Introduction
In physics we study motion (change in position).At the same time, we try to discover physical quantities, which do not change in a physical process. The laws of momentum and energy conservation are typical examples.In this section we shall apply these laws to a commonly encountered phenomena, namely collisions.Several games such as billiards, marbles or carrom involve collisions.We shall study the collision of two masses in an idealised form.
Consider two masses m1 and m2.The particle m3 is moving with speed v1i , the subscript ‘i’ implying initial.We can cosider m2 to be at rest.No loss of generality is involved in making such a selection.In this situation the mass m1 collides with the stationary mass m2 and this is depicted in Fig. 6.10.
fig.6.10.Collision of mass m1, with a stationary mass m2
The masses m1 and m2 fly-off in different directions.We shall see that there are relationships,which connect the masses,the velocities and the angles.
Elastic and Inelastic Collisions
In all collisions the total linear momentum is conserved; the initial momentum of the system is equal to the final momentum of the system.One can argue this as follows. When two objects collide, the mutual impulsive forces acting over the collision time \(\Delta\)t cause a change in their respective momenta :
\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacqGHuo % arcaWGWbWaaSbaaSqaaiaaigdaaeqaaOGaeyypa0JaamOramaaBaaa % leaacaaIXaGaaGOmaaqabaGccqGHuoarcaWG0baabaGaeyiLdqKaam % iCamaaBaaaleaacaaIYaaabeaakiabg2da9iaadAeadaWgaaWcbaGa % aGOmaiaaigdaaeqaaOGaeyiLdqKaamiDaaaaaa!4854! \begin{array}{l} \Delta {p_1} = {F_{12}}\Delta t\\ \Delta {p_2} = {F_{21}}\Delta t \end{array}\)
where F12 is the force exerted on the first particle by the second particle. F21 is likewise the force exerted on the second particle by the first particle. Now from Newton’s third law, F12 = - F21. This implies
\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyiLdqKaam % iCamaaBaaaleaacaaIXaaabeaakiabgUcaRiabgs5aejaadchadaWg % aaWcbaGaaGOmaaqabaGccqGH9aqpcaaIWaaaaa!3F33! \Delta {p_1} + \Delta {p_2} = 0\)
The above conclusion is true even though the forces vary in a complex fashion during the collision time \(\Delta\)t. Since the third law is true at every instant, the total impulse on the first object is equal and opposite to that on the second.
On the other hand, the total kinetic energy of the system is not necessarily conserved.The impact and deformation during collision may generate heat and sound.Part of the initial kinetic energy is transformed into other forms of energy.A useful way to visualise the deformation during collision is in terms of a ‘compressed spring’. If the ‘spring’ connecting the two masses regains its original shape without loss in energy, then the initial kinetic energy is equal to the final kinetic energy but the kinetic energy during the collision time \(\Delta\)t is not constant.Such a collision is called an elastic collision.On the other hand the deformation may not be relieved and the two bodies could move together after the collision.A collision in which the two particles move together after the collision is called a completely inelastic collision.The intermediate case where the deformation is partly relieved and some of the initial kinetic energy is lost is more common and is appropriately called an inelastic collision.
