A particle of mass m_1 moving along the X-axi | vedclass

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(A) Let the initial velocity of $m_1$ be $u$ and the final velocities of $m_1$ and $m_2$ be $v_1$ and $v_2$ respectively. The particle $m_1$ moves at

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(A) Let the initial velocity of $m_1$ be $u$ and the final velocities of $m_1$ and $m_2$ be $v_1$ and $v_2$ respectively. The particle $m_1$ moves at $90^{\circ}$ to the $X$-axis and $m_2$ moves at $30^{\circ}$ to the $X$-axis.Applying conservation of linear momentum along the $X$-axis:$m_1 u = m_2 v_2 \cos 30^{\circ} \quad \dots (i)$Applying conservation of linear momentum along the $Y$-axis:$0 = m_2 v_2 \sin 30^{\circ} - m_1 v_1 \quad \Rightarrow \quad m_1 v_1 = m_2 v_2 \sin 30^{\circ} \quad \dots (ii)$Given that the kinetic energy reduces by $50 \%$,the final kinetic energy is half of the initial kinetic energy:$\frac{1}{2} (\frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2) = \frac{1}{2} (\frac{1}{2} m_1 u^2) \quad \Rightarrow \quad m_1 v_1^2 + m_2 v_2^2 = \frac{1}{2} m_1 u^2 \quad \dots (iii)$From $(i)$,$v_2 \cos 30^{\circ} = \frac{m_1 u}{m_2} \Rightarrow v_2 = \frac{2 m_1 u}{\sqrt{3} m_2}$.From $(ii)$,$v_1 = \frac{m_2 v_2 \sin 30^{\circ}}{m_1} = \frac{m_2}{m_1} \cdot \frac{2 m_1 u}{\sqrt{3} m_2} \cdot \frac{1}{2} = \frac{u}{\sqrt{3}}$.Substituting $v_1$ and $v_2$ into $(iii)$:$m_1 (\frac{u^2}{3}) + m_2 (\frac{4 m_1^2 u^2}{3 m_2^2}) = \frac{1}{2} m_1 u^2$Dividing by $m_1 u^2$:$\frac{1}{3} + \frac{4 m_1}{3 m_2} = \frac{1}{2} \quad \Rightarrow \quad \frac{4 m_1}{3 m_2} = \frac{1}{2} - \frac{1}{3} = \frac{1}{6}$$\frac{m_1}{m_2} = \frac{3}{24} = \frac{1}{8} \quad \Rightarrow \quad \frac{m_2}{m_1} = 8$.

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