Question

Ball Drop A ball with uniform density ρb is placed on the surface of a pool with depth d and liquid density ρp < ρb. Another identical ball is lifted a height h above the pool, and then both balls are released at the same time. In order for both balls to touch the bottom of the pool at the same time, the condition d = nh must be met for some dimensionless n that depends on the values of ρp and ρb. If we define r = (ρb − ρp) / ρb Then we can express n as n = (Ar^3 + Br^2 + Cr) / (Dr^2 + Er + F) Where A, B, C, D, E, F are all nonzero integers, gcd(A, B, C, D, E, F) = 1, and A > 0. What is A + B + C + D + E + F? You may assume that the only forces present are gravity and the buoyant force from the pool. The airborne ball retains all of its energy as it enters the pool.

• A. 1

Answer 1

Answer: (A)

1

Answer 2

Let's analyze the motion of each ball.

1. Motion of the ball placed on the surface (Ball 1):

The ball starts at rest on the surface of the pool. Since $\rho_b > \rho_p$, it will sink. The forces acting on the ball inside the water are gravity ($mg$) downwards and buoyant force ($F_B$) upwards. Mass of the ball $m = V\rho_b$, where $V$ is the volume of the ball. Buoyant force $F_B = V\rho_p g$. Net force $F_{net} = mg - F_B = V\rho_b g - V\rho_p g = Vg(\rho_b - \rho_p)$. The acceleration of the ball in water, $a_w$, is constant: $a_w = \frac{F_{net}}{m} = \frac{Vg(\rho_b - \rho_p)}{V\rho_b} = g\left(1 - \frac{\rho_p}{\rho_b}\right)$. Given $r = \frac{\rho_b - \rho_p}{\rho_b} = 1 - \frac{\rho_p}{\rho_b}$, we have $a_w = gr$. The ball travels a distance $d$ (depth of the pool) from rest. Using the kinematic equation $d = v_0 t + \frac{1}{2}at^2$: $d = 0 \cdot t_1 + \frac{1}{2}(gr)t_1^2 \implies t_1^2 = \frac{2d}{gr} \implies t_1 = \sqrt{\frac{2d}{gr}}$.

2. Motion of the ball dropped from height h (Ball 2):

This motion has two parts: free fall in air and motion in water.

• Part A: Free fall in air (from height h to the water surface)

  The ball starts from rest ($v_0=0$) and falls a distance $h$ under gravity ($g$). Time taken $t_{2A}$: $h = 0 \cdot t_{2A} + \frac{1}{2}gt_{2A}^2 \implies t_{2A} = \sqrt{\frac{2h}{g}}$. Velocity just before hitting the water surface $v_{in}$: $v_{in}^2 = v_0^2 + 2gh = 0 + 2gh \implies v_{in} = \sqrt{2gh}$. The problem states the ball retains all its energy, meaning $v_{in}$ is the initial velocity for the motion in water.

• Part B: Motion in water (from surface to bottom)

  The ball enters the water with velocity $v_{in} = \sqrt{2gh}$. The acceleration in water is $a_w = gr$ (same as for Ball 1). The ball travels a distance $d$. Using the kinematic equation $d = v_{in} t_{2B} + \frac{1}{2}a_w t_{2B}^2$: $d = \sqrt{2gh} t_{2B} + \frac{1}{2}(gr)t_{2B}^2$. This is a quadratic equation for $t_{2B}$: $\frac{1}{2}gr t_{2B}^2 + \sqrt{2gh} t_{2B} - d = 0$. Using the quadratic formula, $t_{2B} = \frac{-\sqrt{2gh} \pm \sqrt{(2gh) - 4(\frac{1}{2}gr)(-d)}}{2(\frac{1}{2}gr)} = \frac{-\sqrt{2gh} \pm \sqrt{2gh + 2grd}}{gr}$. Since $t_{2B}$ must be positive, we take the positive root: $t_{2B} = \frac{-\sqrt{2gh} + \sqrt{2gh + 2grd}}{gr}$.

