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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

Which of the following forces is conservative? (a) \(\mathbf{F}=k(x, 2 y, 3 z)\) where \(k\) is a constant. (b) \(\mathbf{F}=k(y, x, 0) .\) (c) \(\mathbf{F}=k(-y, x, 0)\). For those which are conservative, find the corresponding potential energy \(U,\) and verify by direct differentiation that \(\mathbf{F}=-\nabla U\).

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.

Near to the point where I am standing on the surface of Planet \(X\), the gravitational force on a mass \(m\) is vertically down but has magnitude \(m \gamma y^{2}\) where \(\gamma\) is a constant and \(y\) is the mass's height above the horizontal ground. (a) Find the work done by gravity on a mass \(m\) moving from \(\mathbf{r}_{1}\) to \(\mathbf{r}_{2}\), and use your answer to show that gravity on Planet \(X,\) although most unusual, is still conservative. Find the corresponding potential energy. (b) Still on the same planet, I thread a bead on a curved, frictionless, rigid wire, which extends from ground level to a height \(h\) above the ground. Show clearly in a picture the forces on the bead when it is somewhere on the wire. (Just name the forces so it's clear what they are; don't worry about their magnitude.) Which of the forces are conservative and which are not? (c) If I release the bead from rest at a height \(h\), how fast will it be going when it reaches the ground?

In one dimension, it is obvious that a force obeying Hooke's law is conservative (since \(F=-k x\) depends only on the position \(x,\) and this is sufficient to guarantee that \(F\) is conservative in one dimension). Consider instead a spring that obeys Hooke's law and has one end fixed at the origin, but whose other end is free to move in all three dimensions. (The spring could be fastened to a point in the ceiling and be supporting a bouncing mass \(m\) at its other end, for instance.) Write down the force \(\mathbf{F}(\mathbf{r})\) exerted by the spring in terms of its length \(r\) and its equilibrium length \(r_{\mathrm{o}} .\) Prove that this force is conservative. [Hints: Is the force central? Assume that the spring does not bend.]

(a) Consider a mass \(m\) in a uniform gravitational field \(\mathbf{g},\) so that the force on \(m\) is \(m \mathbf{g},\) where \(\mathbf{g}\) is a constant vector pointing vertically down. If the mass moves by an arbitrary path from point 1 to point \(2,\) show that the work done by gravity is \(W_{\mathrm{grav}}(1 \rightarrow 2)=-m g h\) where \(h\) is the vertical height gained between points 1 and 2. Use this result to prove that the force of gravity is conservative (at least in a region small enough so that \(\mathrm{g}\) can be considered constant). (b) Show that, if we choose axes with \(y\) measured vertically up, the gravitational potential energy is \(U=m g y\) (if we choose \(U=0\) at the origin).

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