Two skaters A and B of weights in the ratio 5 | vedclass

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(D) Since there is no external force acting on the system,the center of mass remains stationary. The ratio of the distances moved by the skaters is in

Vedclass · Accepted Answer

$A$ moves $3.5\,m$ and $B$ moves $2.5\,m$

Vedclass · Answer

(D) Since there is no external force acting on the system,the center of mass remains stationary. The ratio of the distances moved by the skaters is inversely proportional to their masses (or weights).Let $m_A$ and $m_B$ be the weights of skaters $A$ and $B$ respectively,where $m_A/m_B = 5/7$.Let $x_A$ and $x_B$ be the distances moved by $A$ and $B$ respectively. Since they meet,$x_A + x_B = 6\,m$.From the conservation of momentum,$m_A x_A = m_B x_B$.Substituting the ratio,$5 x_A = 7 x_B$.Since $x_B = 6 - x_A$,we have $5 x_A = 7(6 - x_A)$.$5 x_A = 42 - 7 x_A \implies 12 x_A = 42 \implies x_A = 3.5\,m$.Then $x_B = 6 - 3.5 = 2.5\,m$.

Two skaters $A$ and $B$ of weights in the ratio $5 : 7$ start facing each other $6\,m$ apart on a horizontal smooth surface. They pull on a rope stretched between them. How far has each moved when they meet?

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