Total time for Ball 2: $t_2 = t_{2A} + t_{2B} = \sqrt{\frac{2h}{g}} + \frac{-\sqrt{2gh} + \sqrt{2gh + 2grd}}{gr}$.

3. Equating the times:

Both balls touch the bottom at the same time, so $t_1 = t_2$. $\sqrt{\frac{2d}{gr}} = \sqrt{\frac{2h}{g}} + \frac{-\sqrt{2gh} + \sqrt{2gh + 2grd}}{gr}$. We are given the condition $d = nh$. Substitute this into the equation: $\sqrt{\frac{2nh}{gr}} = \sqrt{\frac{2h}{g}} + \frac{-\sqrt{2gh} + \sqrt{2gh + 2gr(nh)}}{gr}$. Divide the entire equation by $\sqrt{\frac{2h}{g}}$: $\sqrt{\frac{n}{r}} = 1 + \frac{-\sqrt{g} + \sqrt{g + grn}}{gr/\sqrt{2gh} \cdot \sqrt{2gh}} = 1 + \frac{-\sqrt{2gh} + \sqrt{2gh(1+nr)}}{gr} \cdot \frac{\sqrt{g}}{\sqrt{2h}}$ $\sqrt{\frac{n}{r}} = 1 + \frac{\sqrt{2gh}(-\sqrt{1} + \sqrt{1+nr})}{gr} \cdot \frac{\sqrt{g}}{\sqrt{2h}}$ $\sqrt{\frac{n}{r}} = 1 + \frac{\sqrt{g}(-1 + \sqrt{1+nr})}{gr/\sqrt{g}\sqrt{r}} \cdot \frac{\sqrt{g}}{\sqrt{g}\sqrt{r}}$ $\sqrt{\frac{n}{r}} = 1 + \frac{-1 + \sqrt{1+nr}}{\sqrt{r}}$. Multiply by $\sqrt{r}$: $\sqrt{n} = \sqrt{r} - 1 + \sqrt{1+nr}$. Rearrange to isolate the square root term: $\sqrt{1+nr} = \sqrt{n} - \sqrt{r} + 1$. Square both sides: $1+nr = (\sqrt{n} - \sqrt{r} + 1)^2$. Expand the right side: $1+nr = (\sqrt{n})^2 + (\sqrt{r})^2 + 1^2 - 2(\sqrt{n})(\sqrt{r}) + 2(\sqrt{n})(1) - 2(\sqrt{r})(1)$. $1+nr = n + r + 1 - 2\sqrt{nr} + 2\sqrt{n} - 2\sqrt{r}$. Subtract 1 from both sides: $nr = n + r - 2\sqrt{nr} + 2\sqrt{n} - 2\sqrt{r}$. Rearrange to group terms with $\sqrt{nr}$: $2\sqrt{nr} = n + r - nr + 2\sqrt{n} - 2\sqrt{r}$. We need to solve for $n$. Let's try to isolate $n$. $2\sqrt{n}\sqrt{r} - 2\sqrt{n} = n + r - nr - 2\sqrt{r}$. $2\sqrt{n}(\sqrt{r} - 1) = n(1-r) + r - 2\sqrt{r}$. Square both sides again to eliminate $\sqrt{n}$: $(2\sqrt{n}(\sqrt{r} - 1))^2 = (n(1-r) + r - 2\sqrt{r})^2$. $4n(\sqrt{r} - 1)^2 = (n(1-r) + r - 2\sqrt{r})^2$. $4n(r - 2\sqrt{r} + 1) = (n(1-r) + r - 2\sqrt{r})^2$.

