Question

A solid sphere rolling on a rough horizontal surface with linear speed \(v\) collides elastically with a fixed, smooth vertical wall. The speed of the sphere after it has started pure rolling in the backward direction is:

• A. \(\dfrac{5v}{7}\)
• B. \(\dfrac{2v}{7}\)
• C. \(\dfrac{7v}{5}\)
• D. \(\dfrac{3v}{7}\)

Hint:

The smooth wall reverses only the linear velocity, not the spin. Conserve angular momentum about the floor contact point until pure rolling returns.

Answer 1

The Correct Option is D

Set the backward direction (away from the wall) as positive. Coming in, the sphere translates at $v$ toward the wall while spinning at $\omega = v/r$ so that it rolls without slipping.

The wall is smooth and the hit is elastic, so only the horizontal velocity flips sign; the spin is untouched. Right after the bounce the centre moves backward at speed $v$, but the sphere is still spinning the way it did before, which now opposes backward rolling.

The cleanest bookkeeping is angular momentum about the point of contact with the floor, because the floor friction produces no torque there. Using $I_{cm} = \tfrac{2}{5}mr^2$ for a solid sphere:

Just after the bounce, translation ($+v$) and the old spin ($-v/r$) give\[L_i = m v r - \tfrac{2}{5}m v r = \tfrac{3}{5}m v r.\]When it finally rolls backward cleanly at speed $v'$ (with $\omega' = v'/r$),\[L_f = m v' r + \tfrac{2}{5}m v' r = \tfrac{7}{5}m v' r.\]Equating $L_i = L_f$:\[\tfrac{7}{5}m v' r = \tfrac{3}{5}m v r \;\Rightarrow\; v' = \frac{3v}{7}.\]\[\boxed{v' = \frac{3v}{7}}\]