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Question Number 50912 by peter frank last updated on 22/Dec/18
Show that in collision where kinetic energy is conserved linear momemtum is also conserved
$${Show}\:{that}\:{in}\:{collision} \\ $$$${where}\:{kinetic}\:{energy}\:{is} \\ $$$${conserved}\:{linear}\:{momemtum} \\ $$$${is}\:{also}\:{conserved} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 22/Dec/18
K.E=(1/2)mu^2 =(1/2)×(p^2 /m) p^2 =2mE [E=K.E and p=momentum] 2p(dp/dE)=2m 2pdp=2mdE pdp=mdE dE=0 [K.E conserved] so dp=0 momentum conserved
$${K}.{E}=\frac{\mathrm{1}}{\mathrm{2}}{mu}^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}×\frac{{p}^{\mathrm{2}} }{{m}} \\ $$$${p}^{\mathrm{2}} =\mathrm{2}{mE}\:\:\:\left[{E}={K}.{E}\:{and}\:{p}={momentum}\right] \\ $$$$\mathrm{2}{p}\frac{{dp}}{{dE}}=\mathrm{2}{m} \\ $$$$\mathrm{2}{pdp}=\mathrm{2}{mdE} \\ $$$${pdp}={mdE} \\ $$$${dE}=\mathrm{0}\:\:\left[{K}.{E}\:{conserved}\right]\:\:{so}\:{dp}=\mathrm{0} \\ $$$${momentum}\:{conserved} \\ $$$$ \\ $$
Commented by peter frank last updated on 22/Dec/18
thanks
$${thanks} \\ $$
Answered by peter frank last updated on 22/Dec/18
from (1/(2 ))m_(1 ) u_(1 ) ^2 +(1/2)m_2 u_(2 ) ^2 =(1/2)m_1 v_(1 ) ^2 +(1/2)m_2 v_(2 ) ^2 m_1 (u_1 ^2 −v_1 ^2 )=m_2 (v_(2 ) ^2 −u_(2 ) ^2 ) m_1 (u_1 −v_1 )(u_1 +v_1 )=m_2 (v_(2 ) −u_(2 ) )(v_2 +u_2 ).....(i) from ((v_1 −u_1 )/(u_1 −u_2 ))=e but e=1 v_2 −v_(1 ) =u_1 −u_2 u_1 +v_1 =v_(2 ) +u_(2 ) ......(ii) take eqn i)÷(ii) and simplify m_1 (u_(1 ) −v_1 )=m_2 (v_2 −u_2 ) m_1 u_1 +m_2 u_2 =m_1 v_1 +m_2 v_2
$${from} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}\:\:}{m}_{\mathrm{1}\:\:} {u}_{\mathrm{1}\:\:} ^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}{m}_{\mathrm{2}} {u}_{\mathrm{2}\:} ^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}{m}_{\mathrm{1}} {v}_{\mathrm{1}\:} ^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}{m}_{\mathrm{2}} {v}_{\mathrm{2}\:} ^{\mathrm{2}} \\ $$$${m}_{\mathrm{1}} \left({u}_{\mathrm{1}} ^{\mathrm{2}} −{v}_{\mathrm{1}} ^{\mathrm{2}} \right)={m}_{\mathrm{2}} \left({v}_{\mathrm{2}\:} ^{\mathrm{2}} −{u}_{\mathrm{2}\:\:} ^{\mathrm{2}} \right) \\ $$$${m}_{\mathrm{1}} \left({u}_{\mathrm{1}} −{v}_{\mathrm{1}} \right)\left({u}_{\mathrm{1}} +{v}_{\mathrm{1}} \right)={m}_{\mathrm{2}} \left({v}_{\mathrm{2}\:} −{u}_{\mathrm{2}\:\:} \right)\left({v}_{\mathrm{2}} +{u}_{\mathrm{2}} \right)…..\left({i}\right) \\ $$$${from} \\ $$$$\frac{{v}_{\mathrm{1}} −{u}_{\mathrm{1}} }{{u}_{\mathrm{1}} −{u}_{\mathrm{2}} }={e}\:\:\:\:\:\:\:\:{but}\:{e}=\mathrm{1} \\ $$$${v}_{\mathrm{2}} −{v}_{\mathrm{1}\:} ={u}_{\mathrm{1}} −{u}_{\mathrm{2}} \\ $$$${u}_{\mathrm{1}} +{v}_{\mathrm{1}} ={v}_{\mathrm{2}\:} +{u}_{\mathrm{2}\:\:\:\:} ……\left({ii}\right) \\ $$$$\left.{take}\:{eqn}\:{i}\right)\boldsymbol{\div}\left({ii}\right)\:{and}\:{simplify} \\ $$$${m}_{\mathrm{1}} \left({u}_{\mathrm{1}\:} −{v}_{\mathrm{1}} \right)={m}_{\mathrm{2}} \left({v}_{\mathrm{2}} −{u}_{\mathrm{2}} \right) \\ $$$${m}_{\mathrm{1}} {u}_{\mathrm{1}} +{m}_{\mathrm{2}} {u}_{\mathrm{2}} ={m}_{\mathrm{1}} {v}_{\mathrm{1}} +{m}_{\mathrm{2}} {v}_{\mathrm{2}} \\ $$$$ \\ $$

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