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, B, C, D

Explanation

Since the collision is elastic in nature applying conservation
of linear momentum and conservation of kinetic energy
`mv=(4m)u+mv'`
where u is the velocity of mass 4m after collision and v' is
the velocity of mass 2m
`rArr v'=v-4u....(i)`
Also, `1/2mv^2=1/2(4m)u^2+1/2mv'^2`
`rArr v^2=4u^2+v^2....(ii)`
From (i) to (ii)
`v^2=4u^2+(v-4u)^2 rArr u=(2v)/5`

Block B starts moving but the block A remains at rest as
there is no friction between A and B
For block A to topple, block B should move a distance 2d.
Let the retardation produced in B due to friction force
between B and the table be a
`F=muN rArr (4m)a= mu(6mg) rArr a=1.5mug`
For the motion of B,
`u=(2v)/5, v=0, s=2d, a=-1.5 mug`
Now, `v^2-u^2=2as rArr (0)^2-((2v)/5)^2=2(-1.5mug)2d`
`rArr v=5/2sqrt(6mugd)`
For elastic collision between two bodies
`v_1=((m_1-m_2)u_1)/(m_1+m_2)+(2m_2u_2)/(m_1+m_2)`
Here, `m_1=m, m_2=4m, u_1=5sqrt(6mugd),u_2=0`
`rArr v_1=((m-4m)5sqrt(6mugd))/(m+4m)+0`

`=-3xx5(sqrt6mugd)/5=-3sqrt6mugd`
The negative sigh shows that the mass m rebounds.
It then follows a projectile motion.
Considering the vetical direction motion of this projectile.
`u_y=0, s_y=d, a_y=g, t_y=?`
`s=ut+1/2at^2 rArr d=1/2gt^2 rArr t=sqrt((2d)/g)`
The horizontal distance travelled by mass m during this
time t
`x=3sqrt(6mugd)xxsqrt((2d)/g)=6sqrt(3mud^2)=6dsqrt(3mu)`