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A solid sphere is rolling without slipping on  a rough horizontal plane. It collides elastically with another identical solid sphere at rest. If the surfaces of the spheres are smooth; and radius of each sphere is " r " and mass " m ". then 


Total kinetic energy of the first sphere after it starts pure rolling.
 

Option: 1

\frac{1}{35} m r^2 \omega^2


Option: 2

\frac{2}{35} m r^2 \omega^2


Option: 3

\frac{3}{35} m r^2 \omega^2


Option: 4

\frac{4}{35} m r^2 \omega^2


Answers (1)

best_answer

Applying conservation of angular momentum

m v r+\frac{2}{5} m r^2 \omega=m v^{\prime} r+\frac{2}{5} m r^2\frac{v\prime}{r}+mvr

v^{\prime}=\frac{2}{7} r \omega

Kinetic energy of the first sphere-

K_T =\frac{1}{2} m v^{\prime 2}+\frac{1}{2} I_{c m} \omega^{\prime 2} \\

=\frac{1}{2} m v^{\prime 2}+\frac{1}{2} \times \frac{2}{5} m r^2 \omega^{^{\prime2}}

=\frac{1}{2} m v^{\prime 2}+\frac{1}{5} m v^{\prime 2} \\

=\frac{7}{10} m v^{\prime 2} \\

=\frac{7}{10} m \times \frac{4}{45} r^2 \omega^2 \\

=\frac{2 m r^2 \omega^2}{35}

Posted by

sudhir.kumar

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