Collisions in One Dimension
Consider first a completely inelastic collision in one dimension. Then, in Fig. 6.10,
\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacqaH4o % qCdaWgaaWcbaGaaGymaaqabaGccqGHsislcqaH4oqCdaWgaaWcbaGa % aGOmaaqabaGccqGH9aqpcaaIWaaabaGaamyBamaaBaaaleaacaaIXa % aabeaakiaadAhadaWgaaWcbaGaaGymaiaadMgaaeqaaOGaeyypa0Za % aeWaaeaacaWGTbWaaSbaaSqaaiaaigdaaeqaaOGaey4kaSIaamyBam % aaBaaaleaacaaIYaaabeaaaOGaayjkaiaawMcaaiaadAhadaWgaaWc % baGaamOzaaqabaGcdaqadaqaaiaad2gacaWGVbGaamyBaiaadwgaca % WGUbGaamiDaiaadwhacaWGTbGaam4yaiaad+gacaWGUbGaam4Caiaa % dwgacaWGYbGaamODaiaadggacaWG0bGaamyAaiaad+gacaWGUbaaca % GLOaGaayzkaaaabaGaamODamaaBaaaleaacaWGMbaabeaakiabg2da % 9maalaaabaGaamyBamaaBaaaleaacaaIXaaabeaaaOqaaiaad2gada % WgaaWcbaGaaGymaaqabaGccqGHRaWkcaWGTbWaaSbaaSqaaiaaikda % aeqaaaaakiaadAhadaWgaaWcbaGaaGymaiaadMgaaeqaaOWaaeWaae % aacaaI2aGaaiOlaiaaikdacaaIZaaacaGLOaGaayzkaaaaaaa!7191! \begin{array}{l} {\theta _1} - {\theta _2} = 0\\ {m_1}{v_{1i}} = \left( {{m_1} + {m_2}} \right){v_f}\left( {momentumconservation} \right)\\ {v_f} = \frac{{{m_1}}}{{{m_1} + {m_2}}}{v_{1i}}\left( {6.23} \right) \end{array}\)
The loss in kinetic energy on collision is
\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacqGHuo % arcaWGlbGaeyypa0ZaaSaaaeaacaaIXaaabaGaaGOmaaaacaWGTbWa % aSbaaSqaaiaaigdaaeqaaOGaamODamaaDaaaleaacaaIXaGaamyAaa % qaaiaaikdaaaGccqGHsisldaWcaaqaaiaaigdaaeaacaaIYaaaamaa % bmaabaGaamyBamaaBaaaleaacaaIXaaabeaakiabgUcaRiaad2gada % WgaaWcbaGaaGOmaaqabaaakiaawIcacaGLPaaacaWG2bWaa0baaSqa % aiaadAgaaeaacaaIYaaaaaGcbaGaeyypa0ZaaSaaaeaacaaIXaaaba % GaaGOmaaaacaWGTbWaaSbaaSqaaiaaigdaaeqaaOGaamODamaaDaaa % leaacaaIXaGaamyAaaqaaiaaikdaaaGccqGHsisldaWcaaqaaiaaig % daaeaacaaIYaaaamaalaaabaGaamyBamaaDaaaleaacaaIXaaabaGa % aGOmaaaaaOqaaiaad2gadaWgaaWcbaGaaGymaaqabaGccqGHRaWkca % WGTbWaaSbaaSqaaiaaikdaaeqaaaaakiaadAhadaqhaaWcbaGaaGym % aiaadMgaaeaacaaIYaaaaOWaamWaaeaacaWG1bGaci4CaiaacMgaca % GGUbGaam4zaiaadweacaWGXbGaaiOlamaabmaabaGaaGOnaiaac6ca % caaIYaGaaG4maaGaayjkaiaawMcaaaGaay5waiaaw2faaaqaaiabg2 % da9maalaaabaGaaGymaaqaaiaaikdaaaGaamyBamaaBaaaleaacaaI % XaaabeaakiaadAhadaqhaaWcbaGaaGymaiaadMgaaeaacaaIYaaaaO % WaamWaaeaacaaIXaGaeyOeI0YaaSaaaeaacaWGTbWaaSbaaSqaaiaa % igdaaeqaaaGcbaGaamyBamaaBaaaleaacaaIXaaabeaakiabgUcaRi % aad2gadaWgaaWcbaGaaGOmaaqabaaaaaGccaGLBbGaayzxaaaabaGa % eyypa0ZaaSaaaeaacaaIXaaabaGaaGOmaaaadaWcaaqaaiaad2gada % WgaaWcbaGaaGymaaqabaGccaWGTbWaaSbaaSqaaiaaikdaaeqaaaGc % baGaamyBamaaBaaaleaacaaIXaaabeaakiabgUcaRiaad2gadaWgaa % WcbaGaaGOmaaqabaaaaOGaamODamaaDaaaleaacaaIXaGaamyAaaqa % aiaaikdaaaaaaaa!8F94! \begin{array}{l} \Delta K = \frac{1}{2}{m_1}v_{1i}^2 - \frac{1}{2}\left( {{m_1} + {m_2}} \right)v_f^2\\ = \frac{1}{2}{m_1}v_{1i}^2 - \frac{1}{2}\frac{{m_1^2}}{{{m_1} + {m_2}}}v_{1i}^2\left[ {u\sin gEq.\left( {6.23} \right)} \right]\\ = \frac{1}{2}{m_1}v_{1i}^2\left[ {1 - \frac{{{m_1}}}{{{m_1} + {m_2}}}} \right]\\ = \frac{1}{2}\frac{{{m_1}{m_2}}}{{{m_1} + {m_2}}}v_{1i}^2 \end{array}\)
which is a positive quantity as expected.
