Question

The CM of an empty 1250-kg car is 2.40 m behind the front of the car. How far from the front of the car will the CM be when two people sit in the front seat 2.80 m from the front of the car, and three people sit in the back seat 3.90 m from the front? Assume that each person has a mass of 65.0 kg.

Short answer

The center of mass of the car is \(2.62\;{\rm{m}}\) behind the front of the car.

Step-by-step solution

Given data

The center of mass of a system is a point where you can assume all the masses of the system to be situated.

To find the center of mass, you have to consider the two people at the front as one particle and the three people in the back seat as another particle.

The mass of the car is \({m_1} = 1250\;{\rm{kg}}\).

The mass of the people is \(m = 65.0\;{\rm{kg}}\).

The distance of the CM of the empty car is \({x_1} = 2.40\;{\rm{m}}\) behind the front of the car.

The distance of the CM of the people sitting in the front seat is \({x_2} = 2.80\;{\rm{m}}\) behind the car's front.

The distance of the CM of the people sitting in the back seat is \({x_3} = 3.90\;{\rm{m}}\) behind the car's front.

Calculation of the center of mass

Now, the center of mass of the mass system is:

\(\begin{array}{c}{x_{{\rm{CM}}}} = \frac{{\left( {{m_1} \times {x_1}} \right) + \left( {2m \times {x_2}} \right) + \left( {3m \times {x_3}} \right)}}{{{m_1} + 2m + 3m}}\\ = \frac{{\left( {1250\;{\rm{kg}} \times 2.40\;{\rm{m}}} \right) + \left( {2 \times 65.0\;{\rm{kg}} \times 2.80\;{\rm{m}}} \right) + \left( {3 \times 65.0\;{\rm{kg}} \times 3.90\;{\rm{m}}} \right)}}{{1250\;{\rm{kg}} + \left( {2 \times 65.0\;{\rm{kg}}} \right) + \left( {3 \times 65.0\;{\rm{kg}}} \right)}}\\ = 2.62\;{\rm{m}}\end{array}\)

Hence, the center of mass of the car is \(2.62\;{\rm{m}}\) behind the front of the car.