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NCERT Exemplar · Q8

Q.Three particles are placed at rest on a single straight line: a particle of mass 2M2M at point A, a particle of mass mm at point B, and a particle of mass MM at point C, with B lying between A and C (order A, B, C). The separation AB is half of the separation BC, i.e. AB=12 BCAB = \tfrac{1}{2}\,BC. The mass mm is much, much smaller than MM, and at time t=0t = 0 all three particles are released from rest. Considering the instants just after release (before any collision takes place):

(a) mm will remain at rest.
(b) mm will move towards MM.
(c) mm will move towards 2M2M.
(d) mm will have oscillatory motion.
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Both heavy masses attract the small mass mm. Because 2M2M is heavier and closer (its distance is half that of MM), the pull from 2M2M overwhelms the pull from MM by a factor of eight. So the net force on mm is directed towards 2M2M, and mm begins to move in that direction. There is no restoring arrangement, so the motion is not oscillatory.

Concept

Each heavy mass exerts an attractive gravitational force on mm given by Newton's law of gravitation, F=G m1m2d2F = \dfrac{G\,m_1 m_2}{d^2}. The net force on mm is the vector sum of the two pulls, which here act along the same line in opposite directions.

Setting up distances

Since AB=12BCAB = \tfrac{1}{2}BC, let AB=dAB = d and BC=2dBC = 2d. Then:

  • Distance from mm (at B) to 2M2M (at A) =d= d.
  • Distance from mm (at B) to MM (at C) =2d= 2d.

Comparing the two forces

Force on mm towards 2M2M (to the left, towards A):

F2M=G(2M)md2=2GMmd2.F_{2M} = \frac{G(2M)m}{d^2} = \frac{2GMm}{d^2}.

Force on mm towards MM (to the right, towards C):

FM=GMm(2d)2=GMm4d2.F_{M} = \frac{GMm}{(2d)^2} = \frac{GMm}{4d^2}.

Taking the ratio:

F2MFM=2GMm/d2GMm/4d2=8.\frac{F_{2M}}{F_{M}} = \frac{2GMm/d^2}{GMm/4d^2} = 8.

The pull towards 2M2M is eight times larger, so the resultant force on mm is directed towards 2M2M. …

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