Consider next an elastic collision.Using the above nomenclature with \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqiUde3aaS % baaSqaaiaaigdaaeqaaOGaeyypa0JaeqiUde3aaSbaaSqaaiaaikda % aeqaaOGaeyypa0JaaGimaiaacYcaaaa!3EBB! {\theta _1} = {\theta _2} = 0,\) the momentum and kinetic energy conservation equations are
\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacaWGTb % WaaSbaaSqaaiaaigdaaeqaaOGaamODamaaBaaaleaacaaIXaGaamyA % aaqabaGccqGH9aqpcaWGTbWaaSbaaSqaaiaaigdaaeqaaOGaamODam % aaBaaaleaacaaIXaGaamOzaaqabaGccqGHRaWkcaWGTbWaaSbaaSqa % aiaaikdaaeqaaOGaamODamaaBaaaleaacaaIYaGaamOzaaqabaGcda % qadaqaaiaaiAdacaGGUaGaaGOmaiaaisdaaiaawIcacaGLPaaaaeaa % caWGTbWaaSbaaSqaaiaaigdaaeqaaOGaamODamaaDaaaleaacaaIXa % GaamyAaaqaaiaaikdaaaGccqGH9aqpcaWGTbWaaSbaaSqaaiaaigda % aeqaaOGaamODamaaDaaaleaacaaIXaGaamOzaaqaaiaaikdaaaGccq % GHRaWkcaWGTbWaaSbaaSqaaiaaikdaaeqaaOGaamODamaaDaaaleaa % caaIYaGaamOzaaqaaiaaikdaaaGcdaqadaqaaiaaiAdacaGGUaGaaG % OmaiaaiwdaaiaawIcacaGLPaaaaaaa!6155! \begin{array}{l} {m_1}{v_{1i}} = {m_1}{v_{1f}} + {m_2}{v_{2f}}\left( {6.24} \right)\\ {m_1}v_{1i}^2 = {m_1}v_{1f}^2 + {m_2}v_{2f}^2\left( {6.25} \right) \end{array}\)
From Eqs. (6.24) and (6.25) it follows that,
\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacaWGTb % WaaSbaaSqaaiaaigdaaeqaaOGaamODamaaBaaaleaacaaIXaGaamyA % aaqabaGcdaqadaqaaiaadAhadaWgaaWcbaGaaGOmaiaadAgaaeqaaO % GaeyOeI0IaamODamaaBaaaleaacaaIXaGaamyAaaqabaaakiaawIca % caGLPaaacqGH9aqpcaWGTbWaaSbaaSqaaiaaigdaaeqaaOGaamODam % aaBaaaleaacaaIXaGaamOzaaqabaGcdaqadaqaaiaadAhadaWgaaWc % baGaaGOmaiaadAgaaeqaaOGaeyOeI0IaamODamaaBaaaleaacaaIXa % GaamOzaaqabaaakiaawIcacaGLPaaaaeaacaWGVbGaamOCaiaacYca % caWG2bWaaSbaaSqaaiaaikdacaWGMbaabeaakmaabmaabaGaamODam % aaBaaaleaacaaIXaGaamyAaaqabaGccqGHsislcaWG2bWaaSbaaSqa % aiaaigdacaWGMbaabeaaaOGaayjkaiaawMcaaiabg2da9iaadAhada % qhaaWcbaGaaGymaiaadMgaaeaacaaIYaaaaOGaeyOeI0IaamODamaa % DaaaleaacaaIXaGaamOzaaqaaiaaikdaaaaakeaacqGH9aqpdaqada % qaaiaadAhadaWgaaWcbaGaaGymaiaadMgaaeqaaOGaeyOeI0IaamOD % amaaBaaaleaacaWGPbGaamOzaaqabaaakiaawIcacaGLPaaadaqada % qaaiaadAhadaWgaaWcbaGaaGymaiaadMgaaeqaaOGaey4kaSIaamOD % amaaBaaaleaacaaIXaGaamOzaaqabaaakiaawIcacaGLPaaaaeaaca % WGibGaamyzaiaad6gacaWGJbGaamyzaiaacYcacqGH0icxcaWG2bWa % aSbaaSqaaiaaikdacaWGMbaabeaakiabg2da9iaadAhadaWgaaWcba % GaaGymaiaadMgaaeqaaOGaey4kaSIaamODamaaBaaaleaacaaIXaGa % amOzaaqabaGcdaqadaqaaiaaiAdacaGGUaGaaGOmaiaaiAdaaiaawI % cacaGLPaaaaaaa!8E4F! \begin{array}{l} {m_1}{v_{1i}}\left( {{v_{2f}} - {v_{1i}}} \right) = {m_1}{v_{1f}}\left( {{v_{2f}} - {v_{1f}}} \right)\\ or,{v_{2f}}\left( {{v_{1i}} - {v_{1f}}} \right) = v_{1i}^2 - v_{1f}^2\\ = \left( {{v_{1i}} - {v_{if}}} \right)\left( {{v_{1i}} + {v_{1f}}} \right)\\ Hence,\therefore {v_{2f}} = {v_{1i}} + {v_{1f}}\left( {6.26} \right) \end{array}\)
Substituting this in Eq. (6.24), we obtain
\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacaWG2b % WaaSbaaSqaaiaaigdacaWGMbaabeaakiabg2da9maalaaabaWaaeWa % aeaacaWGTbWaaSbaaSqaaiaaigdaaeqaaOGaeyOeI0IaamyBamaaBa % aaleaacaaIYaaabeaaaOGaayjkaiaawMcaaaqaaiaad2gadaWgaaWc % baGaaGymaaqabaGccqGHRaWkcaWGTbWaaSbaaSqaaiaaikdaaeqaaa % aakiaadAhadaWgaaWcbaGaaGymaiaadMgaaeqaaOWaaeWaaeaacaaI % 2aGaaiOlaiaaikdacaaI3aaacaGLOaGaayzkaaaabaGaamyyaiaad6 % gacaWGKbGaamODamaaBaaaleaacaaIYaGaamOzaaqabaGccqGH9aqp % daWcaaqaaiaaikdacaWGTbWaaSbaaSqaaiaaigdaaeqaaOGaamODam % aaBaaaleaacaaIXaGaamyAaaqabaaakeaacaWGTbWaaSbaaSqaaiaa % igdaaeqaaOGaey4kaSIaamyBamaaBaaaleaacaaIYaaabeaaaaGcda % qadaqaaiaaiAdacaGGUaGaaGOmaiaaiIdaaiaawIcacaGLPaaaaaaa % !616D! \begin{array}{l} {v_{1f}} = \frac{{\left( {{m_1} - {m_2}} \right)}}{{{m_1} + {m_2}}}{v_{1i}}\left( {6.27} \right)\\ and{v_{2f}} = \frac{{2{m_1}{v_{1i}}}}{{{m_1} + {m_2}}}\left( {6.28} \right) \end{array}\)
Thus, the ‘unknowns’ {v1f, v2f} are obtained in terms of the ‘knowns’ {m1, m2, \(v\)1i }. Special cases of our analysis are interesting.
Case I : If the two masses are equal
\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacaWG2b % WaaSbaaSqaaiaaigdacaWGMbaabeaakiabg2da9iaaicdaaeaacaWG % 2bWaaSbaaSqaaiaaikdacaWGMbaabeaakiabg2da9iaadAhadaWgaa % WcbaGaaGymaiaadMgaaeqaaaaaaa!4142! \begin{array}{l} {v_{1f}} = 0\\ {v_{2f}} = {v_{1i}} \end{array}\)
The first mass comes to rest and pushes off the second mass with its initial speed on collision.
