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A-particle-of-mass-m-moving-with-a-speed-v-hits-elastically-another-stationary-particle-of-mass-2m-on-a-smooth-horizontal-circular-tube-of-radius-r-The-time-in-which-the-next-collision-will-take-plac




Question Number 24945 by Tinkutara last updated on 29/Nov/17
A particle of mass m moving with a  speed v hits elastically another  stationary particle of mass 2m on a  smooth horizontal circular tube of  radius r. The time in which the next  collision will take place is equal to
$$\mathrm{A}\:\mathrm{particle}\:\mathrm{of}\:\mathrm{mass}\:{m}\:\mathrm{moving}\:\mathrm{with}\:\mathrm{a} \\ $$$$\mathrm{speed}\:{v}\:\mathrm{hits}\:\mathrm{elastically}\:\mathrm{another} \\ $$$$\mathrm{stationary}\:\mathrm{particle}\:\mathrm{of}\:\mathrm{mass}\:\mathrm{2}{m}\:\mathrm{on}\:\mathrm{a} \\ $$$$\mathrm{smooth}\:\mathrm{horizontal}\:\mathrm{circular}\:\mathrm{tube}\:\mathrm{of} \\ $$$$\mathrm{radius}\:{r}.\:\mathrm{The}\:\mathrm{time}\:\mathrm{in}\:\mathrm{which}\:\mathrm{the}\:\mathrm{next} \\ $$$$\mathrm{collision}\:\mathrm{will}\:\mathrm{take}\:\mathrm{place}\:\mathrm{is}\:\mathrm{equal}\:\mathrm{to} \\ $$
Commented by Tinkutara last updated on 30/Nov/17
Thank you Sir!
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{Sir}! \\ $$
Commented by mrW1 last updated on 29/Nov/17
since the hit is without energy lost,  the departing velocity after the hit  is equal to the approching velocity  before the hit, which is v.  i.e. after the hit both particles  depart from each other with v in  the tube. the time after which they hit  again is hence ((2πr)/v).
$${since}\:{the}\:{hit}\:{is}\:{without}\:{energy}\:{lost}, \\ $$$${the}\:{departing}\:{velocity}\:{after}\:{the}\:{hit} \\ $$$${is}\:{equal}\:{to}\:{the}\:{approching}\:{velocity} \\ $$$${before}\:{the}\:{hit},\:{which}\:{is}\:{v}. \\ $$$${i}.{e}.\:{after}\:{the}\:{hit}\:{both}\:{particles} \\ $$$${depart}\:{from}\:{each}\:{other}\:{with}\:{v}\:{in} \\ $$$${the}\:{tube}.\:{the}\:{time}\:{after}\:{which}\:{they}\:{hit} \\ $$$${again}\:{is}\:{hence}\:\frac{\mathrm{2}\pi{r}}{{v}}. \\ $$
Commented by mrW1 last updated on 29/Nov/17
I think it doesn′t matter which mass  they have.