Consider the equation $n + r - 2\sqrt{nr} + 2\sqrt{n} - 2\sqrt{r} - nr = 0$. Let's rearrange it to solve for $\sqrt{n}$: $n(1-r) + 2\sqrt{n} + (r - 2\sqrt{r} - 2\sqrt{nr}) = 0$. This is a quadratic in $\sqrt{n}$. Let $X = \sqrt{n}$. $X^2(1-r) + 2X + (r - 2\sqrt{r} - 2X\sqrt{r}) = 0$. $X^2(1-r) + 2X(1-\sqrt{r}) + r - 2\sqrt{r} = 0$. We want to find $n = X^2$. Using the quadratic formula for $X$: $X = \frac{-2(1-\sqrt{r}) \pm \sqrt{(2(1-\sqrt{r}))^2 - 4(1-r)(r-2\sqrt{r})}}{2(1-r)}$. $X = \frac{-2(1-\sqrt{r}) \pm \sqrt{4(1-\sqrt{r})^2 - 4(1-\sqrt{r})(1+\sqrt{r})(r-2\sqrt{r})}}{2(1-\sqrt{r})(1+\sqrt{r})}$. $X = \frac{-(1-\sqrt{r}) \pm \sqrt{(1-\sqrt{r})^2 - (1-\sqrt{r})(1+\sqrt{r})(r-2\sqrt{r})}}{1-r}$.

Let's retry from $\sqrt{n} - \sqrt{r} + 1 = \sqrt{1+nr}$. This equation is of the form $A = \sqrt{B}$. We need $A \ge 0$. So $\sqrt{n} - \sqrt{r} + 1 \ge 0$. Since $n$ and $r$ are positive, $\sqrt{n} > 0$ and $\sqrt{r} > 0$. The term $\sqrt{r}-1$ can be negative. $r = \frac{\rho_b - \rho_p}{\rho_b}$. Since $\rho_p < \rho_b$, $r$ is between 0 and 1. So $0 < r < 1$. This means $\sqrt{r} < 1$, so $\sqrt{r}-1$ is negative. Thus, $\sqrt{n} - (\sqrt{r}-1) = \sqrt{n} + (1-\sqrt{r})$ is always positive since $1-\sqrt{r} > 0$. So the condition $\sqrt{n} - \sqrt{r} + 1 \ge 0$ is naturally satisfied for $n>0$.

Let's go back to $n + r + 1 - 2\sqrt{nr} + 2\sqrt{n} - 2\sqrt{r} = 1 + nr$. $n + r - 2\sqrt{nr} + 2\sqrt{n} - 2\sqrt{r} - nr = 0$. This is a quadratic equation in $\sqrt{n}$ if we treat $r$ as a constant. $n(1-r) + 2\sqrt{n}(1-\sqrt{r}) + (r - 2\sqrt{r}) = 0$. Let $X = \sqrt{n}$. $X^2(1-r) + 2X(1-\sqrt{r}) + (r - 2\sqrt{r}) = 0$. Since $1-r = (1-\sqrt{r})(1+\sqrt{r})$, we can divide by $(1-\sqrt{r})$ (since $r<1$, $1-\sqrt{r} \ne 0$). $X^2(1+\sqrt{r}) + 2X + \frac{r - 2\sqrt{r}}{1-\sqrt{r}} = 0$. $\frac{r - 2\sqrt{r}}{1-\sqrt{r}} = \frac{\sqrt{r}(\sqrt{r}-2)}{1-\sqrt{r}}$. So, $X^2(1+\sqrt{r}) + 2X + \frac{\sqrt{r}(\sqrt{r}-2)}{1-\sqrt{r}} = 0$. Let's use the quadratic formula for $X = \sqrt{n}$: $X = \frac{-2 \pm \sqrt{4 - 4(1+\sqrt{r})\left(\frac{\sqrt{r}(\sqrt{r}-2)}{1-\sqrt{r}}\right)}}{2(1+\sqrt{r})}$. $X = \frac{-1 \pm \sqrt{1 - \frac{\sqrt{r}(\sqrt{r}-2)(1+\sqrt{r})}{1-\sqrt{r}}}}{1+\sqrt{r}}$. $X = \frac{-1 \pm \sqrt{1 - \frac{\sqrt{r}(\sqrt{r}-2)(1+\sqrt{r})^2}{(1-\sqrt{r})(1+\sqrt{r})}}}{1+\sqrt{r}}$.