Case II : If one mass dominates, e.g. m2 > > m1 \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamODamaaBa % aaleaacaaIXaGaamOzaaqabaGccqWIdjYocqGHsislcaWG2bWaaSba % aSqaaiaaigdacaWGPbaabeaakiaacYcacaWG2bWaaSbaaSqaaiaaik % dacaWGMbaabeaakiabloKi7iaaicdaaaa!4338! {v_{1f}} \simeq - {v_{1i}},{v_{2f}} \simeq 0\)
The heavier mass is undisturbed while the
EXAMPLE 12
Slowing down of neutrons: In a nuclear reactor a neutron of high speed (typically 107 m s–1) must be slowed to 103 m s–1 so that it can have a high probability of interacting with isotope \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa0raaSqaai % aaiMdacaaIYaaabaGaaGOmaiaaiodacaaI1aaaaOGaamyvaaaa!3ABF! {}_{92}^{235}U\) and causing it to fission. Show that a neutron can lose most of its kinetic energy in an elastic collision with a light nuclei like deuterium or carbon which has a mass of only a few times the neutron mass. The material making up the light nuclei, usually heavy water (D2O) or graphite, is called a moderator.
ANSWER
The initial kinetic energy of the neutron is \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4samaaBa % aaleaacaaIXaGaamyAaaqabaGccqGH9aqpdaWcaaqaaiaaigdaaeaa % caaIYaaaaiaad2gadaWgaaWcbaGaaGymaaqabaGccaWG2bWaa0baaS % qaaiaaigdacaWGPbaabaGaaGOmaaaaaaa!40A2! {K_{1i}} = \frac{1}{2}{m_1}v_{1i}^2\)
while its final kinetic energy from Eq. (6.27)
\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4samaaBa % aaleaacaaIXaGaamyAaaqabaGccqGH9aqpdaWcaaqaaiaaigdaaeaa % caaIYaaaaiaad2gadaWgaaWcbaGaaGymaaqabaGccaWG2bWaa0baaS % qaaiaaigdacaWGMbaabaGaaGOmaaaakiabg2da9maalaaabaGaaGym % aaqaaiaaikdaaaGaamyBamaaBaaaleaacaaIXaaabeaakmaabmaaba % WaaSaaaeaacaWGTbWaaSbaaSqaaiaaigdaaeqaaOGaeyOeI0IaamyB % amaaBaaaleaacaaIYaaabeaaaOqaaiaad2gadaWgaaWcbaGaaGymaa % qabaGccqGHRaWkcaWGTbWaaSbaaSqaaiaaikdaaeqaaaaaaOGaayjk % aiaawMcaamaaCaaaleqabaGaaGOmaaaakiaadAhadaqhaaWcbaGaaG % ymaiaadMgaaeaacaaIYaaaaaaa!548F! {K_{1i}} = \frac{1}{2}{m_1}v_{1f}^2 = \frac{1}{2}{m_1}{\left( {\frac{{{m_1} - {m_2}}}{{{m_1} + {m_2}}}} \right)^2}v_{1i}^2\)
The fractional kinetic energy lost is \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzamaaBa % aaleaacaaIXaaabeaakiabg2da9maalaaabaGaam4samaaBaaaleaa % caaIXaGaamOzaaqabaaakeaacaWGlbWaaSbaaSqaaiaaigdacaWGPb % aabeaaaaGccqGH9aqpdaqadaqaamaalaaabaGaamyBamaaBaaaleaa % caaIXaaabeaakiabgkHiTiaad2gadaWgaaWcbaGaaGOmaaqabaaake % aacaWGTbWaaSbaaSqaaiaaigdaaeqaaOGaey4kaSIaamyBamaaBaaa % leaacaaIYaaabeaaaaaakiaawIcacaGLPaaadaahaaWcbeqaaiaaik % daaaaaaa!4B28! {f_1} = \frac{{{K_{1f}}}}{{{K_{1i}}}} = {\left( {\frac{{{m_1} - {m_2}}}{{{m_1} + {m_2}}}} \right)^2}\)
while the fractional kinetic energy gained by the moderating nuclei K2f /K1i is
\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacaWGMb % WaaSbaaSqaaiaaikdaaeqaaOGaeyypa0JaaGymaiabgkHiTiaadAga % daWgaaWcbaGaaGymaaqabaGcdaqadaqaaiaadwgacaWGSbGaamyyai % aadohacaWG0bGaamyAaiaadogacaWGJbGaam4BaiaadYgacaWGSbGa % amyAaiaadohacaWGPbGaam4Baiaad6gaaiaawIcacaGLPaaaaeaacq % GH9aqpdaWcaaqaaiaaisdacaWGTbWaaSbaaSqaaiaaigdaaeqaaOGa % amyBamaaBaaaleaacaaIYaaabeaaaOqaamaabmaabaGaamyBamaaBa % aaleaacaaIXaaabeaakiabgUcaRiaad2gadaWgaaWcbaGaaGOmaaqa % baaakiaawIcacaGLPaaadaahaaWcbeqaaiaaikdaaaaaaaaaaa!59A4! \begin{array}{l} {f_2} = 1 - {f_1}\left( {elasticcollision} \right)\\ = \frac{{4{m_1}{m_2}}}{{{{\left( {{m_1} + {m_2}} \right)}^2}}} \end{array}\)
One can also verify this result by substituting from Eq. (6.28).
