Question

Particles of masses 2M, m and M are respectively at points A, B and C with AB = 1/2 (BC). m is much-much smaller than M and at time t = 0, they are all at rest. At subsequent times before any collision takes place

• A. M will remain at rest.
• B. M will move towards M.
• C. M will move towards 2M.
• D. M will have oscillatory motion.

Answer 1

Answer: (C)

M will move towards 2M.

Answer 2

Force on mass m at B due to mass 2M at A is

Force on mass m at B due to mass M at C is

$\therefore$ resultant force on mass m at B due to masses at A and C is

$\left( \because \mathrm { F } _ { 1 } \right.$ and $\left. \mathrm { F } _ { 2 } \right)$ are acting in opposite directions

$= \frac { 2 \mathrm { GmM } } { ( \mathrm { AB } ) ^ { 2 } } - \frac { \mathrm { GmM } } { ( \mathrm { BC } ) ^ { 2 } }$

Therefore, m will move towards 2 M.