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Consider a head-on elastic collision between two particles. (since the collision is head-on, the motion is confined to a single straight line and is therefore one-dimensional.) Prove that the relative velocity after the collision is equal and opposite to that before. That is, \(v_{1}-v_{2}=-\left(v_{1}^{\prime}-v_{2}^{\prime}\right),\) where \(v_{1}\) and \(v_{2}\) are the initial velocities and \(v_{1}^{\prime}\) and \(v_{2}^{\prime}\) the corresponding final velocities.

Short Answer

Expert verified
The relative velocity after is equal and opposite to that before: \(v_1-v_2=-\left(v_1'-v_2'\right).\)

Step by step solution

01

Understand Elastic Collision

In an elastic collision, both momentum and kinetic energy are conserved. We will use these properties to solve the exercise.
02

Write Momentum Conservation Equation

The total initial momentum is equal to the total final momentum: \( m_1 v_1 + m_2 v_2 = m_1 v_1' + m_2 v_2' \).
03

Write Kinetic Energy Conservation Equation

The total initial kinetic energy equals the total final kinetic energy: \( \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 = \frac{1}{2} m_1 v_1'^2 + \frac{1}{2} m_2 v_2'^2 \).
04

Simplify Kinetic Energy Equation

Cancel out the \( \frac{1}{2} \) terms: \( m_1 v_1^2 + m_2 v_2^2 = m_1 v_1'^2 + m_2 v_2'^2 \).
05

Derive Relative Velocity Equation

From the momentum equation, express \(v_1'\) in terms of \(v_2'\) by isolating terms: \[ v_1 - v_1' = - (v_2 - v_2') \]. This simplifies to the relative velocity equation: \[ v_1 - v_2 = - (v_1' - v_2') \].

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Key Concepts

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

Momentum Conservation
In an elastic collision, momentum conservation is a fundamental principle that ensures the total momentum of the system remains constant before and after the collision. Momentum (\( p \)) is defined as the product of mass (\( m \)) and velocity (\( v \)), and this principle can be mathematically represented for two colliding objects as:
  • Initial Total Momentum: \( m_1 v_1 + m_2 v_2 \)
  • Final Total Momentum: \( m_1 v_1' + m_2 v_2' \)
By conserving momentum, we can set these two expressions equal:\[m_1 v_1 + m_2 v_2 = m_1 v_1' + m_2 v_2'\]This equation helps us understand how the individual velocities of the masses change during the collision while keeping the total momentum constant. It's crucial in predicting the outcome of the collision for both velocities and determining the final state of the system after an elastic collision.
Momentum conservation is foundational in physics because it applies to all systems, regardless of energy changes. This principle allows us to solve complex collision problems by keeping track of linear momentum.
Kinetic Energy Conservation
Kinetic energy conservation is another key aspect of elastic collisions. Unlike inelastic collisions, elastic collisions maintain both the total kinetic energy and momentum. Kinetic energy (\( KE \)) is related to the motion of an object and is given by the formula:
  • Initial Kinetic Energy: \( \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 \)
  • Final Kinetic Energy: \( \frac{1}{2} m_1 v_1'^2 + \frac{1}{2} m_2 v_2'^2 \)
The principle of kinetic energy conservation tells us that in the absence of external forces, the kinetic energies before and after the collision must be equal:\[\frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 = \frac{1}{2} m_1 v_1'^2 + \frac{1}{2} m_2 v_2'^2\]Canceling out the \( \frac{1}{2} \) terms simplifies our calculations, giving us:\[m_1 v_1^2 + m_2 v_2^2 = m_1 v_1'^2 + m_2 v_2'^2\]This equation confirms that the energy used in the collision does not disappear but is instead transferred either within each object or from one object to another.
Understanding kinetic energy conservation enables us to delve deeper into how energy dynamics are maintained and distributed among the colliding bodies during elastic collisions.
Relative Velocity
Relative velocity is a concept that helps us understand the velocity of one object in relation to another. In elastic collisions, a particularly interesting property emerges: the relative velocity of two colliding objects post-collision is equal and opposite to their relative velocity pre-collision. Now, let's break it down:
  • Initially, the relative velocity before collision: \( v_1 - v_2 \)
  • Finally, the relative velocity after collision: \( v_1' - v_2' \)
When you hear that these velocities are equal and opposite, it means:\[v_1 - v_2 = -(v_1' - v_2')\]This relationship is derived from the conservation laws of both momentum and kinetic energy.
Why is this important? It tells us how the velocities shift in perspective to each other after the collision and reaffirms that energy and momentum are both internally consistent and behave predictably under the conservation laws during elastic collisions.
By grasping this relative change in velocities, students can better predict how objects interact in a collision and provide insight into the effectiveness of conservation laws in predicting outcomes.

