Question

A block $A$ of mass $2 \,m$ is placed on another block $B$ of mass $4\, m$ which in turn is placed on a fixed table. The two blocks have a same length $4 d$ and they are place ${ }^{-1}$ is shown in figure. 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 $A$ to topple?

(b) If $v=2 v_{0}$, 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 1

If $v_{1}$ and $v_{2}$ are the velocities of object of mass $m$ and block of mass $4 m$, just after collision then by conservation of momentum,
$m v=m v_{1}+4 m v_{2} $
i.e. $ v=v_{1}+4 v_{2}$
Further, as collision is elastic
$\frac{1}{2} m v^{2}=\frac{1}{2} m v_{1}^{2}+\frac{1}{2} 4 m v_{2}^{2}, $
i.e. $ v^{2}=v_{1}^{2}+4 v_{2}^{2}$
Solving, these two equations we get either
$v_{2}=0 $ or $ v_{2}=\frac{2}{5} v$
Therefore, $v_{2}=\frac{2}{5} v$
Substituting in Eq. (i) $v_{1}=\frac{3}{5} v$
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
$s >2 d$
However, the motion of $B$ is retarded by frictional force $f=\mu(4 m+2 m) g$ between table and its lower surface.
So, the distance moved by $B$ till it stops
$0=v_{2}^{2}-2\left(\frac{6 \mu m g}{4 m}\right) s, $
i.e. $ s=\frac{v_{2}^{2}}{3 \mu g}$
Substituting this value of $s$ in Eq. (iii), we find that for toppling of $A$
$v_{2}^{2} >6 \mu g d$
or $\frac{2}{5} v>\sqrt{6 \mu g d}$
$\left[\right.$ as $\left.v_{2}=2 v / 5\right]$
i.e. $v >\frac{5}{2} \sqrt{6 \mu g d} $
or $v_{\min }=v_{0}=\frac{5}{2} \sqrt{6 \mu g d}$
(b) If $v=2 v_{0}=5 \sqrt{6 \mu g d}$, the object will rebound with speed $v_{1}=\frac{3}{5} v=3 \sqrt{6 \mu g d}$
and as time taken by it to fall down
$t=\sqrt{\frac{2 h}{g}}=\sqrt{\frac{2 d}{g}}$
$[$ as $h=d]$
The horizontal distance moved by it to the left of $P$ in this time $x=v_{1} t=6 d \sqrt{3 \mu}$
- Toppling will take place if line of action of weight does not pass through the base area in contact.
- $v_{1}$ and $v_{2}$ can be obtained by using the equations of head on elastic collision
$v_{1}^{\prime}=\left(\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\right) v_{2}+\left(\frac{2 m_{2}}{m_{1}+m_{2}}\right) v_{2} $
and $v_{2}^{\prime}=\left(\frac{m_{2}-m_{1}}{m_{1}+m_{2}}\right) v_{2}+\left(\frac{2 m_{1}}{m_{1}+m_{2}}\right) v_{1}$