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Standard XII

Physics

Error and Uncertainty

Two particles...

Question

Two particles $A$ and $B$ of mass $1 k g$ and $2 k g$ respectively are kept $1 m$ apart and are released to move.

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Solution

$| M_{A} V_{A} | = | M_{A} V_{B} |$
$1 \times V_{A} = 2 \times 3.6$
$V_{A} = 7.2 c m / h r$
energy conservation
$(K . E); + (P E); (K E) + (P . E) Ω$
$P E = \frac{G M_{a} M_{b}}{r}$
$P E = - 2 G$
final $K E = \frac{1}{2} M_{a} V_{a}^{2} + \frac{1}{2} M_{b} V_{b}^{2}$
$0 - 2 G = \frac{G M_{a} M_{b}}{r} + \frac{M_{a} V_{b}^{2}}{2} + \frac{M_{b} V_{b}^{2}}{2}$
$- 2 G = \frac{- 2 G}{r} + \frac{(v a^{2} + 2 V b^{2}}{2}$
$- 2 G = - \frac{- 2 G}{r} + 3 v_{b}^{2}$
$r = 30.5 c m$

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Two particles A and B of mass 1kg and 2kg respectively are kept 1m apart and are released to move.

|MAVA|=|MAVB|1×VA=2×3.6VA=7.2cm/hrenergy conservation (K.E);+(PE);(KE)+(P.E)ΩPE=GMaMbrPE=−2Gfinal KE=12MaV2a+12MbV2b0−2G=GMaMbr+MaV2b2+MbV2b2−2G=−2Gr+(va2+2Vb22−2G=−−2Gr+3v2br=30.5cm

Two particles $A$ and $B$ of mass $1 k g$ and $2 k g$ respectively are kept $1 m$ apart and are released to move.