Consider the initial equation $n + r - 2\sqrt{nr} + 2\sqrt{n} - 2\sqrt{r} - nr = 0$. We need to get $n$ in the form $\frac{Ar^3 + Br^2 + Cr}{Dr^2 + Er + F}$. Look at the powers of $r$. The expression for $n$ suggests that the equation should be polynomial in $r$. The presence of $\sqrt{r}$ and $\sqrt{n}$ is problematic. Let's try to arrange the equation $n + r - 2\sqrt{nr} + 2\sqrt{n} - 2\sqrt{r} - nr = 0$ as a perfect square if possible. This is $(\sqrt{n})^2 - 2\sqrt{n}\sqrt{r} + (\sqrt{r})^2 + 2\sqrt{n} - 2\sqrt{r} - nr = 0$. $(\sqrt{n}-\sqrt{r})^2 + 2(\sqrt{n}-\sqrt{r}) - nr = 0$. Let $Y = \sqrt{n}-\sqrt{r}$. $Y^2 + 2Y - nr = 0$. Substitute $Y = \sqrt{n}-\sqrt{r}$ back: $(\sqrt{n}-\sqrt{r})^2 + 2(\sqrt{n}-\sqrt{r}) - nr = 0$. $n + r - 2\sqrt{nr} + 2\sqrt{n} - 2\sqrt{r} - nr = 0$. This is the same equation.

Let's try to isolate $\sqrt{n}$ from $2\sqrt{n}(\sqrt{r} - 1) = n(1-r) + r - 2\sqrt{r}$. $2\sqrt{n}(1-\sqrt{r}) = n(r-1) - r + 2\sqrt{r}$. $2\sqrt{n}(1-\sqrt{r}) = -n(1-r) - (r - 2\sqrt{r})$. $2\sqrt{n}(1-\sqrt{r}) = -n(1-\sqrt{r})(1+\sqrt{r}) - \sqrt{r}(\sqrt{r}-2)$. Divide by $(1-\sqrt{r})$: $2\sqrt{n} = -n(1+\sqrt{r}) - \frac{\sqrt{r}(\sqrt{r}-2)}{1-\sqrt{r}}$. $2\sqrt{n} = -n(1+\sqrt{r}) - \frac{r-2\sqrt{r}}{1-\sqrt{r}}$. This is $n(1+\sqrt{r}) + 2\sqrt{n} + \frac{r-2\sqrt{r}}{1-\sqrt{r}} = 0$. This is the same quadratic in $\sqrt{n}$ as before: $X^2(1+\sqrt{r}) + 2X + \frac{r - 2\sqrt{r}}{1-\sqrt{r}} = 0$. Let's solve it again more carefully for $X = \sqrt{n}$. $X = \frac{-2 \pm \sqrt{4 - 4(1+\sqrt{r})\left(\frac{r - 2\sqrt{r}}{1-\sqrt{r}}\right)}}{2(1+\sqrt{r})}$. $X = \frac{-1 \pm \sqrt{1 - \frac{(1+\sqrt{r})(r - 2\sqrt{r})}{1-\sqrt{r}}}}{1+\sqrt{r}}$. $X = \frac{-1 \pm \sqrt{\frac{1-\sqrt{r} - (1+\sqrt{r})(r - 2\sqrt{r})}{1-\sqrt{r}}}}{1+\sqrt{r}}$. $X = \frac{-1 \pm \sqrt{\frac{1-\sqrt{r} - (r - 2\sqrt{r} + r\sqrt{r} - 2r)}{1-\sqrt{r}}}}{1+\sqrt{r}}$. $X = \frac{-1 \pm \sqrt{\frac{1-\sqrt{r} - (3r - 2\sqrt{r} + r\sqrt{r})}{1-\sqrt{r}}}}{1+\sqrt{r}}$. $X = \frac{-1 \pm \sqrt{\frac{1-\sqrt{r} - 3r + 2\sqrt{r} - r\sqrt{r}}{1-\sqrt{r}}}}{1+\sqrt{r}}$. $X = \frac{-1 \pm \sqrt{\frac{1+\sqrt{r} - 3r - r\sqrt{r}}{1-\sqrt{r}}}}{1+\sqrt{r}}$.

