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Chapter 9: Problem 45

A 2.00 -kg particle has a velocity \((2.00 \hat{\mathbf{i}}-3.00 \hat{\mathbf{j}}) \mathrm{m} / \mathrm{s},\) and a \(3.00-\mathrm{kg}\) particle has a velocity \((1.00 \hat{\mathbf{i}}+6.00 \hat{\mathbf{j}}) \mathrm{m} / \mathrm{s}\) Find (a) the velocity of the center of mass and (b) the total momentum of the system.

Short Answer

Expert verified
The velocity of the center of mass is \((1.8 \hat{\mathbf{i}} + 1.2 \hat{\mathbf{j}})\, \mathrm{m/s}\) and the total momentum of the system is \((8.00 \hat{\mathbf{i}} + 12.00 \hat{\mathbf{j}})\, \mathrm{kg \cdot m/s}\).

Step by step solution

01

Calculate the total mass of the system

To find the velocity of the center of mass, first we need to determine the total mass of the system by summing the masses of the two particles. The total mass is given by the equation: \(m_{total} = m_1 + m_2\), where \(m_1\) is the mass of the first particle and \(m_2\) is the mass of the second particle.
02

Calculate the momentum of each particle

Next, we calculate the momentum of each particle using the equation \(p = m * v\), where \(p\) represents the momentum, \(m\) is the mass of the particle, and \(v\) is its velocity. We need to compute this for each particle both for the \(i\) (horizontal) and \(j\) (vertical) components separately.
03

Find the total momentum for the system

To find the total momentum of the system, we sum the individual momenta of the particles along each axis. The total momentum along the \(i\)-axis will be the sum of the \(i\)-components of the two momenta, and similarly for the \(j\)-axis.
04

Find the center of mass velocity components

Now, we can calculate the velocity components of the center of mass using the equation \(V_{\text{cm}} = \frac{p_{\text{total}}}{m_{total}}\) for each axis separately.
05

Combine the components to find the center of mass velocity

Once we have the center of mass velocity components for both axes, we can combine them to express the velocity of the center of mass as a vector.

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Key Concepts

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

Momentum
Understanding momentum is essential when analyzing movements within a system. Momentum is defined as the product of a particle's mass and its velocity. Mathematically, it is expressed as \(p = m \times v\), where \(p\) represents momentum, \(m\) is mass, and \(v\) is velocity.

In the context of a two-particle system, each particle's momentum is a vector quantity. This means it has both magnitude and direction, represented respectively by the mass and velocity's magnitude and the direction the particle travels. To find the combined momentum of such a system, one must sum the momentum vectors for each particle.

Combining Momenta in Two Dimensions

For the mentioned exercise, we have a two-dimensional scenario involving the horizontal (\(i\)) and vertical (\(j\)) components of momentum. By calculating each particle's momentum separately and then adding the momentum vectors, we obtain the system's total momentum. This vector addition is crucial to finding the center of mass velocity, which is determined by dividing the total momentum by the total mass of the system.
Mass of a System
The mass of a system simply refers to the sum of the masses of all the individual components, or particles, within that system. For our two-particle system, the total mass is found by the equation \(m_{total} = m_1 + m_2\), with \(m_1\) and \(m_2\) representing the individual masses of the particles.

This total mass is a scalar quantity, meaning it has magnitude but no direction. Understanding how to calculate the total mass is fundamental, as it influences the center of mass velocity and the system's overall response to external forces.

Importance in Center of Mass Calculations

When considering the motion of multiple bodies as a single entity, the total mass becomes integral in calculating the center of mass. It acts as a denominator when determining the center of mass velocity, a concept merging the individual velocities of particles through the lens of the entire system's mass.
Velocity of a Particle
The velocity of a particle is a vector quantity, indicating not just how fast the particle is moving but also in which direction. For each particle in a given system, velocity can be represented by components in multiple dimensions; in our case, along the horizontal (\(i\)) and vertical (\(j\)) axes.

To express this, we use vector notation, such as \(2.00 \hat{\mathbf{i}} - 3.00 \hat{\mathbf{j}}\) m/s for the first particle. This indicates that the particle is moving at 2.00 m/s in the horizontal direction and 3.00 m/s in the opposite vertical direction.

Relevance to Center of Mass

Each particle's velocity contributes to the velocity of the center of mass for the complete system. By calculating the momentum of each particle, which inherently includes velocity, and synthesizing these values, we derive the center of mass velocity. This overall velocity is crucial as it represents the movement of the mass system's center as if all mass were concentrated at that point, reflecting the system's motion through space.

