Question Number 104987 by mr W last updated on 25/Jul/20

Commented bymr W last updated on 25/Jul/20

a block with mass m_1  and velocity v_1   collides with an other block with  mass m_2  in rest. after the collision  the block m_2  has the velocity v_2 .  if the collision is elastic, find v_2 .    if we place an additional block between  m_1  and m_2 , find the maximal  possible velocity which the block m_2   may get.  what is the answer if we place n (n≥1)  additional blocks between m_1  and m_2 ?

Commented byajfour last updated on 25/Jul/20

Great question Sir, i can only hope  to comprehend the solution when  you shall post it, tough one for me  to even try and attempt..

Answered by mr W last updated on 28/Jul/20

collision m_1  with m_2  directly:  let u_1 =speed of m_1  after collision  v_1 =v_2 −u_1   m_1 v_1 =m_1 u_1 +m_2 v_2   m_1 v_1 =m_1 (v_2 −v_1 )+m_2 v_2   2v_1 =(1+(m_2 /m_1 ))v_2   ⇒v_2 =((2v_1 )/(1+(m_2 /m_1 )))    with an additional block M_1 :  V_1 =velocity of M_1  after collision  V_1 =((2v_1 )/(1+(M_1 /m_1 )))  v_2 =((2V_1 )/(1+(m_2 /M_1 )))=((2^2 v_1 )/((1+(M_1 /m_1 ))(1+(m_2 /M_1 ))))    in general with n additional blocks:  M_1 ,M_2 ,...,M_n   v_2 =((2^(n+1) v_1 )/((1+(M_1 /m_1 ))(1+(M_2 /M_1 ))(1+(M_3 /M_2 ))...(1+(m_2 /M_n ))))    v_2  is maximum, if  (M_1 /m_1 )=(M_2 /M_1 )=(M_3 /M_2 )=...=(m_2 /M_n )=k  ⇒(m_2 /m_1 )=k^(n+1)   ⇒k=((m_2 /m_1 ))^(1/(n+1))   ⇒M_1 =km_1   ⇒M_2 =k^2 m_1   ⇒M_n =k^n m_1   ⇒v_(2, max) =((2^(n+1) v_1 )/((1+k)^(n+1) ))=((2/(1+((m_2 /m_1 ))^(1/(n+1)) )))^(n+1) v_1