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A collision occurs between a \(2.00 \mathrm{~kg}\) particle traveling with velocity \(\vec{v}_{1}=(-4.00 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{i}}+(-5.00 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{j}}\) and a \(4.00 \mathrm{~kg}\) particle traveling with velocity \(\vec{v}_{2}=(6.00 \mathrm{~m} / \mathrm{s}) \mathrm{i}+(-2.00 \mathrm{~m} / \mathrm{s}) \mathrm{j} .\) The colli- sion connects the two particles. What then is their velocity in (a) unit-vector notation and as a (b) magnitude and (c) angle?

Short Answer

Expert verified
Final velocity: (2.67 m/s)\hat{i} - (3.00 m/s)\hat{j}, Magnitude: 4.02 m/s, Angle: 311.63°.

Step by step solution

01

Understand the System

This is a collision problem where two particles stick together after the collision, which means they undergo an inelastic collision. We need to find their final velocity.
02

Initial Momentum Calculation

Calculate the initial momentum of each particle. For the first particle: \( \vec{p}_1 = m_1 \vec{v}_1 = 2.00 \times [(-4.00) \hat{i} + (-5.00) \hat{j}] = (-8.00 \hat{i} - 10.00 \hat{j}) \, \text{kg m/s} \).For the second particle: \( \vec{p}_2 = m_2 \vec{v}_2 = 4.00 \times [(6.00) \hat{i} + (-2.00) \hat{j}] = (24.00 \hat{i} - 8.00 \hat{j}) \, \text{kg m/s} \).
03

Total Initial Momentum

Add the initial momenta of the two particles to find the total initial momentum:\( \vec{p}_{\text{total}} = \vec{p}_1 + \vec{p}_2 = (-8.00 \hat{i} - 10.00 \hat{j}) + (24.00 \hat{i} - 8.00 \hat{j}) = (16.00 \hat{i} - 18.00 \hat{j}) \, \text{kg m/s} \).
04

Total Mass of the System

Find the total mass of the system, which is the sum of the masses of the two particles:\( m_{\text{total}} = m_1 + m_2 = 2.00 + 4.00 = 6.00 \, \text{kg} \).
05

Final Velocity in Unit Vector Notation

Use conservation of momentum to find the final velocity of the system. \( \vec{v}_{\text{final}} = \frac{\vec{p}_{\text{total}}}{m_{\text{total}}} = \frac{(16.00 \hat{i} - 18.00 \hat{j})}{6.00} = (2.67 \hat{i} - 3.00 \hat{j}) \, \text{m/s} \).
06

Find Magnitude of Final Velocity

Calculate the magnitude of the final velocity using the Pythagorean theorem:\( \| \vec{v}_{\text{final}} \| = \sqrt{(2.67)^2 + (-3.00)^2} = \sqrt{7.13 + 9.00} = \sqrt{16.13} \approx 4.02 \, \text{m/s} \).
07

Calculate the Angle of the Final Velocity

Find the angle \( \theta \) that the velocity vector makes with the positive x-axis. Use the arctangent function:\( \theta = \tan^{-1}\left(\frac{-3.00}{2.67}\right) \approx \tan^{-1}(-1.12) \approx -48.37^\circ \). Since the velocity vector is in the fourth quadrant, convert it to positive angle reference by adding 360 degrees, giving \( \theta \approx 311.63^\circ \).

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Key Concepts

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

Momentum Conservation
In physics, momentum conservation is a fundamental concept crucial to understanding collisions. Momentum is defined as the product of an object's mass and its velocity. An inelastic collision, like the one described in the exercise, is a type of collision where the colliding objects stick together post-collision. While kinetic energy is not conserved, momentum is conserved. This means the total momentum of a system before collision equals the total momentum after collision.

To conserve momentum in our exercise, we begin by calculating the initial momentum of each particle. The momentum of the first particle is calculated using its mass and velocity vector. Similarly, the momentum of the second particle is calculated. Total initial momentum is then determined by summing these two momentum vectors. By understanding and applying this key concept, we can deduce the final momentum of the combined particle system, which directly leads to discovering the final velocity of these particles once they unite.
Final Velocity Calculation
Final velocity calculation is the process of determining the velocity of two combined masses after a collision has occurred. In the context of our exercise, after we have conserved the momentum, the next step is to find the velocity of the merged particles.

