Question

Two skaters A and B of mass 50 and 60 kg, respectively, stand facing each other, 11m apart
on a horizontal smooth surface. They pull a rope stretched between them. How far each has
moved when they meet ?

Answer 1

A moves 3.5 metres and B moves 2.5 metres.

Here no any external force is applied taking two stakers A and B as a system.
then center of mass of both stakers will not change therefore they meat at center of mass, taking axis as shown in figure 1.
X
com
​
=
50+70
50×0+70×6
​
=
120
70×6
​
=3.5m from 50kg sketers.
X
com
​
= X-coordinate of center of mass.
So A mover 3.5cm
B mover (6 - 3.5) = 2.5m

Answer 2

A moves 3.5 meters and B moves 7.5 meters.

GIVEN

Mass of A = 50 kg

Mass of B = 60 kg

Distance between A and B = 11 meter

TO FIND

The distance each skater has moved.

SOLUTION

We can simply solve the above problem as follows-

Since there is no external force acting on A and B, so the center of mass of A and B will not change.

We can assume the whole scenario as a system.

So, A and B will meet at the center of mass.

So, X - coordinate of the center of mass = X(com)

X(com) = 50×0 + 70×6 /50+70

= 420/120

= 3.5 meter

B moves = 11-3.5 = 7.5 meters.

Hence, A moves 3.5 meters and B moves 7.5 meters.

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