Two balls of masses m_1 and m_2 are separated | vedclass

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(C) Let the velocities acquired by the masses $m_1$ and $m_2$ after the explosion be $v_1$ and $v_2$ respectively. By the law of conservation of linea

Vedclass · Accepted Answer

$s_2 = \frac{m_1^2}{m_2^2} s_1$

Vedclass · Answer

(C) Let the velocities acquired by the masses $m_1$ and $m_2$ after the explosion be $v_1$ and $v_2$ respectively. By the law of conservation of linear momentum,$m_1 v_1 = m_2 v_2$,which implies $v_1/v_2 = m_2/m_1$.When a mass $m$ moves on a surface with coefficient of friction $\mu$,the work done by friction is $W = -\mu m g s = -\frac{1}{2} m v^2$. This gives $s = \frac{v^2}{2 \mu g}$.Since $\mu$ and $g$ are constant,$s \propto v^2$.Therefore,$\frac{s_2}{s_1} = \frac{v_2^2}{v_1^2} = \left( \frac{v_2}{v_1} \right)^2$.Substituting the ratio from momentum conservation,$\frac{s_2}{s_1} = \left( \frac{m_1}{m_2} \right)^2$.Thus,$s_2 = \left( \frac{m_1}{m_2} \right)^2 s_1$.

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