### Overarch 2

Bricks are 20cm long and 10cm high. How high could an arch be built without mortar on a flat horizontal surface, to overhang by 1 metre? How big an overhang is it possible to make like this?

### Stonehenge

Explain why, when moving heavy objects on rollers, the object moves twice as fast as the rollers. Try a similar experiment yourself.

### Maximum Flow

Given the graph of a supply network and the maximum capacity for flow in each section find the maximum flow across the network.

# Powerfully Fast

##### Stage: 5 Challenge Level:

The specific power of the 'plane is
$$\frac{2 \times 12000 \mathrm{\ W\ kg^{-1}} \times 4000 \mathrm{\ kg}}{140000 \mathrm{\ kg}} = 685.714 \mathrm{\ W\ kg^{-1}}$$
For the next part, equate the energy used to the kinetic energy.
$$Pt = \frac{1}{2}mv^2 \quad \therefore \quad t = \frac{mv^2}{2P}$$
$$60 \mathrm{\ mph} = 26.8224 \mathrm{\ m s^{-1}}\Rightarrow t = 0.5246 \mathrm{\ s}$$

Then we look at the change in energy.

120 mph to 180 mph is $3 \times 60 \mathrm{\ mph} - 2 \times 60 \mathrm{\ mph}$, and energy is proportional to the square of velocity, so this requires $(9-4) \times 0.5246 \mathrm{\ s} = 2.623 \mathrm{\ s}$ at the constant power.

180 mph to 240 mph by the same logic requires $(16 - 9) \times 0.5246 \mathrm{\ s} = 3.672 \mathrm{\ s}$

For the rocket,
$$\textrm{[Energy]} = 8 \mathrm{\ hr} \times 3600 \mathrm{\ s\ hr^{-1}} \times 2 \textrm{[engines]} \times 12000 \mathrm{\ W\ kg^{-1}} \times 4000 \mathrm{\ kg \ \textrm{[engine]}^{-1}} = 2.7648 \times 10^{12} \mathrm{\ J}$$
Equate this to $\frac{1}{2}mv^2$, to find $V_{max} = 7509.4 \mathrm{\ m\ s^{-1}}$

For the horse, find a foot-pound in Joules, which the energy requires to lift a pound of mass through one foot, which equates to $1.356 \mathrm{\ J}$. 32400 of these divided by 60 seconds gives $732.4 \mathrm{\ W}$. Divide this through by about $500 \mathrm{\ kg}$, which is approximately the mass of a large horse, to get $1.465 \mathrm{\ W\ kg^{-1}}$.

To find the time taken to accelerate to the given speed, apply the same equations as with the aircraft to find that $t = 4 \mathrm{\ min}\ 6 \mathrm{\ s}$.

Extension: find the formula for acceleration in terms of time for a mass m accelerated with a constant power $P$. How large is the force applied to the mass when the acceleration starts?

Extension solution:
$$Pt = \frac{1}{2}mv^2 \quad \therefore \quad v = \sqrt{\frac{2Pt}{m}}$$
$$a = \frac{\mathrm{d}v}{\mathrm{d}t} = \sqrt{\frac{P}{2mt}}$$

When $t = 0$, the acceleration, and so the force, is theoretically infinite. We can get close to constant power acceleration, but in reality it is not reached, and thus forces are never infinite. For example a hard insect hitting a car windscreen can be thought of as a near constant-power acceleration, because the speed of the car does not change measurably during the collision. The glass is very hard, and so a bug will thus splat at even moderate speeds. The non constant-power element in that example comes from the fact that the insect is not infinitely hard, and so takes some time to compress during the collision.