Question

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 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 $v=2v_0$, 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 1

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,
$mv=m{v}_1+4m{v}_2$
i.e. $v=v_1+4v_2$
Further, as collision is elastic
$\frac{1}{2}m{v}^2=\frac{1}{2}m{v}_1^2+\frac{1}{2}4m{v}_2^2,$
i.e. $v^2=v_1^2+4v_2^2$
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
$s>2d$
However, the motion of $B$ is retarded by frictional force $f=\mu (4m+2m)g$ between table and its lower surface.
So, the distance moved by $B$ till it stops
$0=v_2^2-2\left(\frac{6\mu mg}{4m}\right)s,$
i.e. $s=\frac{v_2^2}{3\mu g}$
Substituting this value of $s$ in Eq. (iii), we find that for toppling of $A$
$v_2^2>6\mu gd$
or $\frac{2}{5}v>\sqrt{6\mu gd}$
$[$ as $v_2=2v/5]$
i.e. $v>\frac{5}{2}\sqrt{6\mu gd}$
or $v_{\min }=v_0=\frac{5}{2}\sqrt{6\mu gd}$
(b) If $v=2v_0=5\sqrt{6\mu gd}$, the object will rebound with speed $v_1=\frac{3}{5}v=3\sqrt{6\mu gd}$
and as time taken by it to fall down
$t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2d}{g}}$
$[$ as $h=d]$
The horizontal distance moved by it to the left of $P$ in this time $x=v_{1}t=6d\sqrt{3\mu }$
- 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
$v_1^{\prime }=\left(\frac{m_1-m_2}{m_1+m_2}\right)v_2+\left(\frac{2m_2}{m_1+m_2}\right)v_2$
and $v_2^{\prime }=\left(\frac{m_2-m_1}{m_1+m_2}\right)v_2+\left(\frac{2m_1}{m_1+m_2}\right)v_1$