Conservation of Linear Momentum and Mechanical Energy

Problems from IIT JEE

Problem (IIT JEE 1997): An isolated particle of mass $m$ is moving in horizontal plane ($x\text{-}y$), along the $x$-axis, at a certain height above the ground. It suddenly explodes into two fragment of masses $m/4$ and $3m/4$. An instant later, the smaller fragment is at $y={15}\;\mathrm{cm}$. The larger fragment at this instant is at,

  1. $y={-5}\;\mathrm{cm}$
  2. $y={20}\;\mathrm{cm}$
  3. $y={5}\;\mathrm{cm}$
  4. $y={-20}\;\mathrm{cm}$

Solution: There is no external force on the system in the $x\text{-}y$ plane. Hence, linear momentum of the system is conserved along $x$ and $y$ directions, separately. Initially, the particle was moving along $x$ axis with a speed $v_x$. Hence, coordinates of its centre of mass varies with time $t$ as, \begin{align} x_{c}=v_x t, \qquad y_{c}=0.\nonumber \end{align} By conservation of linear momentum, $y_{c}$ remains constant (here zero) after the collision i.e., \begin{align} \label{ppb:eqn:1} y_{c}=\frac{({m}/{4}) 15+({3m}/{4})y}{{m}/{4}+{3m}/{4}}=0. \end{align} Solve to get $y={-5}\;\mathrm{cm}$.

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