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Consider a gravity free container as shown. System is initially at rest and electric potential in the regon is $V = (y^3+2)\ J/C$. A ball of charge $q$ and mass $m$ is released from rest from base starts to move up due to electric field and collides with the shaded face as shown.If its speed just after collision is $1.5\ m/s$ and time for which ball is in contact with shaded face is $0.1\ sec$, find external force required to hold the container fixed in its position during collision assuming ball exerts constant force on wall during entire span of collision.......$N$
$70$
$72$
$74$
$76$
Solution
$\mathrm{V}=\mathrm{y}^{3}+2$
$\Rightarrow \mathrm{E}=\frac{-\delta V}{\delta y} \hat{i}=-3 y^{2} \hat{j}$
$\mathrm{V}_{\mathrm{A}}=2 \mathrm{\,volt}$
$\mathrm{V}_{\mathrm{B}}=10$ volt $\left[\mathrm{V}=\mathrm{y}^{3}+2\right]$
$\mathrm{q}\left(\mathrm{V}_{\mathrm{B}}-\mathrm{V}_{\mathrm{A}}\right)=\frac{1}{2} m v^{2} \Rightarrow \frac{1}{2}(8)=\frac{1}{2}(2) \,V^{2}$
$\Rightarrow \mathrm{V}=2 \mathrm{\,m} / \mathrm{s}$
So, velocity of ball before collision $ = (2{\rm{\,m}}/{\rm{s}})j$
So, velocity of ball after collision $ = – (1.5\,{\rm{m}}/{\rm{s}})j$
change in momentum $ = m\left( {{{\overrightarrow V }_F} – {{\overrightarrow V }_i}} \right) = ( – 7\,N.S)j$
Net force $ = ( – 7)/(0.1) = ( – 70\,{\rm{N}}){\rm{j}}$
from $\mathrm{FBD}$ of ball during collision
${{\rm{F}}_{{\rm{net}} = }}{\rm{ = }}{{\rm{F}}_{{\rm{wall}}}} – {\rm{qE}}$
$\mathrm{F}_{\mathrm{wall}}=\mathrm{F}_{\mathrm{net}}+\mathrm{q} \mathrm{E}$
$=(70+6)=76 \mathrm{\,N}$
${\rm{[E}}$ at top face $ = 3{{\rm{y}}^2} = 3{(2)^2} = 12\,{\mkern 1mu} {\rm{N}}/{\rm{C]}}$