For deuterium m 2 = 2m1 and we obtain f1 = 1/9 while f 2 = 8/9. Almost 90% of the neutron’s energy is transferred to deuterium. For carbon f1 = 71.6% and f2 = 28.4%. In practice, however, this number is smaller since head-on collisions are rare.
If the initial velocities and final velocities of both the bodies are along the same straight line, then it is called a one-dimensional collision, or head-on collision. In the case of small spherical bodies,this is possible if the direction of travel of body 1 passes through the centre of body 2 which is at rest. In general, the collision is two-dimensional, where the initial velocities and the final velocities lie in a plane.
Collisions in Two Dimension
Fig. 6.10 also depicts the collision of a moving mass m1 with the stationary mass m2.Linear momentum is conserved in such a collision.
Since momentum is a vector this implies three equations for the three directions {x, y, z}.Consider the plane determined by the final velocity directions of m1 and m2 and choose it to be the x-y plane.The conservation of the z-component of the linear momentum implies that the entire collision is in the x-y plane. The x- and y-component equations are
\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacaWGTb % WaaSbaaSqaaiaaigdaaeqaaOGaamODamaaBaaaleaacaaIXaGaamyA % aaqabaGccqGH9aqpcaWGTbWaaSbaaSqaaiaaigdaaeqaaOGaamODam % aaBaaaleaacaaIXaGaamOzaaqabaGcciGGJbGaai4BaiaacohacqaH % 4oqCdaWgaaWcbaGaaGymaaqabaGccqGHRaWkcaWGTbWaaSbaaSqaai % aaikdaaeqaaOGaamODamaaBaaaleaacaaIYaGaamOzaaqabaGcciGG % JbGaai4BaiaacohacqaH4oqCdaWgaaWcbaGaaGOmaaqabaGcdaqada % qaaiaaiAdacaGGUaGaaGOmaiaaiMdaaiaawIcacaGLPaaaaeaacaaI % WaGaeyypa0JaamyBamaaBaaaleaacaaIXaaabeaakiaadAhadaWgaa % WcbaGaaGymaiaadAgaaeqaaOGaci4CaiaacMgacaGGUbGaeqiUde3a % aSbaaSqaaiaaigdaaeqaaOGaeyOeI0IaamyBamaaBaaaleaacaaIYa % aabeaakiaadAhadaWgaaWcbaGaaGOmaiaadAgaaeqaaOGaci4Caiaa % cMgacaGGUbGaeqiUde3aaSbaaSqaaiaaikdaaeqaaOWaaeWaaeaaca % aI2aGaaiOlaiaaiodacaaIWaaacaGLOaGaayzkaaaaaaa!711B! \begin{array}{l} {m_1}{v_{1i}} = {m_1}{v_{1f}}\cos {\theta _1} + {m_2}{v_{2f}}\cos {\theta _2}\left( {6.29} \right)\\ 0 = {m_1}{v_{1f}}\sin {\theta _1} - {m_2}{v_{2f}}\sin {\theta _2}\left( {6.30} \right) \end{array}\)
One knows {m1, m2, v1i} in most situations. There are thus four unknowns {\(v_{1f}\) , \(v_{2f}\) ,\(\theta_1\) and \(\theta_2\) }, and only two equations.If \(\theta_1\) = \(\theta_2\)= 0, we regain Eq. (6.24) for one dimensional collision. If, further the collision is elastic,
