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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
Showthatincollisionwherekineticenergyisconservedlinearmomemtumisalsoconserved
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=12mu2=12×p2mp2=2mE[E=K.Eandp=momentum]2pdpdE=2m2pdp=2mdEpdp=mdEdE=0[K.Econserved]sodp=0momentumconserved
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
from12m1u12+12m2u22=12m1v12+12m2v22m1(u12v12)=m2(v22u22)m1(u1v1)(u1+v1)=m2(v2u2)(v2+u2)..(i)fromv1u1u1u2=ebute=1v2v1=u1u2u1+v1=v2+u2(ii)takeeqni)÷(ii)andsimplifym1(u1v1)=m2(v2u2)m1u1+m2u2=m1v1+m2v2

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