Question

Particles of masses 2M, m and M are respectively at points A, B and C with `AB=(1)/(2)` (BC). M is much smaller than M and at time t = 0, they are all at rest given in the figure. 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 2 M
D. m will have oscillatory motion

Answer 1

Correct Answer - C
Gravitational force on m due to `2M = F_(BA)`
`= (G(2M)m)/((AB)^(2))` along `vec(BA)" "` …. (1)
Gravitational force on m due to `M = F_(BC)`
`=(G(M)m)/((BC)^(2))` along `vec(BC)`
`because BC=2AB " " therefore " " F_(BC)=(GMm)/(4AB^(2))" "` ...... (2)
Thus from (1) and (2) we find that `F_(BA)gt F_(BC)`
`therefore` m will move towards 2M along `vec(BA)`.