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

When a bismuth-208 nucleus at rest decays, thallium-204 is produced, along with an alpha particle (helium- 4 nucleus). The mass numbers of bismuth-208, thallium- 204 , and helium- 4 are 208,204 , and 4 , respectively. (The mass number represents the total number of protons and neutrons in the nucleus.) The kinetic energy of the thallium nucleus is a) equal to that of the alpha particle. b) less than that of the alpha particle. c) greater than that of the alpha particle.

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The Saturn \(V\) rocket, which was used to launch the Apollo spacecraft on their way to the Moon, has an initial mass \(M_{0}=2.80 \cdot 10^{6} \mathrm{~kg}\) and a final mass \(M_{1}=8.00 \cdot 10^{5} \mathrm{~kg}\) and burns fuel at a constant rate for \(160 .\) s. The speed of the exhaust relative to the rocket is about \(v=2700 . \mathrm{m} / \mathrm{s}\). a) Find the upward acceleration of the rocket, as it lifts off the launch pad (while its mass is the initial mass). b) Find the upward acceleration of the rocket, just as it finishes burning its fuel (when its mass is the final mass). c) If the same rocket were fired in deep space, where there is negligible gravitational force, what would be the net change in the speed of the rocket during the time it was burning fuel?

Many nuclear collisions studied in laboratories are analyzed in a frame of reference relative to the laboratory. A proton, with a mass of \(1.6605 \cdot 10^{-27} \mathrm{~kg}\) and traveling at a speed of \(70.0 \%\) of the speed of light, \(c\), collides with a tin\(116\left({ }^{116} \mathrm{Sn}\right)\) nucleus with a mass of \(1.9096 \cdot 10^{-25} \mathrm{~kg} .\) What is the speed of the center of mass with respect to the laboratory frame? Answer in terms of \(c\), the speed of light.

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