Consider two pool balls constrained to move along the x-axis. There are two obse

Question

Consider two pool balls constrained to move along the x-axis. There are two observers watching these balls as they collide and fly apart from each other. One observer is at rest in frame S while the other observer is moving at a constant speed in frame S 0 . The balls have masses m1 and m2, initial velocities u0-1i and u0-2i , and final velocities u0-1f and u0-2f as measured from the moving observer. Show that the conservation of momentum holds under the Galilean transformations (i.e. is invariant under the transformation). In other words, conservation of momentum works in both reference frames. (Hint: Start with p0i = p0f)

Explanation / Answer

http://www.newagepublishers.com/samplechapter/000485.pdf

Suppose the two balls moving with initial velocities u-1i and u2-i , are so far apart that they do not possess any potential energy of interaction. Then their total energy of the system of two particles is only the total kinetic energies of the particles.

=1/2m1(u-1i)2+1/2m2(u-2i)2

After they collide and separate far apart ,according to law of conservation of energy in frame S, we get

1/2m1(u-1i)2+1/2m2(u-21)2=1/2m1(u-1f)2+1/2m2(u-2f)2+dE.....(1)

where dE is change in in the change (loss or gain) in internal energy of the particles during collision.

If dE= 0, collision is elastic; otherwise inelastic.

Now let us view the process from another inertial frameS0 moving with velocity V relative to S

. According to Galilean transformation, the velocities of particles as observed from S0 are u0-1i=u-1i-V and u0-2i = u-2i-V before collision .....(A)

& u0-1f=u-1f-V and u0-2f=u-2f-V after collision.......(B)

Now we introduce the idea that principle of conservation of energy is frame independ-

ent. In fact, conservation of mechanical energy being a consequence of homogeneity of

time, is fundamentally an invariant principle in Newtonian mechanics. As shown by experiments, we further assume that internal energy change dE also remains invariant under Galilean transformations.

Hence, conservation of energy in collision process as observed in S0 gives

1/2m1(u0-1i)2+1/2m2(u0-21)2=1/2m1(u0-1f)2+1/2m2(u0-2f)2+dE.....(2)

Using (A) and (B) we get (u0-1i)2= (u-1i)2 +V2-2(u-1i).V similarly subsituting for for (u0-21)2, (u0-1f)2, (u0-2f)2 we get

1/2m1(u-1i)2-(m1.(u-1i) +m2(u-2i))V+1/2m2(u-21)2=

1/2m1(u-1f)2+1/2m2(u-2f)2-(m1.(u-1f) +m2(u-2f)).V+dE

Comparing with equation (1) we get m1.(u-1i)+m2.(u-2i)= m1.(u-1f)+m2.(u-2f)

which is the law of conservation of momentum in collision process. Thus, the principle of

conservation of energy and the concept of Galilean invariance shows that momentum

remains conserved during collision, whether elastic or inelastic

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.