A particle of mass $4 m$ explodes into three pieces of masses $m, m$ and $2 \mathrm{~m}$. The equal masses move along $X$-axis and $Y$-axis with velocities $4 \mathrm{~ms}^{-1}$ and $6 \mathrm{~ms}^{-1}$ respectively. The magnitude of the velocity of the heavier mass is

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Aicharya · Accepted Answer

Let third mass particle $(2 \mathrm{~m}$ ) moves making angle $\theta$ with $X$-axis. The horizontal component of velocity of $2 \mathrm{~m}$ mass particle $=u \cos \theta$ and vertical component $=u \sin \theta$ From conservation of linear momentum in $X$-direction $\begin{array}{rlrl}\text { or } & m_1 u_1+m_2 u_2 =m_1 v_1+m_2 v_2 \\ \text { or } & 0 =m \times 4+2 m(u \cos \theta) \\ \text { or } & -4 =2 u \cos \theta\end{array}$

Again, applying law of conservation of linear momentum in $Y$-direction. $\begin{aligned} 0 & =m \times 6+2 m(u \sin \theta) \\ \Rightarrow \quad-\frac{6}{2} & =u \sin \theta \end{aligned}$ or $\quad-3=u \sin \theta \quad \ldots$ (ii) Squaring Eqs. (i) and(ii) and adding, we get $\begin{aligned} \text { (4) } +(9) & =u^2 \cos ^2 \theta+u^2 \sin ^2 \theta \\ & =u^2\left(\cos ^2 \theta+\sin ^2 \theta\right) \end{aligned}$ $\begin{aligned} \text { or } 13=u^2 \\ \text { or } u=\sqrt{13} \mathrm{~ms}^{-1}\end{aligned}$

A particle of mass $4 m$ explodes into three pieces of masses $m, m$ and $2 \mathrm{~m}$. The equal masses…

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