# Vertically falling sphere striking a wedge and rolling without slipping

I have a question which states that a sphere of mass $$m$$ falls vertically and strikes a wedge of mass $$M$$ and inclination $$\alpha$$, at rest on a smooth horizontal plane. There is sufficient friction on the wedge's surface which cause the sphere to start pure rolling, i.e, rolling without slipping. If the spheres's pre-impact velocity is $$v$$, then what is the vertical component of its post-impact velocity?

I let the final velocity of the sphere be $$v'$$ and its radius be $$r$$.

I thought that this question will be most easily solved with angular momentum conservation, but, mistakenly, I conserved it about the point at which the sphere strikes. This is invalid since the wedge accelerates and torque cannot be calculated about that point.

Then, I thought of linear and angular impulse-momentum theorem, using which, I get, \begin {align} N\Delta t &= mv\cos \alpha \qquad \text{Linear momentum perpendicular to wedge}\\ \mu N\Delta t &= mv' - mv\sin \alpha \qquad \text{Linear momentum parallel to wedge}\\ \mu N\Delta tr &= I\omega \qquad \text{Angular momentum about Center of Mass} \end {align}

I do not know the answer to the assignment, but from the above I get, $$v' = \dfrac{5v\sin \alpha}{7}$$

Is this correct? If not, where did I go wrong? Any help or hint will be appreciated.

PS: I know high school mechanics problems are more easily solved using advanced techniques like Lagrangian mechanics, but I would appreciate it, if the answers were given using Newtonian mechanics only. Thanks.

• If the wedge is movable, then the "Linear Momentum perpendicular to wedge" equation is wrong and it will have some final velocity perpendicular to the wedge – Archimedesprinciple May 29 at 1:11
• @Archimedesprinciple Then how should I proceed? – Aaratrick May 29 at 1:52