Question

A 2120 kg car traveling at 13.4 m/s collides with a 2810 kg car that is initally at rest at a stoplight. The cars stick together and move 1.97 m before friction causes them to stop. Determine the coefficient of kinetic friction between the cars and the road, assuming that the negative acceleration is constant and all wheels on both cars lock at the time of impact.

Answer 1

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:

$m_{A}\cdot v_{A} + m_{B}\cdot v_{B} = (m_{A} + m_{B})\cdot v$ (1)

Where:

$m_{A}$, $m_{B}$ - Masses of the cars, in kilograms.

$v_{A}$, $v_{B}$ - Initial velocities of the cars, in meters per second.

$v$ - Velocity of the resulting system, in meters per second.

If we know that $m_{A} = 2120\,kg$, $v_{A} = 13.4\,\frac{m}{s }$, $m_{B} = 2810\,kg$ and $v_{B} = 0\,\frac{m}{s}$, then the  velocity of the resulting system:

$v = \frac{m_{A}\cdot v_{A}+m_{B}\cdot v_{B}}{m_{A}+m_{B}}$

$v = \frac{(2120\,kg)\cdot \left(13.4\,\frac{m}{s} \right)+(2810\,kg)\cdot \left(0\,\frac{m}{s} \right)}{2120\,kg + 2810\,kg}$

$v = 5.762\,\frac{m}{s}$

By Principle of Energy Conservation and Work-Energy Theorem, we understand that the initial translational kinetic energy ($K$), in joules, is dissipated due to work done by friction ($W_{f}$), in joules, that is to say:

$K = W_{f}$ (2)

$\frac{1}{2}\cdot (m_{A}+m_{B})\cdot v^{2} = \mu\cdot (m_{A}+m_{B})\cdot g \cdot s$

$\frac{1}{2}\cdot v^{2} = \mu \cdot g\cdot s$ (2b)

Where:

$\mu$ - Coefficient of friction, no unit.

$g$ - Gravitational acceleration, in meters per square second.

$s$- Travelled distance, in meters.

If we know that $v = 5.762\,\frac{m}{s}$, $g = 9.807\,\frac{m}{s^{2}}$ and $s = 1.97\,m$, then the coefficient of friction is:

$\mu = \frac{v^{2}}{2\cdot g\cdot s}$

$\mu = \frac{\left(5.762\,\frac{m}{s} \right)^{2}}{2\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (1.97\,m)}$

$\mu = 0.859$

The coefficient of friction between the cars and the road is 0.859.