A solid uniform sphere of radius R and mass M rolls without slipping with angular velocity $\omega_0$

when it encounters a step of height $0.4 R$. Find the angular velocity immediately after inelastic impact with the rough step.

I tried to conserve momentum

As after collision the linear velocity is zero .

$\frac{3}{5}mR^2\omega_0 +(2/5)mR^2\omega_0 = (2/5)mR^2\omega_{final}$

By this I m getting the answer as $\frac{5}{2}\omega_0$

But the answer given as $\frac{5}{7}\omega_0$

My thinking is that the collision with the step imparts a momentum impulse $\Delta P$ to the sphere. This impulse is sufficient to reduce the linear momentum of the sphere to zero, so $\Delta P=mv=mR\omega_0$.

At the same time this impulse $\Delta P$ creates an angular impulse which increases the angular momentum of the sphere. The angular impulse equals $\Delta P \times (R-h)$, which is also equal to (the tangential component of $\Delta P) \times R$.

edited Dec 23, 2016 by sammy gerbil