[tex]v[/tex] - Velocity of the resulting system, in meters per second.
If we know that [tex]m_{A} = 2120\,kg[/tex], [tex]v_{A} = 13.4\,\frac{m}{s }[/tex], [tex]m_{B} = 2810\,kg[/tex] and [tex]v_{B} = 0\,\frac{m}{s}[/tex], then the velocity of the resulting system:
[tex]v = \frac{m_{A}\cdot v_{A}+m_{B}\cdot v_{B}}{m_{A}+m_{B}}[/tex]
[tex]v = \frac{(2120\,kg)\cdot \left(13.4\,\frac{m}{s} \right)+(2810\,kg)\cdot \left(0\,\frac{m}{s} \right)}{2120\,kg + 2810\,kg}[/tex]
[tex]v = 5.762\,\frac{m}{s}[/tex]
By Principle of Energy Conservation and Work-Energy Theorem, we understand that the initial translational kinetic energy ([tex]K[/tex]), in joules, is dissipated due to work done by friction ([tex]W_{f}[/tex]), in joules, that is to say:
[tex]K = W_{f}[/tex] (2)
[tex]\frac{1}{2}\cdot (m_{A}+m_{B})\cdot v^{2} = \mu\cdot (m_{A}+m_{B})\cdot g \cdot s[/tex]
[tex]\frac{1}{2}\cdot v^{2} = \mu \cdot g\cdot s[/tex] (2b)
Where:
[tex]\mu[/tex] - Coefficient of friction, no unit.
[tex]g[/tex] - Gravitational acceleration, in meters per square second.
[tex]s[/tex]- Travelled distance, in meters.
If we know that [tex]v = 5.762\,\frac{m}{s}[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex] and [tex]s = 1.97\,m[/tex], then the coefficient of friction is:
[tex]\mu = \frac{v^{2}}{2\cdot g\cdot s}[/tex]
[tex]\mu = \frac{\left(5.762\,\frac{m}{s} \right)^{2}}{2\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (1.97\,m)}[/tex]
[tex]\mu = 0.859[/tex]
The coefficient of friction between the cars and the road is 0.859.
