The Workbook

1. Torricelli: \(v = \sqrt{2gh} = \sqrt{2 \times 10 \times 1.8} = 6\) m/s, horizontal.

2. The hole is \(H - h = 3.2\) m above the ground. \(3.2 = \dfrac{1}{2}gt^2\) gives \(t = \sqrt{0.64} = 0.8\) s.

3. \(R = vt = 6 \times 0.8 = 4.8\) m.

\(v = \sqrt{2gh}\) and the fall of \(H - h\) takes \(t = \sqrt{\dfrac{2(H - h)}{g}}\).

\(R = vt = \sqrt{2gh}\sqrt{\dfrac{2(H - h)}{g}} = 2\sqrt{h(H - h)}\). The value of \(g\) cancels.

Check: \(2\sqrt{1.8 \times 3.2} = 2\sqrt{5.76} = 2 \times 2.4 = 4.8\) m.

Put \(h' = H - h\) in the formula: \(R' = 2\sqrt{(H - h)(H - (H - h))} = 2\sqrt{(H - h)h} = R\).

The upper hole gives a faster fall from a smaller speed; the lower hole gives a faster jet with a shorter fall. The two effects balance.

\(h = 1\) m: \(R = 2\sqrt{1 \times 4} = 4\) m. \(h = 4\) m: \(R = 2\sqrt{4 \times 1} = 4\) m.

\(R^2 = 4h(H - h) = 4(Hh - h^2)\). This is largest when \(\dfrac{d}{dh}(Hh - h^2) = H - 2h = 0\), that is, \(h = \dfrac{H}{2}\).

\(R_{\max} = 2\sqrt{\dfrac{H}{2} \cdot \dfrac{H}{2}} = H = 5\) m. A hole halfway down throws the water as far as the water is deep.

1. The jet leaves at \(\sqrt{2gh}\) and falls a height \(H - h + d\): \(t = \sqrt{\dfrac{2(H - h + d)}{g}}\).

   \(R = 2\sqrt{h(H - h + d)}\).

2. \(R = 2\sqrt{1 \times (2 - 1 + 3)} = 2\sqrt{4} = 4\) m.

\(4.8 = 2\sqrt{h(5 - h)}\), so \(h(5 - h) = 5.76\), that is, \(h^2 - 5h + 5.76 = 0\).

\(h = \dfrac{5 \pm \sqrt{25 - 23.04}}{2} = \dfrac{5 \pm 1.4}{2} = 3.2\) m or \(1.8\) m. The two answers are the symmetric pair of Question 3.

1. The lower jet (\(3.2\) m deep) is only \(1.8\) m above the ground: \(t = \sqrt{\dfrac{2 \times 1.8}{10}} = 0.6\) s. The upper jet takes \(0.8\) s (Question 1). The lower jet lands first.

2. Energy: the water starts from rest at the surface and lands \(H\) below it, with no energy lost. So \(v^2 = 2gH\) for both jets: \(v = \sqrt{2 \times 10 \times 5} = 10\) m/s.

   Check for the upper jet: horizontal \(6\) m/s, vertical \(gt = 8\) m/s, resultant \(\sqrt{36 + 64} = 10\) m/s.

\(x = vt\) with \(v = \sqrt{2gh}\), and \(y = \dfrac{1}{2}gt^2\).

\(t = \dfrac{x}{v}\), so \(y = \dfrac{g x^2}{2v^2} = \dfrac{g x^2}{2 \times 2gh} = \dfrac{x^2}{4h}\). The jet is a parabola whose shape depends only on \(h\).

Time to reach the wall: \(t = \dfrac{2.4}{6} = 0.4\) s. Drop in that time: \(\dfrac{1}{2} \times 10 \times 0.16 = 0.8\) m.

Height of impact \(= 3.2 - 0.8 = 2.4\) m above the ground. (Or from Question 8: \(y = \dfrac{2.4^2}{4 \times 1.8} = 0.8\) m.)

1. The base is \(5\) m below the surface: \(v = \sqrt{2 \times 10 \times 5} = 10\) m/s, straight down.

2. The jet leaves at \(\sqrt{2gh}\) upward and rises until its speed is zero: a height \(\dfrac{v^2}{2g} = h = 1.8\) m, exactly back to the level of the surface.

