A block 'A' of mass 2m is placed on another block 'B' of mass 4m which in turn is placed on a fixed table. The two blocks have a same length 4d and they are placed as shown in fig The coefficient of friction (both static and kinetic) between the block 'B' and table is `mu`. There is no friction between the two blocks. A small object of mass m moving horizontally along a line passing through the centre of mass (cm.) of the block B and perpendicular to its face with a speed v collides elastically with the block B at a height d above the table.

(a) What is the minimum value of v(call it `v_0`) required to make the block A topple?
(b) If `v=2v_0`,find the distance (from the point P in the figure ) at which the mass m falls on the table after collision. (Ignore the role of friction during the collision).

Answer

(a) `(5)/(2) sqrt(6 mu gd)` , (b) 6d sqrt(3 mu)`

Explanation

If `v_(1)` and `v_(2)` are the velocities of object of mass `m` and block of mass `4m`, just after collision then by conservation of momentum,
`m v = m v_(1) + 4m v_(2)`,
i.e., `v = v_(1)+4v_(2)` …(1)
Further, as collision is elastic
`(1)/(2) mv_(2) = (1)/(2)m v_(1)^(2)+(1)/(2)4mv_(2)^(2)`, i.e.,
`v^(2) = v_(1)^(2) + 4v_(2)^(2)` ...(2)
Soliving, these two equations we get either
`v_(2) = 0` or `v_(2) = (2)/(5)v`
Therefore, `v_(2) = (2)/(5)v` , Substituting in Eq. (1)
`v_(1) = (3)/(5) v`
When `v_(2) = 0, v-(1) = v_(2)`, but it is physically unacceptable.
(a) Now, after collision the block `B` will start moving with velocity `v_(2)` to the right. Since, there is no friction between blocks `A` and `B` the upper block `A` will stay at its position and will topple if `B` moves a distance `s` such that `s gt 2d` ....(3)
However, the motion of `B` is retarded by frictional force `f = m(4m+2m)g` between table and iis lower surface. So, the distance moved by `B` till stops
`0 = v_(2)^(2) gt 6 gd` or `(2)/(5)v gt sqrt( 6 mu gd) [as v_(2) = (2v)/(5)]`
i.e., `v gt (5)/(2) sqrt(6m)` or `v_("min") = v_(0) = (5)/(6) sqrt(mu g d)`
(b) If `v = 2v_(0) = 5 sqrt(6 mu gd)`, the object will rebound
with speed , `v_(1) = (3)/(5) v = 3 sqrt(6 mu gd)`
and as time taken by it to fall down ltbgt `t = sqrt((2h)/(g)) = sqrt((2d)/(g)) [as h = d]`
The horizontal distance moved by it to the left of `P` in this time `x = v_(1)t = 6d sqrt(3 mu)`