Let's check the square of $\sqrt{n} - \sqrt{r} + 1 = \sqrt{1+nr}$. $n + r + 1 - 2\sqrt{nr} + 2\sqrt{n} - 2\sqrt{r} = 1 + nr$. $n + r - 2\sqrt{nr} + 2\sqrt{n} - 2\sqrt{r} - nr = 0$. Let's rearrange as: $n(1-r) + 2\sqrt{n}(1-\sqrt{r}) + (r-2\sqrt{r}) = 0$. This is $n(1-\sqrt{r})(1+\sqrt{r}) + 2\sqrt{n}(1-\sqrt{r}) + \sqrt{r}(\sqrt{r}-2) = 0$. Divide by $(1-\sqrt{r})$: $n(1+\sqrt{r}) + 2\sqrt{n} + \frac{\sqrt{r}(\sqrt{r}-2)}{1-\sqrt{r}} = 0$.

Let's go back to $t_1 = t_{2A} + t_{2B}$. $\sqrt{\frac{2d}{gr}} = \sqrt{\frac{2h}{g}} + \frac{-\sqrt{2gh} + \sqrt{2gh + 2grd}}{gr}$. Substitute $d=nh$. $\sqrt{\frac{2nh}{gr}} = \sqrt{\frac{2h}{g}} + \frac{-\sqrt{2gh} + \sqrt{2gh + 2gr(nh)}}{gr}$. Divide by $\sqrt{2h/g}$: $\sqrt{\frac{n}{r}} = 1 + \frac{-\sqrt{g} + \sqrt{g(1+nr)}}{gr/\sqrt{2gh}} \cdot \frac{\sqrt{g}}{\sqrt{2h}}$. $\frac{-\sqrt{2gh} + \sqrt{2gh(1+nr)}}{gr} = \frac{\sqrt{2gh}(-1 + \sqrt{1+nr})}{gr}$. So, $\sqrt{\frac{n}{r}} = 1 + \frac{\sqrt{2gh}(-1 + \sqrt{1+nr})}{gr}$. $\sqrt{\frac{n}{r}} = 1 + \frac{\sqrt{2gh}(-1 + \sqrt{1+nr})}{gr} = 1 + \frac{\sqrt{2h}\sqrt{g}(-1 + \sqrt{1+nr})}{gr}$. Divide by $\sqrt{2h/g}$: $\sqrt{\frac{n}{r}} = 1 + \frac{\sqrt{g}(-1 + \sqrt{1+nr})}{gr/\sqrt{g}}$. Let's divide by $\sqrt{2h/g}$ more carefully: $\frac{\sqrt{2nh/gr}}{\sqrt{2h/g}} = \frac{\sqrt{2h/g}}{\sqrt{2h/g}} + \frac{\frac{\sqrt{2gh}(-1 + \sqrt{1+nr})}{gr}}{\sqrt{2h/g}}$. $\sqrt{\frac{n}{r}} = 1 + \frac{\sqrt{2gh}(-1 + \sqrt{1+nr})}{gr} \cdot \sqrt{\frac{g}{2h}}$. $\sqrt{\frac{n}{r}} = 1 + \frac{\sqrt{2gh}(-1 + \sqrt{1+nr})}{gr} \cdot \frac{\sqrt{g}}{\sqrt{2h}}$. $\sqrt{\frac{n}{r}} = 1 + \frac{\sqrt{g}\sqrt{g}(-1 + \sqrt{1+nr})}{gr}$. $\sqrt{\frac{n}{r}} = 1 + \frac{g(-1 + \sqrt{1+nr})}{gr}$. $\sqrt{\frac{n}{r}} = 1 + \frac{-1 + \sqrt{1+nr}}{r}$. Multiply by $r$: $r\sqrt{\frac{n}{r}} = r + (-1 + \sqrt{1+nr})$. $r\frac{\sqrt{n}}{\sqrt{r}} = r - 1 + \sqrt{1+nr}$. $\sqrt{r}\sqrt{n} = r - 1 + \sqrt{1+nr}$. $\sqrt{nr} = r - 1 + \sqrt{1+nr}$. Let $X = \sqrt{nr}$. $X = r - 1 + \sqrt{1+X^2}$. Rearrange: $\sqrt{1+X^2} = X - r + 1$. Square both sides: $1+X^2 = (X - r + 1)^2$. $1+X^2 = X^2 + (r-1)^2 - 2X(r-1)$. $1 = (r-1)^2 - 2X(r-1)$. $1 = r^2 - 2r + 1 - 2X(r-1)$. $0 = r^2 - 2r - 2X(r-1)$. $2X(r-1) = r^2 - 2r$. $2X(r-1) = r(r-2)$. $X = \frac{r(r-2)}{2(r-1)}$. Recall $X = \sqrt{nr}$. $\sqrt{nr} = \frac{r(r-2)}{2(r-1)}$. Square both sides to find $nr$: $nr = \left(\frac{r(r-2)}{2(r-1)}\right)^2$. $nr = \frac{r^2(r-2)^2}{4(r-1)^2}$. Divide by $r$ (since $r \ne 0$): $n = \frac{r(r-2)^2}{4(r-1)^2}$. Expand the terms: $n = \frac{r(r^2 - 4r + 4)}{4(r^2 - 2r + 1)}$. $n = \frac{r^3 - 4r^2 + 4r}{4r^2 - 8r + 4}$. This is in the form $\frac{Ar^3 + Br^2 + Cr}{Dr^2 + Er + F}$. Comparing coefficients: $A = 1$ $B = -4$ $C = 4$ $D = 4$ $E = -8$ $F = 4$

