A system consists of two particles. Particle 1 with mass \(2.0 \mathrm{~kg}\) is located at \((2.0 \mathrm{~m}, 6.0 \mathrm{~m})\) and has a velocity of \((4.0 \mathrm{~m} / \mathrm{s}, 2.0 \mathrm{~m} / \mathrm{s}) .\) Particle 2 with mass \(3.0 \mathrm{~kg}\) is located at \((4.0 \mathrm{~m}, 1.0 \mathrm{~m})\) and has a velocity of \((0,4.0 \mathrm{~m} / \mathrm{s})\) a) Determine the position and the velocity of the center of mass of the system. b) Sketch the position and velocity vectors for the individual particles and for the center of mass.

Short Answer

Expert verified
Question: Determine the position and the velocity of the center of mass of the system and sketch the position and velocity vectors for the individual particles and for the center of mass. Solution: a) The position of the center of mass is calculated as: $$\vec{R}_{cm} = \begin{pmatrix}3.2\\2.2\end{pmatrix} \mathrm{m}$$. The velocity of the center of mass is calculated as: $$\vec{V}_{cm} = \begin{pmatrix}1.6\\3.2\end{pmatrix} \frac{\mathrm{m}}{\mathrm{s}}$$. b) The position and velocity vectors for the individual particles and the center of mass can be sketched by drawing arrows representing the position vectors from each particle's position to the center of mass position, and arrows representing each particle's velocity vector and the center of mass velocity vector.

Step by step solution

01

To find the position of the center of mass, use the following formula: $$\vec{R}_{cm} = \frac{m_1\vec{r}_1 + m_2\vec{r}_2}{m_1 + m_2}$$ Here, \(\vec{r}_1\) and \(\vec{r}_2\) are the position vectors of particles 1 and 2, and \(m_1\) and \(m_2\) are their respective masses. #Step 2: Calculate center of mass coordinates#

Substitute the given values and calculate the center of mass coordinates: $$\vec{R}_{cm} = \frac{2.0\begin{pmatrix}2.0\\6.0\end{pmatrix} + 3.0\begin{pmatrix}4.0\\1.0\end{pmatrix}}{2.0 + 3.0}$$ $$\vec{R}_{cm} = \frac{1}{5}\begin{pmatrix}16.0\\11.0\end{pmatrix} = \begin{pmatrix}3.2\\2.2\end{pmatrix} \mathrm{m}$$ #Step 3: Find the velocity of the center of mass#
02

To find the velocity of the center of mass, use the following formula: $$\vec{V}_{cm} = \frac{m_1\vec{v}_1 + m_2\vec{v}_2}{m_1 + m_2}$$ Here, \(\vec{v}_1\) and \(\vec{v}_2\) are the velocity vectors of particles 1 and 2, and \(m_1\) and \(m_2\) are their respective masses. #Step 4: Calculate center of mass velocity#

Substitute the given values and calculate the center of mass velocity: $$\vec{V}_{cm} = \frac{2.0\begin{pmatrix}4.0\\2.0\end{pmatrix} + 3.0\begin{pmatrix}0\\4.0\end{pmatrix}}{2.0 + 3.0}$$ $$\vec{V}_{cm} = \frac{1}{5}\begin{pmatrix}8.0\\16.0\end{pmatrix} = \begin{pmatrix}1.6\\3.2\end{pmatrix} \frac{\mathrm{m}}{\mathrm{s}}$$ #b) Sketch the position and velocity vectors for the individual particles and for the center of mass.# #Step 5: Draw the diagram of the position vectors#
03

Using the given positions of the particles and the calculated center of mass position, sketch the position vectors. Draw an arrow from each particle position to the center of mass position. #Step 6: Draw the diagram of the velocity vectors#

Using the given velocities of the particles and the calculated center of mass velocity, sketch the velocity vectors. Draw an arrow for each particle's velocity vector and the center of mass velocity vector. The resulting diagrams will show the position and velocity vectors for both particles and the center of mass, providing a visual representation of the given system.

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Key Concepts

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

Understanding Physics Through the Center of Mass
In physics, center of mass is a fundamental concept that represents the average position of all the parts of the system, weighted by their masses. It plays a crucial role in analyzing the motion of objects, particularly when dealing with systems composed of multiple particles.

The center of mass can be thought of as the balancing point of an object, where if supported, the object would remain in equilibrium without rotating. In the exercise, the center of mass gives us an insight into the overall behavior of the two-particle system, simplifying the way we approach the problem of motion for multiple bodies.

