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Particles of masses 2 M, 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). - Physics

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प्रश्न

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 ______.

विकल्प

  • m will remain at rest.

  • m will move towards M.

  • m will move towards 2M.

  • m will have oscillatory motion.

MCQ
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उत्तर

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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अध्याय 8: Gravitation - Exercises [पृष्ठ ५९]

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एनसीईआरटी एक्झांप्लर Physics [English] Class 11
अध्याय 8 Gravitation
Exercises | Q 8.8 | पृष्ठ ५९

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