Question

A block of mass of $m$ is projected along the surface of a smooth circular wall with velocity $v$. What is the value of average normal force exerted by the circular wall on the block during its motion from $A$ to $B$. Assume uniform speed during the journey.

• A. $\frac{m{v}^2}{R}$
• B. $\frac{m{v}^2}{\pi R}$
• C. $\frac{2m{v}^2}{R}$
• D. $\frac{2m{v}^2}{\pi R}$

Answer 1

The correct option is D $\frac{2m{v}^2}{\pi R}$


Time taken by block to reach $B$ from $A$ at uniform speed $v$:
$\Delta t=\frac{\text{distance}}{\text{speed}}=\frac{\pi R}{v}$
Change in momentum of the block is :
$\overrightarrow{\Delta P}=\overrightarrow{P_f}-\overrightarrow{P_i}$
$\therefore \overrightarrow{\Delta P}=(-mv)\hat{i}-(+mv)\hat{i}=-2mv\hat{i}$

Change in momentum is along $-ve$ x-direction, hence direction of average normal force exerted by wall will also be along $-ve$ x-direction. Since there is no change in magnitude in velocity, net force on the particle is in the radial direction, which can only be provided by the normal force from wall.
${\overrightarrow{F}}_{\text{avg}}=\frac{\overrightarrow{\Delta P}}{\Delta t}\Rightarrow |{\overrightarrow{F}}_{\text{avg}}|=\frac{|\overrightarrow{\Delta P}|}{\Delta t}$
($\because$ Magnitude of change in momentum$=2mv$)
$|{\overrightarrow{F}}_{\text{avg}}|=\frac{2mv}{\left(\frac{\pi R}{v}\right)}=\frac{2m{v}^2}{\pi R}$
is the magnitude of average normal force exerted by the wall of the surface on the block