By identifying the center of mass, one can often make accurate predictions about a system's movement without needing to calculate the behavior of every individual component. This is especially useful in situations involving gravity, where the force acts as though it were applied at the center of mass.
Velocity Vectors: Visualizing Motion
Velocity vectors are graphical representations of an object's speed and direction, an essential aspect of kinematics in physics. In the given exercise, velocity vectors help us understand how each particle in the system is moving and at what rate.

A velocity vector is characterized by its magnitude, which indicates the object's speed, and its direction, showing the way the object is headed. For particle 1, the velocity vector is (4.0 m/s, 2.0 m/s), meaning it moves 4.0 meters per second horizontally and 2.0 meters per second vertically.

When calculating the velocity of the center of mass, the velocity vectors of individual particles are combined, considering their masses, to create a single vector. This composite vector reflects the overall motion of the system. Visualizing these vectors can make it easier to grasp both individual and collective movements within the system.
Position Vectors: Determining Locations
Position vectors are as vital to the study of physics as velocity vectors, but instead of representing motion, they indicate the specific location of an object in space. In the exercise, position vectors are used to locate particles 1 and 2, and enable us to calculate the center of mass's position.

Each position vector has an x- and y-component corresponding to the object's position on the horizontal and vertical axes, respectively. The position vector for particle 1, for example, is (2.0 m, 6.0 m), indicating it is 2.0 meters to the right and 6.0 meters above the origin.

Combining the position vectors of the particles according to their masses allows for the determination of the system's center of mass. This process helps students visualize not just where the particles are, but also where the 'average' position or balancing point of the whole system resides.

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Most popular questions from this chapter

A uniform chain with a mass of \(1.32 \mathrm{~kg}\) per meter of length is coiled on a table. One end is pulled upward at a constant rate of \(0.47 \mathrm{~m} / \mathrm{s}\). a) Calculate the net force acting on the chain. b) At the instant when \(0.15 \mathrm{~m}\) of the chain has been lifted off the table, how much force must be applied to the end being raised?

A cart running on frictionless air tracks is propelled by a stream of water expelled by a gas-powered pressure washer stationed on the cart. There is a \(1.00-\mathrm{m}^{3}\) water tank on the cart to provide the water for the pressure washer. The mass of the cart, including the operator riding it, the pressure washer with its fuel, and the empty water tank, is \(400 . \mathrm{kg} .\) The water can be directed, by switching a valve, either backward or forward. In both directions, the pressure washer ejects \(200 .\) L of water per min with a muzzle velocity of \(25.0 \mathrm{~m} / \mathrm{s}\). a) If the cart starts from rest, after what time should the valve be switched from backward (forward thrust) to forward (backward thrust) for the cart to end up at rest? b) What is the mass of the cart at that time, and what is its velocity? (Hint: It is safe to neglect the decrease in mass due to the gas consumption of the gas-powered pressure washer!) c) What is the thrust of this "rocket"? d) What is the acceleration of the cart immediately before the valve is switched?

Two masses, \(m_{1}=2.0 \mathrm{~kg}\) and \(m_{2}=3.0 \mathrm{~kg}\), are moving in the \(x y\) -plane. The velocity of their center of mass and the velocity of mass 1 relative to mass 2 are given by the vectors \(v_{\mathrm{cm}}=(-1.0,+2.4) \mathrm{m} / \mathrm{s}\) and \(v_{\mathrm{rel}}=(+5.0,+1.0) \mathrm{m} / \mathrm{s} .\) Determine a) the total momentum of the system b) the momentum of mass 1 , and c) the momentum of mass 2 .

Young acrobats are standing still on a circular horizontal platform suspended at the center. The origin of the two-dimensional Cartesian coordinate system is assumed to be at the center of the platform. A 30.0 -kg acrobat is located at \((3.00 \mathrm{~m}, 4.00 \mathrm{~m})\), and a 40 - \(\mathrm{kg}\) acrobat is located at \((-2.00 \mathrm{~m}\) \(-2.00 \mathrm{~m})\). Assuming that the acrobats stand still in their positions, where must a 20.0 -kg acrobat be located so that the center of mass of the system consisting of the three acrobats is at the origin and the platform is balanced?

The density of a \(1.00-\mathrm{m}\) long rod can be described by the linear density function \(\lambda(x)=\) \(100 \cdot \mathrm{g} / \mathrm{m}+10.0 x \mathrm{~g} / \mathrm{m}^{2}\) One end of the rod is positioned at \(x=0\) and the other at \(x=1.00 \mathrm{~m} .\) Determine (a) the total mass of the rod, and (b) the center-of-mass coordinate.

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