   A real jet loses energy to viscosity and air resistance, so it stops a little below the surface level.

No. The speed \(\sqrt{2gh}\) and the fall time depend only on the heights, so \(R = 2\sqrt{h(H - h)}\) holds for any small hole. A larger hole gives a thicker jet, not a longer one.

No. The density cancels in Torricelli's law: a heavier liquid has more pressure at the hole and more mass to push.

Now \(h = 4 - 3.2 = 0.8\) m and the hole is still \(3.2\) m above the ground: \(R = 2\sqrt{0.8 \times 3.2} = 2\sqrt{2.56} = 3.2\) m.

1. Mass per second \(= \rho a v = 1000 \times 10^{-4} \times 6 = 0.6\) kg/s. Momentum per second \(= 0.6 \times 6 = 3.6\) kg m/s\(^2\), that is, \(3.6\) N.

2. Force \(=\) momentum per second \(= \rho a v^2 = \rho a (2gh) = 2\rho g h a\). The tank is pushed backward with \(2 \times 1000 \times 10 \times 1.8 \times 10^{-4} = 3.6\) N, twice the hydrostatic force \(\rho g h a\) on an area \(a\).

3. The wall must destroy the same momentum per second: \(3.6\) N.

1. \(v = 6\) m/s. Horizontal: \(6\cos 30^\circ = 3\sqrt{3} \approx 5.2\) m/s. Vertical: \(6\sin 30^\circ = 3\) m/s.

2. Greatest height above the nozzle \(= \dfrac{3^2}{2g} = 0.45\) m.

3. Take downward as positive: \(3.2 = -3t + 5t^2\), so \(5t^2 - 3t - 3.2 = 0\) and \(t = \dfrac{3 + \sqrt{9 + 64}}{10} = \dfrac{3 + 8.54}{10} = 1.15\) s.

   Range \(= 5.2 \times 1.15 \approx 6\) m, more than the \(4.8\) m of the horizontal jet.

\(h = 1\) m: \(R = 2\sqrt{1 \times 4} = 4\) m. \(h = 2\) m: \(R = 2\sqrt{2 \times 3} = 2\sqrt{6} \approx 4.9\) m.

The deeper hole lands farther, because \(2\) m is closer to the best depth \(H/2 = 2.5\) m.

\(3 = 2\sqrt{h(5 - h)}\), so \(h(5 - h) = 2.25\) and \(h^2 - 5h + 2.25 = 0\).

\(h = \dfrac{5 \pm \sqrt{25 - 9}}{2} = \dfrac{5 \pm 4}{2} = 4.5\) m or \(0.5\) m.

1. In the lift the effective gravity is \(g + a\). Speed of efflux \(v = \sqrt{2(g + a)h}\); time of fall \(t = \sqrt{\dfrac{2(H - h)}{g + a}}\).

2. \(R = vt = 2\sqrt{h(H - h)}\): the factor \((g + a)\) cancels. The range is the same. On the Moon \(g\) is smaller, the jet is slower and falls more slowly, and the range is again unchanged.

1. The peak of the curve is \(R = 5\) m at \(h = 2.5\) m, as in Question 4.

2. The line \(R = 4\) m cuts the curve at \(h = 1\) m and \(h = 4\) m, as in Question 3.

3. \(R = 2\sqrt{h(H - h)}\) is unchanged when \(h\) is replaced by \(H - h\), so depths equally far above and below \(H/2\) give equal ranges.

\(R = 2\sqrt{0.2 \times 4.8} = 2\sqrt{0.96} = 2 \times 0.98 = 1.96\) m.

The symmetric hole is at depth \(H - h = 4.8\) m, that is, \(20\) cm above the base.

\(R = 2\sqrt{h(H - h + d)}\): \(6 = 2\sqrt{1 \times (1 + d)}\), so \(1 + d = 9\) and \(d = 8\) m.

1. Continuity: \(A_1v_1 = A_2v_2\), so \(v_2 = \dfrac{4 \times 1}{2} = 2\) m/s.

2. \(Q = A_1v_1 = 4 \times 10^{-4} \times 1 = 4 \times 10^{-4}\) m\(^3\)/s \(= 0.4\) L/s.

