(a) Show that m2 (vi)1 – ÜCM Au, mi + m2 and -Aŭ. m1 + m2 That is, m2Aï/(m1 + m2) represents the velocity of particle 1 relative to the centre of (üi)2 – UcM mass, and similarly for particle 2. (b) Starting from P, = m1(j)1 + m2(0,)2, and P, = m1(ü,)ı + m2(ūs)2, show that the total momentum is P = P; = (m1 + m2)ūcM- That is, the uCM terms carry the total momentum, while the momenta of the Au terms add up to zero. (c) Starting from K = }m(v;){ + }m2(v.)3, show that the total kinetic energy of the system can be written: m2 Av mi +m2. m1 Av mi + m2 1 1 K = ; (m, + m2)vồm +m 5m2
(a) Show that m2 (vi)1 – ÜCM Au, mi + m2 and -Aŭ. m1 + m2 That is, m2Aï/(m1 + m2) represents the velocity of particle 1 relative to the centre of (üi)2 – UcM mass, and similarly for particle 2. (b) Starting from P, = m1(j)1 + m2(0,)2, and P, = m1(ü,)ı + m2(ūs)2, show that the total momentum is P = P; = (m1 + m2)ūcM- That is, the uCM terms carry the total momentum, while the momenta of the Au terms add up to zero. (c) Starting from K = }m(v;){ + }m2(v.)3, show that the total kinetic energy of the system can be written: m2 Av mi +m2. m1 Av mi + m2 1 1 K = ; (m, + m2)vồm +m 5m2
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Transcribed Image Text:and then writing the velocities of particles 1 and 2 before and after the collision as
m2
(T:)2 = JCM
-Aŭ;
mị + m2
= ÜCM +
and
mi + m2
m2
m1
(7s)2 = ŪCM +
-Aī.
mi + m2
and
=
mi + m2
(a) Show that
m2
(v)1 – ŪCM
Aī,
mị + m2
and
(v)2 – ÜCM
m1 + m2
That is, m2Aŭ/(m + m2) represents the velocity of particle 1 relative to the centre of
mass, and similarly for particle 2.
(b) Starting from P = m1(v)1 + m2(ū)2, and P; = m1(üs)1 + m2(T;)2, show that the
total momentum is
P = P; = (m1 + m2)īcm-
That is, the ücM terms carry the total momentum, while the momenta of the Au terms
add up to zero.
(c) Starting from K = }m(v;){ + }m2(v;);, show that the total kinetic energy of the
system can be written:
m2 Av \?, 1
+
m Δυ
(m, + m2.
1
1
K = ;(m, + m2)vćm + 2m1 (m1 + m2,
That is, the kinetic energy separates into a part associated with the centre of mass, and
another part that represents motion relative to the centre of mass, with no cross-terms
involving ucMAJ.
(d) Consider now a (partly) inelastic collision, in which the final velocities are (ū;)ı and
(üs)2. Show that the minimum kinetic energy of the particles after the collision is
1
K = (m, + m2)vCM:
For this minimum kinetic energy, What is the velocity of the particles relative to each
other after the collision, and what kind of collision is this?
Transcribed Image Text:1. In class in Lectures 19 and 20 we treated elastic collisions by first calculating the quantities
m1(T;)1+ m2(T;)2
and
Au = (7;)1 – (T;)2,
mi + m2
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