\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % aIXaaabaGaaGOmaaaacaWGTbWaaSbaaSqaaiaaigdaaeqaaOGaamOD % amaaDaaaleaacaaIXaGaamyAaaqaaiaaikdaaaGccqGH9aqpdaWcaa % qaaiaaigdaaeaacaaIYaaaaiaad2gadaWgaaWcbaGaaGymaaqabaGc % caWG2bWaaSbaaSqaaiaaigdaaeqaaOWaa0baaSqaaiaadAgaaeaaca % aIYaaaaOGaey4kaSYaaSaaaeaacaaIXaaabaGaaGOmaaaacaWGTbWa % aSbaaSqaaiaaikdaaeqaaOGaamODamaaBaaaleaacaaIYaaabeaakm % aaDaaaleaacaWGMbaabaGaaGOmaaaakmaabmaabaGaaGOnaiaac6ca % caaIZaGaaGymaaGaayjkaiaawMcaaaaa!51BC! \frac{1}{2}{m_1}v_{1i}^2 = \frac{1}{2}{m_1}{v_1}_f^2 + \frac{1}{2}{m_2}{v_2}_f^2\left( {6.31} \right)\)
We obtain an additional equation. That still leaves us one equation short.At least one of the four unknowns, say \(\theta\)1,must be made known for the problem to be solvable. For example, \(\theta\)1 can be determined by moving a detector in an angular fashion from the x to the y axis. Given \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaiWaaeaaca % WGTbWaaSbaaSqaaiaaigdaaeqaaOGaaiilaiaad2gadaWgaaWcbaGa % aGOmaaqabaGccaGGSaGaamODamaaBaaaleaacaaIXaGaamyAaaqaba % GccaGGSaGaeqiUde3aaSbaaSqaaiaaigdaaeqaaaGccaGL7bGaayzF % aaaaaa!437F! \left\{ {{m_1},{m_2},{v_{1i}},{\theta _1}} \right\}\) we can determine \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaiWaaeaaca % WG2bWaaSbaaSqaaiaaigdacaWGMbaabeaakiaacYcacaWG2bWaaSba % aSqaaiaaikdacaWGMbaabeaakiaacYcacqaH4oqCdaWgaaWcbaGaaG % OmaaqabaaakiaawUhacaGL9baaaaa!41DE! \left\{ {{v_{1f}},{v_{2f}},{\theta _2}} \right\}\) from Eqs. (6.29)-(6.31).
EXAMPLE 13
Consider the collision
depicted in Fig. 6.10 to be between two billiard balls with equal masses m1 = m2.The first ball is called the cue while the second ball is called the target. The billiard player wants to ‘sink’ the target ball in a corner pocket, which is at an angle \(\theta\)= 37°. Assume that the collision is elastic and that friction and rotational motion are not important.Obtain \(\theta\).
ANSWER
From momentum conservation, since the masses are equal
\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacaWG2b % WaaSbaaSqaaiaaigdacaWGPbaabeaakiabg2da9iaadAhadaWgaaWc % baGaaGymaiaadAgaaeqaaOGaey4kaSIaamODamaaBaaaleaacaaIYa % GaamOzaaqabaaakeaacaWGVbGaamOCaiaadAhadaqhaaWcbaGaaGym % aiaadMgaaeaacaaIYaaaaOGaeyypa0ZaaeWaaeaacaWG2bWaaSbaaS % qaaiaaigdacaWGMbaabeaakiabgUcaRiaadAhadaWgaaWcbaGaaGOm % aiaadAgaaeqaaaGccaGLOaGaayzkaaGaaiOlamaabmaabaGaamODam % aaBaaaleaacaaIXaGaamOzaaqabaGccqGHRaWkcaWG2bWaaSbaaSqa % aiaaikdacaWGMbaabeaaaOGaayjkaiaawMcaaaaaaa!57DC! \begin{array}{l} {v_{1i}} = {v_{1f}} + {v_{2f}}\\ orv_{1i}^2 = \left( {{v_{1f}} + {v_{2f}}} \right).\left( {{v_{1f}} + {v_{2f}}} \right) \end{array}\)
\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyypa0Jaam % ODamaaBaaaleaacaaIXaaabeaakmaaDaaaleaacaWGMbaabaGaaGOm % aaaakiabgUcaRiaadAhadaWgaaWcbaGaaGOmaaqabaGcdaqhaaWcba % GaamOzaaqaaiaaikdaaaGccqGHRaWkcaaIYaGaamODamaaBaaaleaa % caaIXaGaamOzaaqabaGccaGGUaGaamODamaaBaaaleaacaaIYaGaam % Ozaaqabaaaaa!4768! = {v_1}_f^2 + {v_2}_f^2 + 2{v_{1f}}.{v_{2f}}\)
