Question

The curved part of the surface shown is semi-circular. All the surfaces are frictionless. Speed of the bigger block when the smaller block reaches the point $A$ of the surface is:

• A. $\frac{Mv}{M+m}$
• B. $\frac{mv}{M-m}$
• C. $\frac{Mv}{M-m}$
• D. $\frac{mv}{M+m}$

Answer 1

The correct option is D $\frac{mv}{M+m}$

Smaller block will continue to rise along the curved surface of bigger block until velocities of both blocks become equal.
The horizontal component of the normal reaction (exerted by smaller block) will accelerate the bigger block.
Since $F_{\text{ext}}=0$ on the system of blocks in horizontal direction, applying momentum conservation :
$P=m_{\text{system}}{v}_{\text{com}}=\text{constant}$
Let the common speed of the blocks be $v_0$.
$\Rightarrow (M+m)v_0=m_{\text{system}}{v}_{\text{com}}$
$\therefore v_0=v_{\text{com}}$
Hence, common speed will be equal to the speed of the centre of mass of the system.


Velocity of centre of mass in horizontal direction will remain constant, as $F_{\text{ext}}=0$
$\Rightarrow v_{\text{com}}=\frac{m_1{v}_1+m_2{v}_2}{m_1+m_2}$
For intial condition, $v_1=v,v_2=0$
$\Rightarrow v_{\text{com}}=\frac{mv+0}{m+M}$
$\Rightarrow v_{\text{com}}=\frac{mv}{M+m}...\left(i\right)$
Speed of the bigger block when the smaller block reaches the point $A$ of the surface is $\left(\frac{mv}{M+m}\right)$

Alternate method:

Initial momentum of system in horizontal direction$P_i=mv+M\left(0\right)=mv\hat{i}$

Let velocity of mass $M$ be $v_0\hat{i}$
Now when mass $m$ reaches A, it will be moving vertically relative to $M$,
Let $V_{mM}=v_y\hat{j}$
$\Rightarrow V_m-V_M=v_y\hat{j}$
$\Rightarrow V_m=v_y\hat{j}+v_0\hat{i}$
Final momentum of system in horizontal direction $P_f=(M+m)v_0\hat{i}$
As there is no external force in horizontal direction,
$P_i=P_f$
$\Rightarrow \left(mv\right)\hat{i}=(M+m)v_0\hat{i}$
$\Rightarrow v_0=\frac{mv}{M+m}$