3. Mass flow \(= \rho Q = 1000 \times 4 \times 10^{-4} = 0.4\) kg/s.

\(Q = \dfrac{0.5 \times 10^{-3}}{5} = 10^{-4}\) m\(^3\)/s. \(v = \dfrac{Q}{A} = \dfrac{10^{-4}}{10^{-4}} = 1\) m/s.

1. Free fall: \(v^2 = v_0^2 + 2gh = 1 + 2 \times 10 \times 0.15 = 4\), so \(v = 2\) m/s.

2. The same volume passes every section each second: \(A_0v_0 = Av\), so \(A = \dfrac{1 \times 1}{2} = 0.5\) cm\(^2\).

   The water speeds up as it falls, so a smaller cross-section carries the same flow. The stream narrows.

1. A streamline is the path of a fluid particle in steady flow. The velocity at each point is along the tangent to the streamline through that point.

   If two streamlines crossed, a particle at the crossing would have two velocities at once, which is impossible.

2. In streamline flow every particle passing a point follows the same path, and the flow pattern does not change with time. In turbulent flow the paths are irregular and eddies form.

   Flow becomes turbulent when the speed exceeds a critical value, which is lower for wider pipes and less viscous fluids.

\(Q = Av = (20 \times 2) \times 0.5 = 20\) m\(^3\)/s.

In the gorge \(A = 10 \times 2 = 20\) m\(^2\), so \(v = \dfrac{20}{20} = 1\) m/s.

\(P + \dfrac{1}{2}\rho v^2 + \rho g h = \) constant along a streamline.

\(P\) is the pressure energy per unit volume, \(\dfrac{1}{2}\rho v^2\) the kinetic energy per unit volume, and \(\rho g h\) the potential energy per unit volume.

Unit check: \(\rho v^2\) is kg m\(^{-3}\) m\(^2\) s\(^{-2}\), that is, N/m\(^2\). This equals J/m\(^3\), the same as \(P\).

It holds for an ideal fluid: incompressible, non-viscous, in steady streamline flow.

Horizontal pipe, so the \(\rho g h\) terms cancel: \(P_2 = P_1 - \dfrac{1}{2}\rho(v_2^2 - v_1^2) = 1.2 \times 10^{5} - 500 \times (4 - 1) = 1.185 \times 10^{5}\) Pa.

The water speeds up as it enters the narrow part. The push that speeds it up comes from the higher pressure behind, so the pressure ahead, in the narrow part, is lower.

The speed is the same at both ends, so only the height term changes: \(P_1 - P_2 = \rho g h = 1000 \times 10 \times 5 = 5 \times 10^{4}\) Pa.

1. \(\Delta P = \rho g\,\Delta h = 1000 \times 10 \times 0.15 = 1500\) Pa.

2. Continuity: \(v_2 = 2v_1\). Bernoulli: \(\Delta P = \dfrac{1}{2}\rho(v_2^2 - v_1^2) = \dfrac{1}{2} \times 1000 \times 3v_1^2 = 1500v_1^2\).

   \(1500v_1^2 = 1500\), so \(v_1 = 1\) m/s and \(v_2 = 2\) m/s.

3. \(Q = A_1v_1 = 10 \times 10^{-4} \times 1 = 10^{-3}\) m\(^3\)/s \(= 1\) L/s.

1. The water entering the bent tube is brought to rest. Its kinetic energy per unit volume, \(\dfrac{1}{2}\rho v^2\), is converted into extra pressure, which supports a taller column.

2. Bernoulli between the moving stream and the stagnant water at the tube mouth: \(P + \dfrac{1}{2}\rho v^2 = P_{\text{stag}}\), so \(P_{\text{stag}} - P = \dfrac{1}{2}\rho v^2\).

   The level difference gives \(P_{\text{stag}} - P = \rho g h\). So \(\dfrac{1}{2}\rho v^2 = \rho g h\) and \(v = \sqrt{2gh} = \sqrt{2 \times 10 \times 0.2} = 2\) m/s.

1. \(\Delta P = \dfrac{1}{2}\rho(v_1^2 - v_2^2) = \dfrac{1}{2} \times 1.2 \times (3600 - 2500) = 660\) Pa.

2. Lift \(= \Delta P \times A = 660 \times 20 = 13200\) N.

3. \(m = \dfrac{13200}{10} = 1320\) kg.