\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyypa0Zaai % WaaeaacaWG2bWaaSbaaSqaaiaaigdaaeqaaOWaa0baaSqaaiaadAga % aeaacaaIYaaaaOGaey4kaSIaamODamaaBaaaleaacaaIYaaabeaakm % aaDaaaleaacaWGMbaabaGaaGOmaaaakiabgUcaRiaaikdacaWG2bWa % aSbaaSqaaiaaigdacaWGMbaabeaakiaadAhadaWgaaWcbaGaaGOmai % aadAgaaeqaaOGaci4yaiaac+gacaGGZbWaaeWaaeaacqaH4oqCdaWg % aaWcbaGaaGymaaqabaGccqGHRaWkcaaIZaGaaG4namaaCaaaleqaba % GaeSigI8gaaaGccaGLOaGaayzkaaaacaGL7bGaayzFaaWaaeWaaeaa % caaI2aGaaiOlaiaaiodacaaIYaaacaGLOaGaayzkaaaaaa!5839! = \left\{ {{v_1}_f^2 + {v_2}_f^2 + 2{v_{1f}}{v_{2f}}\cos \left( {{\theta _1} + {{37}^ \circ }} \right)} \right\}\left( {6.32} \right)\)
Since the collision is elastic and m1=m2 it follow from conservation of kinetic energy that
\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamODamaaBa % aaleaacaaIXaaabeaakmaaDaaaleaacaWGPbaabaGaaGOmaaaakiab % g2da9iaadAhadaWgaaWcbaGaaGymaaqabaGcdaqhaaWcbaGaamOzaa % qaaiaaikdaaaGccqGHRaWkcaWG2bWaaSbaaSqaaiaaikdaaeqaaOWa % a0baaSqaaiaadAgaaeaacaaIYaaaaOWaaeWaaeaacaaI2aGaaiOlai % aaiodacaaIZaaacaGLOaGaayzkaaaaaa!47B5! {v_1}_i^2 = {v_1}_f^2 + {v_2}_f^2\left( {6.33} \right)\)
Comparing Eqs. (6.32) and (6.33), we get
\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaaciGGJb % Gaai4BaiaacohadaqadaqaaiabeI7aXnaaBaaaleaacaaIXaaabeaa % kiabgUcaRiaaiodacaaI3aWaaWbaaSqabeaacqWIyiYBaaaakiaawI % cacaGLPaaacqGH9aqpcaaIWaaabaGaam4BaiaadkhacqaH4oqCdaWg % aaWcbaGaaGymaaqabaGccqGHRaWkcaaIZaGaaG4namaaCaaaleqaba % GaeSigI8gaaOGaeyypa0JaaGyoaiaaicdadaahaaWcbeqaaiablIHi % VbaaaOqaaiaadsfacaWGObGaamyDaiaadohacaGGSaGaeqiUde3aaS % baaSqaaiaaigdaaeqaaOGaeyypa0JaaGynaiaaiodadaahaaWcbeqa % aiablIHiVbaaaaaa!59E1! \begin{array}{l} \cos \left( {{\theta _1} + {{37}^ \circ }} \right) = 0\\ or{\theta _1} + {37^ \circ } = {90^ \circ }\\ Thus,{\theta _1} = {53^ \circ } \end{array}\)
This proves the following result : when two equal masses undergo a glancing elastic collision with one of them at rest, after the collision, they will move at right angles to each other.
The matter simplifies greatly if we consider spherical masses with smooth surfaces, and assume that collision takes place only when the bodies touch each other. This is what happens in the games of marbles, carrom and billiards.
In our everyday world, collisions take place only when two bodies touch each other.But consider a comet coming from far distances to the sun, or alpha particle coming towards a nucleus and going away in some direction.Here we have to deal with forces involving action at a distance.
Such an event is called scattering.The velocities and directions in which the two particles go away depend on their initial velocities as well as the type of interaction between them, their masses, shapes and sizes.