Ncert Exercises & Solutions

Particles of masses 2M, m and M are respectively at points A, B and C with $A B = \frac{1}{2} (B C)$ . 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).

NEET Information

courses

Company

Botany Questions

Chemistry Questions

Physics Questions

Zoology Questions