Two skaters, initially at rest, are $5 m$ apart. They each have one end of a single rope and each pull on the rope with a force of $50 N$ for a period of $1 s$ . One skater weighs $80 kg$ and the other weighs $45 kg$ (Assume no friction between the skates and the ice.)

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the two skaters meet at a distance of $1.8 m$ from the initial position of the heavy skater

the two skaters meet at a distance of $3.2 m$ from the initial position of the heavy skater

the relative velocity of the skaters when they meet is $125 / 72 m / s$

the relative velocity of the skaters when they meet is $25 / 6 m / s$

answer is A, D.

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Detailed Solution

$T h e s k a t e r s w i l l m e e t a t t h e C O M o f t h e s y s t e m w h o s e d i s \tan c e f r o m t h e h e a v i e r p e r s o n i s g i v e n b y r_{c} = \frac{45 \times 5}{45 + 80} m = 1.8 m R e l a t i v e a c c e l e r a t i o n a = (\frac{50}{45} + \frac{50}{80}) m / s = \frac{125}{72} m / s^{2} S o r e l a t i v e v e l o c i t y v_{r} = \sqrt{2 \times \frac{125}{72} \times 5} m / s = \frac{25}{6} m / s$