Ice Objects

The sphere will fall straight downward and displace the wedges, thrusting them outward. The total energy of the system equals the initial energy. We don't have to consider the potential energy of the wedges because their height will not change. Using $m$ as the mass of the sphere and $H$ as the height of the wedges, the initial energy is therefore the potential energy of the sphere, $mgH$ . The energy at each subsequent point in time is

$E(t)=E_0\\ mgh(t)+\frac{1}{2}mv(t)^2+2\times \frac{1}{2}Mv_w(t)^2=mgH\\ m[gh(t)+\frac{1}{2}v(t)^2]+Mv_w(t)^2=mgH,$

where $M$ is the mass of one wedge, $v(t)$ is the downward speed of the sphere, and $v_w(t)$ is the outward speed of a wedge. The downward acceleration $a$ of the sphere will be constant. If there is any doubt about this, we will see in the following equations that $a$ is not a function of time and is therefore constant. Therefore, we can use the equations of motion $v(t)=v_0+at$ and $x(t)=x_0+v_0t+\frac{1}{2}at^2$ , which in this case become

$v(t)=at\\ h(t)=H-\frac{1}{2}at^2,$

respectively. Furthermore, we can relate $v(t)$ and $v_w(t)$ by noting the geometry of the wedges. For each meter the sphere falls, the wedges will each move outward by a certain fraction of a meter. To determine this fraction, it is easiest to consider a single wedge--let's say the one on the right. It is hard to figure out the fraction based on the movement of the sphere in this problem, but we can imagine a different case in which the point of contact of the sphere with the wedge is initially exactly at the top of the wedge (slightly higher than in the problem) and ends at ground level (meaning part of the sphere is below ground, which is impossible, but we just want to look at the math). We can then see that, hypothetically, the sphere would have fallen a height of $H$ , and the bottom acute corner of the wedge would have to be at the same horizontal position as the initial position of the top of the wedge. For every $H$ that the sphere falls, the wedges are each displaced a distance of $B$ , the base width of the wedge. Since the distance traversed by a wedge is $\frac{B}{H}$ the distance traversed by the sphere, it follows that $v_w(t)=\frac{B}{H}v(t)=\frac{Bat}{H}$ . We can insert the three new expressions into the equation of conservation of energy as follows to find the acceleration:

$m[g(H-\frac{1}{2}at^2)+\frac{1}{2}(at)^2]+M(\frac{Bat}{H})^2=mgH\\ \frac{MB^2a^2t^2}{H^2}=\frac{1}{2}at^2m(g-a)\\ \frac{MB^2a}{H^2}=\frac{1}{2}m(g-a)\\ a=\frac{g}{1+2\left(\frac{B}{H}\right)^2\left(\frac{M}{m}\right)}$

As promised, the acceleration is not a function of $t$ and is therefore constant. Finally, we can solve the equation for $h(t)$ using a final height of the unknown radius of the sphere (it is possible but unnecessary to calculate) to get

$t=\sqrt{\frac{2(H-R)}{a}}$ .

Here we see that the time it takes to hit the ground is inversely proportional to the square root of the acceleration. Without the wedges, the acceleration is simply $g$ , so the ratio of times of each case is the inverse of the ratio of square roots of these accelerations. The increase, then, is obtained by subtracting 1 from this ratio:

$\frac{t}{t_0}-1=\frac{\sqrt{g}}{\sqrt{a}}-1=\boxed{\sqrt{1+2\left(\frac{B}{H}\right)^2\left(\frac{M}{m}\right)}-1}$ .

Inserting 2 for $B$ , 10 for $H$ , 64 for $M$ , and 32 for $m$ and multiplying by 100% gives a final answer of 7.70% .