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Question
Particles of masses 2M, m and M are respectively at points A, B and C with AB = ½ (BC). m is much-much smaller than M and at time t = 0, they are all at rest (Figure). At subsequent times before any collision takes place ______.
Options
m will remain at rest.
m will move towards M.
m will move towards 2M.
m will have oscillatory motion.
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Solution
Particles of masses 2M, m and M are respectively at points A, B and C with AB = ½ (BC). m is much-much smaller than M and at time t = 0, they are all at rest (Figure). At subsequent times before any collision takes place m will move towards 2M..
Explanation:
The particle m at B will move towards A with the greater force, due to particle 2M at A.
Force on m at B due to 2M at A is
`F_(BA) = (G(m xx 2M))/(AB)^2` towards `BA`
Force on mass m at B due to mass M at C is
`F_(BC) = (G(M xx M))/(BC)^2` towards `BC`
Therefore, the resultant force on mass m at B due to masses M and 2M is
`F_"net" = F_(BA) - F_(BC)` .....(Because FBA and FBC are acting in opposite directions)
`F_"net" = (2GMm)/(AB)^2 - (GMm)/(BC)^2`
As `(BC) = 2AB`
⇒ `F_"net" = (2GMm)/(AB)^2 - (GMm)/(2AB)^2`
= `(2GMm)/(AB)^2 - (GMm)/(4(AB)^2`
= `(7GMm)/(4(AB)^2` (along BA)
Hence, m will move towards BA (i.e., 2M).
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