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Most popular questions from this chapter

A glider of mass \(m\) is free to slide along a horizontal air track. It is pushed against a launcher at one end of the track. Model the launcher as a light spring of force constant \(k\) compressed by a distance \(x\) . The glider is released from rest. (a) Show that the glider attains a speed of \(v=x(k / m)^{1 / 2} .\) (b) Does a glider of large or of small mass attain a greater speed? (c) Show that the impulse imparted to the glider is given by the expression \(x(k m)^{1 / 2} .\) (d) Is a greater impulse injected into a large or a small mass? (e) Is more work done on a large or a small mass?

A neutron in a nuclear reactor makes an elastic head-on collision with the nucleus of a carbon atom initially at rest. (a) What fraction of the neutron's kinetic energy is transferred to the carbon nucleus? (b) If the initial kinetic energy of the neutron is \(1.60 \times 10^{-13} \mathrm{J}\) , find its final kinetic energy and the kinetic energy of the carbon nucleus after the collision. (The mass of the carbon nucleus is nearly 12.0 times the mass of the neutron.)

Review problem. There are (one can say) three coequal theories of motion: Newton's second law, stating that the total force on an object causes its acceleration; the work-kinetic energy theorem, stating that the total work on an object causes its change in kinetic energy; and the impulse-momentum theorem, stating that the total impulse on an object causes its change in momentum. In this problem, you compare predictions of the three theories in one particular case. A \(3.00-\mathrm{kg}\) object has velocity 7.00\(\hat{\mathrm{j}} \mathrm{m} / \mathrm{s}\) . Then, a total force 12.0\(\hat{\mathrm{i}} \mathrm{N}\) acts on the object for 5.00 \(\mathrm{s}\) . (a) Calculate the object's final velocity, using the impulse-momentum theorem. (b) Calculate its acceleration from \(\mathbf{a}=\left(\mathbf{v}_{f}-\mathbf{v}_{i}\right) / \Delta t\) (c) Calculate its acceleration from \(\mathbf{a}=\Sigma \mathbf{F} / m .\) (d) Find the object's vector displacement from \(\Delta \mathbf{r}=\mathbf{v}_{i} t+\frac{1}{2} \mathbf{a} t^{2}\) (e) Find the work done on the object from \(W=\mathbf{F} \cdot \Delta \mathbf{r} .\) (f) Find the final kinetic energy from \(\frac{1}{2} m v_{f}^{2}=\frac{1}{2} m \mathbf{v}_{f} \cdot \mathbf{v}_{f}\) (g) Find the final kinetic energy from \(\frac{1}{2} m v_{i}^{2}+W .\)

Rocket Science. A rocket has total mass \(M_{i}=360 \mathrm{kg}\) , including 330 \(\mathrm{kg}\) of fuel and oxidizer. In interstellar space it starts from rest, turns on its engine at time \(t=0\) , and puts out exhaust with relative speed \(v_{e}=1500 \mathrm{m} / \mathrm{s}\) at the constant rate \(k=2.50 \mathrm{kg} / \mathrm{s}\) . The fuel will last for an actual burn time of \(330 \mathrm{kg} /(2.5 \mathrm{kg} / \mathrm{s})=132 \mathrm{s},\) but define a "projected depletion time" as \(T_{p}=M_{i} / k=\) 144 s. (This would be the burn time if the rocket could use its payload and fuel tanks as fuel, and even the walls of the combustion chamber.) (a) Show that during the burn the velocity of the rocket is given as a function of time by $$ v(t)=-v_{e} \ln \left[1-\left(t / T_{p}\right)\right] $$ (b) Make a graph of the velocity of the rocket as a function of time for times running from 0 to 132 s. (c) Show that the acceleration of the rocket is $$ a(t)=v_{e} /\left(T_{p}-t\right) $$ (d) Graph the acceleration as a function of time. (e) Show that the position of the rocket is $$ x(t)=v_{e}\left(T_{p}-t\right) \ln \left[1-\left(t / T_{p}\right)\right]+v_{e} t $$ (f) Graph the position during the burn.

High-speed stroboscopic photographs show that the head of a golf club of mass 200 \(\mathrm{g}\) is traveling at 55.0 \(\mathrm{m} / \mathrm{s}\) just before it strikes a \(46.0-\mathrm{g}\) golf ball at rest on a tee. After the collision, the club head travels (in the same direction) at 40.0 \(\mathrm{m} / \mathrm{s}\) . Find the speed of the golf ball just after impact.
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