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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, v1v2=(v1v2), where v1 and v2 are the initial velocities and v1 and v2 the corresponding final velocities.

Short Answer

Expert verified
The relative velocity after is equal and opposite to that before: v1v2=(v1v2).

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: m1v1+m2v2=m1v1+m2v2.
03

Write Kinetic Energy Conservation Equation

The total initial kinetic energy equals the total final kinetic energy: 12m1v12+12m2v22=12m1v12+12m2v22.
04

Simplify Kinetic Energy Equation

Cancel out the 12 terms: m1v12+m2v22=m1v12+m2v22.
05

Derive Relative Velocity Equation

From the momentum equation, express v1 in terms of v2 by isolating terms: v1v1=(v2v2). This simplifies to the relative velocity equation: v1v2=(v1v2).

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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: m1v1+m2v2
  • Final Total Momentum: m1v1+m2v2
By conserving momentum, we can set these two expressions equal:m1v1+m2v2=m1v1+m2v2This 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: 12m1v12+12m2v22
  • Final Kinetic Energy: 12m1v12+12m2v22
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:12m1v12+12m2v22=12m1v12+12m2v22Canceling out the 12 terms simplifies our calculations, giving us:m1v12+m2v22=m1v12+m2v22This 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: v1v2
  • Finally, the relative velocity after collision: v1v2
When you hear that these velocities are equal and opposite, it means:v1v2=(v1v2)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

(a) The force exerted by a one-dimensional spring, fixed at one end, is F=kx, where x is the displacement of the other end from its equilibrium position. Assuming that this force is conservative (which it is) show that the corresponding potential energy is U=12kx2, if we choose U to be zero at the equilibrium position. (b) Suppose that this spring is hung vertically from the ceiling with a mass m suspended from the other end and constrained to move in the vertical direction only. Find the extension xo of the new equilibrium position with the suspended mass. Show that the total potential energy (spring plus gravity) has the same form 12ky2 if we use the coordinate y equal to the displacement measured from the new equilibrium position at x=xo (and redefine our reference point so that U=0 at y=0 ).

In one dimension, it is obvious that a force obeying Hooke's law is conservative (since F=kx 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 F(r) exerted by the spring in terms of its length r and its equilibrium length ro. Prove that this force is conservative. [Hints: Is the force central? Assume that the spring does not bend.]

A mass m is in a uniform gravitational field, which exerts the usual force F=mg vertically down, but with g varying with time, g=g(t). Choosing axes with y measured vertically up and defining U=mgy as usual, show that F=U as usual, but, by differentiating E=12mv2+U with respect to t, show that E is not conserved.

The proof that the condition ×F=0 guarantees the path independence of the work 12Fdr done by 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 12Fdr is equivalent to the statement that the integral TFdr around any closed path Γ is zero. (By tradition, the symbol 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 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 TFdr=(×F)n^dA, where the integral on the right is a surface integral over a surface for which the path Γ is the boundary, and n^ and dA are a unit normal to the surface and an element of area. Show that Stokes's theorem implies that if ×F=0 everywhere, then TFdr=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 Γ 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 ΓFdr=(×F)n^dA where n^=z^ and the integral on the right runs over the flat, rectangular area inside Γ. [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 Fx(x,C+c,z)Fx(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 Fx(x,y,z)/y (and similarly with the other term), and you're home.]

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 mγy2 where γ 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 r1 to r2, 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?

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