A particle of mass $4 M$ which is initially at rest explodes into three pieces of masses $M$, $M$ and $2 M$.…

A particle of mass $4 M$ which is initially at rest explodes into three pieces of masses $M$, $M$ and $2 M$. The equal masses move along $X$ and $Y$-axes with velocities $4 \mathrm{~ms}^{-1}$ and $6 \mathrm{~ms}^{-1}$, respectively. The magnitude of the velocity of the heavier mass is
  1. $\sqrt{17} \mathrm{~ms}^{-1}$
  2. $2 \sqrt{13} \mathrm{~ms}^{-1}$
  3. $\sqrt{13} \mathrm{~ms}^{-1}$
  4. $\frac{\sqrt{13}}{2} \mathrm{~ms}^{-1}$

Solution

Initial mass of particle $=4 M$ Initial velocity of particle $=0$. After explosion into three pieces of masses $M, M$ and $2 M$ particle of equal masses move along $X$ and $Y$ axes with velocities $4 \mathrm{~m} / \mathrm{s}$ and $6 \mathrm{~m} / \mathrm{s}$, respectively. So, according to law of conservation of momentum mass $2 M$ particle will move in the direction shown in the figure.
Law of conservation of momentum, Momentum before collision $=$ Momentum after collision. $ \begin{array}{rlrl} & & 4 M \times 0 & =2 M \mathbf{v}+M \mathbf{v}_x+M \mathbf{v}_y \\ \Rightarrow & & -2 M \mathbf{v} & =M\left(\mathbf{v}_x+\mathbf{v}_y\right) \\ \Rightarrow & & -2 \mathbf{v} & =4 \hat{\mathbf{i}}+6 \hat{\mathbf{j}} \Rightarrow \mathbf{v}=-2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}} \\ \text { So, }|\mathbf{v}|=\sqrt{2^2+3^2}=\sqrt{4+9}=\sqrt{13} \mathrm{~ms}^{-1} \end{array} $

Asked in: AP EAMCET 2018 (23 Apr Shift 1)

Practice more Center of Mass Momentum and Collision questions on Aicharya