Question : Consider a two particle system with particles having masses m1 and m2. If the first particle is pushed towards the centre of mass through a distance d, by what distance should the second particle be moved, so as to keep the centre of mass at the same position?
Doubt by Suhani
Solution :
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m1 O m2
Initially, let us consider that the centre of mass of the two particle system lies at O and the distance of the two particles m1 and m2 from the origin is x1 and x2 then
O(x,y) = [m1x1+m2x2]/[m1+m2] — (1)
m1 O m2
Initially, let us consider that the centre of mass of the two particle system lies at O and the distance of the two particles m1 and m2 from the origin is x1 and x2 then
O(x,y) = [m1x1+m2x2]/[m1+m2] — (1)
When the mass m1 is moved towards the centre of mass with the distance d & mass m2 with the distance d' then then the new distance of masses m1 and m2 from the origin will be x1-d and x2-(-d') i.e. x2+d'
O(x,y) = [m1(x1-d)+m2(x2+d')]/[m1+m2] — (2)
Equating equation (1) and (2)
[m1x1+m2x2]/[m1+m2]=[m1(x1-d)+m2(x2+d')]/[m1+m2]
[m1x1+m2x2]=[m1(x1-d)+m2(x2+d')]
m1x1+m2x2=m1x1-m1d+m2x2+m2d'
0=-m1d+m2d'
m1d=m2d'
(m1/m2)d=d'
d' =(m1/m2)d
m1x1+m2x2=m1x1-m1d+m2x2+m2d'
0=-m1d+m2d'
m1d=m2d'
(m1/m2)d=d'
d' =(m1/m2)d
Hence, in order to keep the CM same we have to move move the mass m2 towards the centre by the distance of (m1/m2)d.