Check conditions:

1. A, B, C, D, E, F are all nonzero integers. Yes.
2. gcd(A, B, C, D, E, F) = 1. gcd(1, -4, 4, 4, -8, 4) = 1. Yes.
3. A > 0. Yes, A=1.

Now, calculate A + B + C + D + E + F: A + B + C + D + E + F = 1 + (-4) + 4 + 4 + (-8) + 4 = 1 - 4 + 4 + 4 - 8 + 4 = (1+4+4+4) - (4+8) = 13 - 12 = 1.

Final check of the condition for squaring: $X - r + 1 \ge 0$. $X = \frac{r(r-2)}{2(r-1)}$. Since $0 < r < 1$, $(r-1)$ is negative. $(r-2)$ is negative. $r$ is positive. So $X = \frac{\text{positive} \cdot \text{negative}}{\text{negative}} = \text{positive}$. So $X > 0$. $X - r + 1 = \frac{r(r-2)}{2(r-1)} - r + 1 = \frac{r^2 - 2r - 2r(r-1) + 2(r-1)}{2(r-1)}$ $= \frac{r^2 - 2r - 2r^2 + 2r + 2r - 2}{2(r-1)}$ $= \frac{-r^2 + 2r - 2}{2(r-1)}$. The numerator is $-(r^2 - 2r + 2) = -((r-1)^2 + 1)$. This is always negative. The denominator $2(r-1)$ is also negative for $0 < r < 1$. So $\frac{\text{negative}}{\text{negative}} = \text{positive}$. Thus, $X - r + 1 > 0$ is satisfied.

The solution is consistent.

The values are: A = 1 B = -4 C = 4 D = 4 E = -8 F = 4

Sum = 1 - 4 + 4 + 4 - 8 + 4 = 1.