Ncert Exercises & Solutions

Particles of masses \(2M, m\) and \(M\) are respectively at points \(A, B\) and \(C\) with \(A B = \dfrac{1}{2} \left( B C \right).\) The mass \(m\) is much-much smaller than \(M\) and at time \(t = 0\), they are all at rest as given in the figure. At subsequent times before any collision takes place,

1. \(m\) will remain at rest.

2. \(m\) will move towards \(M.\)

3. \(m\) will move towards \(2M.\)

4. \(m\) will have oscillatory motion.

(c) Hint: The motion of mass m will depend on the forces acting on it.

Step 1: Find the net force acting on mass m.

Force on B due to

A = F_{B A} = \frac{G (2 M m)}{{(A B)}^{2}}

towards BA

Force on B due to C

= F = \frac{G M m}{{(B C)}^{2}}

towards BC

Step 2: Find the motion of the mass m.

(B C) = 2 A B

\Rightarrow F = \frac{G M m}{{(2 A B)}^{2}} = \frac{G M m}{4 {(A B)}^{2}} < F_{B A}

Hence, m will move towards BA (i.e., 2M).

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