1. Wind moves fast over the roof, so the pressure above it falls below the still-air pressure inside the house. The net upward push can lift the roof.

2. The air between the sheets moves fast, so its pressure drops. The still air outside pushes the sheets inward, towards each other.

3. On the top, the surface of the spinning ball moves with the air stream and drags the air along. So the air there is faster and the pressure lower. The net force is upward, so the ball curves upward. This is the Magnus effect.

1. Take the surface and the hole as two points of a streamline. Both are at atmospheric pressure. The tank is large, so the surface speed is negligible.

   \(P_0 + 0 + \rho g h = P_0 + \dfrac{1}{2}\rho v^2 + 0\), so \(v = \sqrt{2gh}\). Assumptions: ideal fluid, small hole, wide tank.

2. \(v = \sqrt{2 \times 10 \times 5} = 10\) m/s, the speed of a body that has fallen \(5\) m.

\(Q = av = 10^{-4} \times 10 = 10^{-3}\) m\(^3\)/s \(= 1\) litre per second.

Now the pressures at the two points differ: \(P_{\text{gas}} + \rho g h = P_0 + \dfrac{1}{2}\rho v^2\).

\(\dfrac{1}{2}v^2 = \dfrac{P_{\text{gas}} - P_0}{\rho} + gh = \dfrac{1.5 \times 10^{5}}{1000} + 50 = 200\), so \(v = 20\) m/s.

1. When the depth is \(h\), water leaves at \(a\sqrt{2gh}\) per second, so the level falls at \(-\dfrac{dh}{dt} = \dfrac{a}{A}\sqrt{2gh}\).

   \(dt = -\dfrac{A}{a}\dfrac{dh}{\sqrt{2gh}}\). Integrate from \(h = H\) to \(0\): \(t = \dfrac{A}{a\sqrt{2g}}\left[2\sqrt{h}\right]_0^H = \dfrac{A}{a}\sqrt{\dfrac{2H}{g}}\).

2. \(t = \dfrac{1}{10^{-4}}\sqrt{\dfrac{2 \times 5}{10}} = 10^{4} \times 1 = 10^{4}\) s, about \(2.8\) hours. The flow slows as the level falls.

From the integral, the time to fall from \(H\) to \(h\) is \(t = \dfrac{A}{a}\sqrt{\dfrac{2}{g}}\left(\sqrt{H} - \sqrt{h}\right)\).

With \(h = \dfrac{H}{4}\): \(\sqrt{H} - \dfrac{\sqrt{H}}{2} = \dfrac{\sqrt{H}}{2}\), so \(t = \dfrac{1}{2} \times \dfrac{A}{a}\sqrt{\dfrac{2H}{g}} = 5000\) s.

The upper three quarters of the water take half the time. The last quarter takes the other half, because the flow is slow near the end.

No. In \(v = \sqrt{2gh}\) the density has cancelled: a denser liquid has more pressure at the hole but also more mass to accelerate. Mercury and water leave at the same speed from the same depth.

No, as long as the hole is small compared with the tank, so that the surface hardly moves. A larger hole passes more volume per second, but at the same speed.

Neglect the slow inlet speed. \(P_1 = P_2 + \dfrac{1}{2}\rho v_2^2\) with \(P_2 = 0\): \(v_2 = \sqrt{\dfrac{2 \times 1.5 \times 10^{5}}{1000}} = \sqrt{300} \approx 17\) m/s.

A liquid cannot have negative pressure. If the throat is narrowed further the water boils into vapour bubbles at the throat (cavitation), and the smooth flow breaks down.

The fast air lowers the pressure at the top of the tube by \(\dfrac{1}{2}\rho_{\text{air}}v^2\). The water rises when this equals the pressure of a \(10\) cm column, \(\rho_w g h = 1000 \times 10 \times 0.1 = 1000\) Pa.

\(v = \sqrt{\dfrac{2 \times 1000}{1.2}} = \sqrt{1667} \approx 41\) m/s.

Each cubic metre of water gains \(\rho g h + \dfrac{1}{2}\rho v^2 = 1000 \times 10 \times 10 + \dfrac{1}{2} \times 1000 \times 100 = 1.5 \times 10^{5}\) J.

Power \(= Q \times 1.5 \times 10^{5} = 2 \times 10^{-3} \times 1.5 \times 10^{5} = 300\) W.