Question

a solid sphere rolling on a rough horizontal surface with a linear speed v collides elastically with a fixed smooth vertical wall find the speed of the sphere after it has started ​

Answer 1

Answer:

The speed of the sphere is $\frac{3}{7}v$

Explanation:

The complete question is:

A solid sphere rolling on a rough horizontal surface with a linear speed v collides elastically with a fixed smooth vertical wall find the speed of the sphere after it has started ​pure rolling in backwards direction.

Solution:

After the collision, the linear velocity of the solid sphere will be reversed but its angular velocity will be same

Angular momentum w.r.t the centre of mass of the sphere just after collision

$L=mvr-I\omega$

Where, $I$ is the moment of inertia and $\omega$ is the angular velocity, $r$ is the radius of the solid sphere

If the velocity of the sphere is v' when it starts pure rolling then the angular momentum

$L'=mv'r+I\omega'$

The angular momentum around the centre of mass will be conserved as there is no external force

Thus,

$L=L'$

$\implies mvr-\frac{2}{5}m\times r^2\times \frac{v}{r}=mv'r+\frac{2}{5}m\times r^2\times \frac{v'}{r}$

$\implies mvr-\frac{2mvr}{5}=mv'r+\frac{2mv'r}{5}$

$\implies \frac{7mv'r}{5}=\frac{3mvr}{5}$

$\implies v'=\frac{3}{7}v$

Therefore, the speed of the sphere is $\frac{3}{7}v$

Hope this helps.

Answer 2

Answer:

3u/7...

Explanation:

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