A toy car of mass \(2.0 \mathrm{~kg}\) is stationary, and a child rolls a toy truck of mass 3.5 kg straight toward it with a speed of \(4.0 \mathrm{~m} / \mathrm{s}\) a) What is the velocity of the center of mass of the system consisting of the two toys? b) What are the velocities of the truck and the car with respect to the center of mass of the system consisting of the two toys?

Short Answer

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b) What are the velocities of the toy car and toy truck with respect to the center of mass?

Step by step solution

01

a) Velocity of the center of mass

First, find the velocity of the center of mass using the equation mentioned above: \(v_{cm} =\frac{\Sigma m_iv_i}{\Sigma m_i} = \frac{m_1v_1 + m_2v_2}{m_1 + m_2} \) Given, the mass of the toy car \(m_1 = 2.0 \mathrm{~kg}\) and is stationary, so \(v_1 = 0\,\mathrm{m/s}\). The mass of the toy truck \(m_2 = 3.5\,\mathrm{kg}\) and its speed is \(v_2 = 4.0\, \mathrm{m/s}\). Substituting the known values: \(v_{cm} = \frac{(2.0)(0) + (3.5)(4.0)}{2.0 + 3.5}\) \(v_{cm} = \frac{14}{5.5}\,\mathrm{m/s}\) Therefore, the velocity of the center of mass of the system consisting of the two toys is \(\frac{14}{5.5}\,\mathrm{m/s}\).
02

b) Velocities of the truck and car with respect to the center of mass

Next, let's find the velocities of toy car and toy truck with respect to the center of mass. Using the equation: \(v_{i}^{'} = v_i-v_{cm}\) For toy car: \(v_{1}^{'} = v_1 - v_{cm} = 0 - \frac{14}{5.5}\,\mathrm{m/s} = -\frac{14}{5.5}\, \mathrm{m/s}\) So, the velocity of the toy car with respect to the center of mass is \(-\frac{14}{5.5}\,\mathrm{m/s}\). For the toy truck: \(v_{2}^{'} = v_2 - v_{cm} = 4.0\, \mathrm{m/s} - \frac{14}{5.5}\,\mathrm{m/s}\) \(v_{2}^{'} = \frac{22}{5.5}\,\mathrm{m/s}\) Therefore, the velocity of the toy truck with respect to the center of mass is \(\frac{22}{5.5}\,\mathrm{m/s}\). To summarize: a) The velocity of the center of mass of the system consisting of the two toys is \(\frac{14}{5.5}\,\mathrm{m/s}\). b) The velocities of the toy car and toy truck with respect to the center of mass of the system are \(-\frac{14}{5.5}\,\mathrm{m/s}\) and \(\frac{22}{5.5}\,\mathrm{m/s}\), respectively.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Momentum Conservation
Understanding momentum conservation is crucial for solving many problems in physics, especially when analyzing collisions or interactions between objects. Momentum is a measure of the motion of an object and is calculated by multiplying the mass of the object by its velocity. The principle of conservation of momentum states that in an isolated system, where no external forces are acting, the total momentum remains constant throughout any interaction.

In the toy car and truck problem, even though the truck is moving and the car is stationary, the total momentum before and after they interact must remain the same as they are considered an isolated system for the duration of their interaction. When people ask about the center of mass velocity of the system, they are referring to the single velocity that represents the motion of the system as a whole, taking into account the masses and velocities of all components. This is why, using the conservation of momentum, we were able to calculate the velocity of the center of mass.
Velocity Calculations
Velocity calculations in physics often involve finding the speed and direction of moving objects. In our textbook solution, we calculated the center of mass velocity using the formula that expresses the weighted average velocities of the individual masses.

The use of the formula \(v_{cm} =\frac{\Sigma m_iv_i}{\Sigma m_i}\) boils down to a simple weighted average, which in this scenario includes the truck's mass and velocity and the car's mass and stationary position. This formula elegantly handles the otherwise complex interactions within the system.

The second part of the problem involved calculating the velocities of both the car and the truck with respect to this center of mass. By understanding the velocity of the system's center of mass, we can deduce the individual velocities relative to that point. It's much like analyzing how fast you're walking down the aisle of a moving train relative to the train itself.
Newtonian Mechanics
Newtonian mechanics, formulated by Sir Isaac Newton, provides the foundation for classical physics and describes the behavior of objects under the influence of forces. It's based on three fundamental laws of motion that describe the relationship between an object and the forces acting upon it, the object's mass, and its motion.

The analysis of the toy car and truck situation relies on Newton's principles, specifically the first and second laws. The first law, also known as the law of inertia, explains that the toy car remains at rest until acted upon by an external force. In this case, the moving truck could provide such a force upon collision. The second law explains how the force acting on the truck and its resulting acceleration relate to its mass and the change in velocity—namely, force equals mass times acceleration (F=ma). These principles are integral to solving problems in classical mechanics and were implicitly used in calculating the system's center of mass and the velocities related to it.

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