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

Step 1: Before the collision the sphere rolls toward the wall with linear speed \(v\) and angular speed \(\omega = \dfrac{v}{r}\). Take the backward direction (away from the wall) as positive.

Step 2: The wall is smooth, so it exerts no vertical friction and does not change the spin. The collision is elastic, so the normal (horizontal) velocity just reverses. Just after impact the translational velocity is \(+v\) (backward), but the spin is still \(\omega = \dfrac{v}{r}\) in the forward-rolling sense (which is negative in our sign convention).

Step 3: On the rough floor, friction acts at the contact point. Angular momentum about the contact point on the ground is conserved (friction, normal force, and gravity give no torque about that point). For a solid sphere \(I_{cm} = \dfrac{2}{5}mr^2\).

Just after impact:\[L_i = mv\,r + I_{cm}\left(-\frac{v}{r}\right) = mvr - \frac{2}{5}mvr = \frac{3}{5}mvr.\]
Step 4: When pure rolling is reached backward, \(v' = \omega' r\), so\[L_f = mv'r + \frac{2}{5}mr^2\cdot\frac{v'}{r} = \frac{7}{5}mv'r.\]
Step 5: Conserving angular momentum, \(\dfrac{7}{5}mv'r = \dfrac{3}{5}mvr\), giving\[v' = \frac{3v}{7}.\]This is option (D).\[\boxed{v' = \frac{3v}{7}}\]