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Most popular questions from this chapter

The proof that the condition \(\nabla \times \mathbf{F}=0\) guarantees the path independence of the work \(\int_{1}^{2} \mathbf{F} \cdot d \mathbf{r}\) done by \(\mathbf{F}\) is unfortunately too lengthy to be included here. However, the following three exercises capture the main points: \(^{16}\) (a) Show that the path independence of \(\int_{1}^{2} \mathbf{F} \cdot d \mathbf{r}\) is equivalent to the statement that the integral \(\oint_{\mathrm{T}} \mathbf{F} \cdot d \mathbf{r}\) around any closed path \(\Gamma\) is zero. (By tradition, the symbol \(\oint\) is used for integrals around a closed path \(-\) a path that starts and stops at the same point.) [Hint: For any two points 1 and 2 and any two paths from 1 to 2 , consider the work done by \(\mathbf{F}\) going from 1 to 2 along the first path and then back to 1 along the second in the reverse direction. \((\) b) Stokes's theorem asserts that \(\oint_{\mathrm{T}} \mathbf{F} \cdot d \mathbf{r}=\int(\nabla \times \mathbf{F}) \cdot \hat{\mathbf{n}} d A,\) where the integral on the right is a surface integral over a surface for which the path \(\Gamma\) is the boundary, and \(\hat{\mathbf{n}}\) and \(d A\) are a unit normal to the surface and an element of area. Show that Stokes's theorem implies that if \(\nabla \times \mathbf{F}=0\) everywhere, then \(\oint_{\mathrm{T}} \mathbf{F} \cdot d \mathbf{r}=0 .\) (c) While the general proof of Stokes's theorem is beyond our scope here, the following special case is quite easy to prove (and is an important step toward the general proof): Let \(\Gamma\) denote a rectangular closed path lying in a plane perpendicular to the \(z\) direction and bounded by the lines \(x=B, x=B+b, y=C\) and \(y=C+c .\) For this simple path (traced counterclockwise as seen from above), prove Strokes's theorem that \(\oint_{\Gamma} \mathbf{F} \cdot d \mathbf{r}=\int(\mathbf{\nabla} \times \mathbf{F}) \cdot \hat{\mathbf{n}} d A\) where \(\hat{\mathbf{n}}=\hat{\mathbf{z}}\) and the integral on the right runs over the flat, rectangular area inside \(\Gamma\). [Hint: The integral on the left contains four terms, two of which are integrals over \(x\) and two over \(y\). If you pair them in this way, you can combine each pair into a single integral with an integrand of the form \(F_{x}(x, C+c, z)-F_{x}(x, C, z)\) (or a similar term with the roles of \(x\) and \(y\) exchanged). You can rewrite this integrand as an integral over \(y\) of \(\partial F_{x}(x, y, z) / \partial y\) (and similarly with the other term), and you're home.]

A charge \(q\) in a uniform electric field \(\mathbf{E}_{0}\) experiences a constant force \(\mathbf{F}=q \mathbf{E}_{0}\). (a) Show that this force is conservative and verify that the potential energy of the charge at position \(\mathbf{r}\) is \(U(\mathbf{r})=-q \mathbf{E}_{0} \cdot \mathbf{r}\). (b) By doing the necessary derivatives, check that \(\mathbf{F}=-\nabla U\).

Consider a small frictionless puck perched at the top of a fixed sphere of radius \(R\). If the puck is given a tiny nudge so that it begins to slide down, through what vertical height will it descend before it leaves the surface of the sphere? [Hint: Use conservation of energy to find the puck's speed as a function of its height, then use Newton's second law to find the normal force of the sphere on the puck. At what value of this normal force does the puck leave the sphere?]

A mass \(m\) moves in a circular orbit (centered ón the origin) in the field of an attractive central force with potential energy \(U=k r^{n} .\) Prove the virial theorem that \(T=n U / 2\).

Find the curl, \(\nabla \times \mathbf{F},\) for the following forces: \((\mathbf{a}) \mathbf{F}=k \mathbf{r} ;(\mathbf{b}) \mathbf{F}=\left(A x, B y^{2}, C z^{3}\right) ;(\mathbf{c}) \mathbf{F}=\) \(\left(A y^{2}, B x, C z\right),\) where \(A, B, C\) and \(k\) are constants.

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