Using the previously calculated total initial momentum and the total combined mass of the two particles, the final velocity \[\vec{v}_{\text{final}} = \frac{\vec{p}_{\text{total}}}{m_{\text{total}}}\] is determined. This equation expresses that the velocity of the combined particles after collision is the ratio of total momentum to total mass. By dividing the momentum vector components by the total mass, we obtain the final velocity in unit vector notation.
  • The \(i\)-component of velocity: \[2.67 \, \text{m/s}\]
  • The \(j\)-component of velocity: \[-3.00 \, \text{m/s}\]
These components give the final direction of the velocity, representing how the combined masses move post-collision.
Vector Components
Understanding vector components is essential for solving problems involving direction and magnitude like inelastic collisions. Vectors have two main characteristics: magnitude and direction, both of which are crucial in physics.

In the exercise, we describe velocities using unit vector notation, where \(\hat{i}\) and \(\hat{j}\) represent directions along the x and y axes, respectively. Each particle's velocity has been split into these components. Calculating the changes in each component due to the collision allows us to then construct the final velocity vector. To find the magnitude of this vector, we use the Pythagorean theorem:
\[\| \vec{v}_{\text{final}} \| = \sqrt{(2.67)^2 + (-3.00)^2}\]
This gives us the speed of the combined particles.

Furthermore, vector direction is described using an angle. We calculated this angle \(\theta\) by determining the arctangent of the ratio between the \(j\)-component and the \(i\)-component. This angle describes the orientation of our resultant vector relative to the x-axis. Recognizing the relationship between vector components, magnitude, and direction is key in collision analysis and resolving complex physics problems.

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Most popular questions from this chapter

An atomic nucleus at rest at the origin of an \(x y\) coordinate system transforms into three particles. Particle \(1, \operatorname{mass} 16.7 \times 10^{-27}\) kg, moves away from the origin at velocity \(\left(6.00 \times 10^{6} \mathrm{~m} / \mathrm{s}\right) \hat{\text { is }}\), particle 2, mass \(8.35 \times 10^{-27} \mathrm{~kg}\), moves away at velocity \(\left(-8.00 \times 10^{6} \mathrm{~m} / \mathrm{s}\right)\) ) (a) In unit-vector notation, what is the linear momentum of the third particle, mass \(11.7 \times 10^{-27} \mathrm{~kg}\) ? (b) How much kinetic energy appears in this transformation?

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A \(20.0 \mathrm{~kg}\) body is moving through space in the positive dircction of an \(x\) axis with a speed of \(200 \mathrm{~m} / \mathrm{s}\) when, due to an internal explosion, it breaks into three parts. One part, with a mass of \(10.0 \mathrm{~kg}\), moves away from the point of explosion with a speed of \(100 \mathrm{~m} / \mathrm{s}\) in the positive \(y\) direction. A second part, with a mass of \(4.00 \mathrm{~kg}\), moves in the negative \(x\) direction with a speed of \(500 \mathrm{~m} / \mathrm{s}\). (a) In unit-vector notation, what is the velocity of the third part? (b) How much cnergy is released in the explosion? Ignore effects due to the gravitational force.

Particle \(A\) and particle \(B\) are held together with a com- pressed spring between them. When they are released, the spring pushes them apart, and they then fly off in opposite directions, free of the spring. The mass of \(A\) is 2.00 times the mass of \(B,\) and the energy stored in the spring was 60 J. Assume that the spring has negligible mass and that all its stored energy is transferred to the particles. Once that transfer is complete, what are the kinetic energics of (a) particle \(A\) and (b) particle \(B ?\)

Two particles are launched from the origin of the coordinate system at time \(t=0 .\) Particle 1 of mass \(m_{1}=5.00 \mathrm{~g}\) is shot directly along the \(x\) axis on a frictionless floor, with constant speed \(10.0 \mathrm{~m} / \mathrm{s} .\) Particle 2 of mass \(m_{2}=3.00 \mathrm{~g}\) is shot with a velocity of magnitude \(20.0 \mathrm{~m} / \mathrm{s},\) at an upward angle such that it always stays directly above particle \(1 .\) (a) What is the maximum height \(H_{\max }\) reached by the com of the two-particle system? In unit- vector notation, what are the (b) velocity and (c) acceleration of the com when the com reaches \(H_{\max } ?\)

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