USAPhO 1997, Четвъртфинал — Задача 3

• (a) $K_1 = \tfrac{1}{2}I\omega_o^2 + mg(h - R)$
• (b) $v_x = I\omega_o/mR,\quad v_y = [\,2\beta K_1/m - v_x^2\,]^{1/2}$

(a, 3)
By conservation of energy,
$\tfrac{1}{2}I\omega_o^2 + mgh = K_1 + mgR$
so that
$K_1 = \tfrac{1}{2}I\omega_o^2 + mg(h - R)$ (3)

(b, 17)

Let $(v_x, v_y)$ = velocity of the ball immediately after the collision is over.
Given:
$\beta K_1 = \tfrac{1}{2}m(v_x^2 + v_y^2)$ (2)
Since $\beta$ and $K_1$ are known, if either of $v_x$ or $v_y$ is known then so is the other. Let us find $v_x$.
(Note that linear momentum is not conserved.)

In the collision, the angular frequency goes from $\omega_o$ to $0$.
Let $f$ be the average force of friction that acts horizontally on the ball during the collision. Then from torque = $\Delta L/\Delta t$,
$fR = I\omega_o/\Delta t$
where $\Delta t$ is the collision time. (5)

Now apply $F = m\,\Delta v/\Delta t$ in the $x$-direction.
The $x$-component of velocity changes from $0$ to $v_x$ during time $\Delta t$, and the horizontal force is $f$:
$f = m\,v_x/\Delta t$. (5)

From the torque equation we have $f\Delta t = I\omega_o/R$, so combining the torque and force equations, (2)
$I\omega_o/R = m\,v_x$
and solving for $v_x$,
$v_x = I\omega_o / mR$ (2)

Now $v_x$ is known. Thus, so is $v_y$.
$v_y = [\,2\beta K_1/m - v_x^2\,]^{1/2}$ (1)