USAPhO 1997, Четвъртфинал — Задача 2

• (a) 2.2 m
• (b) Yes, the box lands in the flowers ($x = 1.9$ m).

(a, 10) Work energy theorem: $W = \Delta K$ (2)

Setting up the work energy theorem,
$[\,mg\sin\theta - \mu N\,]L = \tfrac{1}{2}mv^2 - 0$, (4)

where $N$ is the normal force. But from $F = ma$ normal to the roof, $N = \mu mg\cos\theta$ (2)

thus, $L = \tfrac{1}{2}v^2 / [g\,(\sin\theta - \mu\cos\theta)]$
and numerically,
$L = \tfrac{1}{2}(3.5\ \mathrm{m/s})^2 / [(9.8\ \mathrm{m/s^2})\,(\sin 30^{\circ} - \tfrac{1}{4}\cos 30^{\circ})\,] = 2.2\ \mathrm{m}$. (2)

(-1 for incorrect units; -1 for gross neglect of significant figures)

Alternative method:
$F = ma$ tangent to the roof gives $a = g\sin\theta - \mu N$ [4]
$= g\sin\theta - \mu g\cos\theta$ [2]
$= 2.8\ \mathrm{m/s^2}$. [1]

Using $v^2 - v_o^2 = 2a(x - x_o)$ [2]

leads to $L = v^2 / 2a = 2.2$ m. [1]

(b, 10) Projectile problem:
$v_{oy} = (3.5\ \mathrm{m/s})\sin 30^{\circ} = 1.75\ \mathrm{m/s}$, (1)
$v_{ox} = (3.5\ \mathrm{m/s})\cos 30^{\circ} = 3.03\ \mathrm{m/s} \approx 3.0\ \mathrm{m/s}$. (1)

For the $x$-direction,
$x = v_{ox}\,t$, (2)

For the $y$-direction (taking down positive)
$y = v_{oy}\,t + \tfrac{1}{2}gt^2$. (2)

Solve the equations simultaneously by some appropriate strategy. For instance, eliminating $t$ in the $y$-equation by using the $x$-equation: $y = v_{oy}\,(x/v_{ox}) + \tfrac{1}{2}g(x/v_{ox})^2$ (2)

Putting in numerical values, solving the quadratic equation correctly, choosing the + root in the quadratic formula (with the − root one gets an $x \lt 0$, which in inside the house), one finds $x = 1.9$ m. (2)
Yes, the box lands in the flowers.