Center of MassEasy
Question
Two skaters A and B of mass 50 kg and 70 kg, respectively, stand facing each other, 6 metres apart on a horizontal smooth surface. They pull a rope stretched between them. How far has each moved when they meet :-
Options
A.Both have moved 3m
B.A moves 4 m and B moves 2m
C.A moves 2.5 m and B moves 3.5 m
D.A moves 3.5 m and B moves 2.5 m
Solution
Let mA and mB be the masses of skaters A and B and aA and aB their respective accelerations, when they pull at each other. From Newton′s third law, action and reaction forces are equal m magnitude, i.e.
mAaA = mBaB or mA
= mB 
or mAvA = mBvB or mA2vA2 = mB2vB2 ...(i)
where vA and vB are their respective speeds and t is the time taken for them to meet. Let sA and sB be the distances travelled by them when they meet, we have,
2aAsA = v2A and 2aBsB = v2B
Using these equations in Eq. (i), and mAaA = mBaB,
we get
Since sA + sB = 6m; sA = 3.5 and sB = 2.5 m.
mAaA = mBaB or mA
= mB or mAvA = mBvB or mA2vA2 = mB2vB2 ...(i)
where vA and vB are their respective speeds and t is the time taken for them to meet. Let sA and sB be the distances travelled by them when they meet, we have,
2aAsA = v2A and 2aBsB = v2B
Using these equations in Eq. (i), and mAaA = mBaB,
we get
Since sA + sB = 6m; sA = 3.5 and sB = 2.5 m.
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