$${I}\:{think}\:{it}\:{doesn}'{t}\:{matter}\:{which}\:{mass} \\ $$$${they}\:{have}. \\ $$
Commented by mrW1 last updated on 30/Nov/17
this is the proof:  let m_2 =λm_1   before the hit:  particle 1: v  particle 2: 0  after hit:  particle 1: u_1   particle 2: u_2   m_1 v=m_1 u_1 +m_2 u_2   ⇒v=u_1 +λu_2    ...(i)  (1/2)m_1 v^2 =(1/2)m_1 u_1 ^2 +(1/2)m_2 u_2 ^2   ⇒v^2 =u_1 ^2 +λu_2 ^2     ...(ii)  (i) into (ii)  ⇒v^2 =v^2 −2λvu_2 +λ^2 u_2 ^2 +λu_2 ^2   ⇒λu_2 [(λ+1)u_2 −2v]=0  ⇒(λ+1)u_2 =2v  ⇒u_2 =(2/(λ+1))×v  ⇒u_1 =v−λ×((2v)/(λ+1))=((1−λ)/(λ+1))×v  ⇒Δu=u_2 −u_1 =((2−1+λ)/(λ+1))×v=v  i.e. after hit the particles depart from  each other with the same velocity  as they approch to each other before  the hit.  the time till next collision is hence  ((2πr)/v)
$${this}\:{is}\:{the}\:{proof}: \\ $$$${let}\:{m}_{\mathrm{2}} =\lambda{m}_{\mathrm{1}} \\ $$$${before}\:{the}\:{hit}: \\ $$$${particle}\:\mathrm{1}:\:{v} \\ $$$${particle}\:\mathrm{2}:\:\mathrm{0} \\ $$$${after}\:{hit}: \\ $$$${particle}\:\mathrm{1}:\:{u}_{\mathrm{1}} \\ $$$${particle}\:\mathrm{2}:\:{u}_{\mathrm{2}} \\ $$$${m}_{\mathrm{1}} {v}={m}_{\mathrm{1}} {u}_{\mathrm{1}} +{m}_{\mathrm{2}} {u}_{\mathrm{2}} \\ $$$$\Rightarrow{v}={u}_{\mathrm{1}} +\lambda{u}_{\mathrm{2}} \:\:\:…\left({i}\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{m}_{\mathrm{1}} {v}^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}{m}_{\mathrm{1}} {u}_{\mathrm{1}} ^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}{m}_{\mathrm{2}} {u}_{\mathrm{2}} ^{\mathrm{2}} \\ $$$$\Rightarrow{v}^{\mathrm{2}} ={u}_{\mathrm{1}} ^{\mathrm{2}} +\lambda{u}_{\mathrm{2}} ^{\mathrm{2}} \:\:\:\:…\left({ii}\right) \\ $$$$\left({i}\right)\:{into}\:\left({ii}\right) \\ $$$$\Rightarrow{v}^{\mathrm{2}} ={v}^{\mathrm{2}} −\mathrm{2}\lambda{vu}_{\mathrm{2}} +\lambda^{\mathrm{2}} {u}_{\mathrm{2}} ^{\mathrm{2}} +\lambda{u}_{\mathrm{2}} ^{\mathrm{2}} \\ $$$$\Rightarrow\lambda{u}_{\mathrm{2}} \left[\left(\lambda+\mathrm{1}\right){u}_{\mathrm{2}} −\mathrm{2}{v}\right]=\mathrm{0} \\ $$$$\Rightarrow\left(\lambda+\mathrm{1}\right){u}_{\mathrm{2}} =\mathrm{2}{v} \\ $$$$\Rightarrow{u}_{\mathrm{2}} =\frac{\mathrm{2}}{\lambda+\mathrm{1}}×{v} \\ $$$$\Rightarrow{u}_{\mathrm{1}} ={v}−\lambda×\frac{\mathrm{2}{v}}{\lambda+\mathrm{1}}=\frac{\mathrm{1}−\lambda}{\lambda+\mathrm{1}}×{v} \\ $$$$\Rightarrow\Delta{u}={u}_{\mathrm{2}} −{u}_{\mathrm{1}} =\frac{\mathrm{2}−\mathrm{1}+\lambda}{\lambda+\mathrm{1}}×{v}={v} \\ $$$${i}.{e}.\:{after}\:{hit}\:{the}\:{particles}\:{depart}\:{from} \\ $$$${each}\:{other}\:{with}\:{the}\:{same}\:{velocity} \\ $$$${as}\:{they}\:{approch}\:{to}\:{each}\:{other}\:{before} \\ $$$${the}\:{hit}. \\ $$$${the}\:{time}\:{till}\:{next}\:{collision}\:{is}\:{hence} \\ $$$$\frac{\mathrm{2}\pi{r}}{{v}} \\ $$

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