Science of Apps! - Part 1: Flappy Bird

For this problem, we use the following formulas: $p=mv$ , $v=\frac{d}{t}$ , and $d=\frac{1}{2}at^2$ , where $p$ is the momentum of the bird, $m$ is the mass in kilograms, $v$ is the velocity in meters per second, $a=9.8 \frac{\text{m}}{\text{s}^2}$ , $d$ is the displacement in meters, and $t$ is time in seconds.

Rewriting the momentum formula, we have $p=mv=m\left(\frac{d}{t}\right)=\frac{m\left(\frac{1}{2}at^2\right)}{t}=\frac{1}{2}mat$ . Therefore, the momentum of the bird right after the click is $p=\frac{1}{2}\times 500 \text{ g} \times \frac{1\text{ kg}}{1000 \text{ g}} \times 9.8 \frac{\text{m}}{\text{s}^2} \times 1 \text{ s}=\boxed{2.45 \frac{\text{kg}\text{ m}}{\text{s}}}.$

for finding initial velocity : V= U + at here V = velocity at top i.e , V=0 U= initial velocity : to be found a= -g t= half of total time i.e t = o.5 sec

so 0 = U + (- 9.8) * (0.5) => U = 9.8/2 = 4.9 m/s

momentum = m * U = (o.5 kg) * (4.9 m/s) = 2.45 kg-m/s (answer)

In one second the bird did the same path twice (going up and returning to its initial position), and since its acceleration is constant (g), the time taken to go up is equal to the time taken to go back down.

So the total time divided by two gives us 0.5 sec. This is the time that was needed to go up and reach a velocity of 0. If the initial velocity was v , then

m(0-v)=mgt

-v=gt

0.5(-v)= (0.5)(-9.8)(0.5)

v=4.9 m/s

momentum=mv=0.5x4.9=2.45 kg.m/s

$J=p_f-p_i=m\cdot{}g\cdot{}\Delta{}t$ $-2p_i=(0.5 \text{ kg})(-9.8 \text{ m} \text{ s}^{-2})(1 \text{ s})\Rightarrow{}p_i=2.45 \text{ N m}$

Time required for upward journey from any point during the journey to the topmost point is always equal to the time required for downward journey from the topmost point to the same point. Hence time taken for total upward journey and total downward journey is equal $t_{u} = t_{d}=0.5 s$ .

Now when the particle reaches the topmost point its velocity becomes zero. Hence $v = v_{0} - gt \\ 0 = v_{0} - g t_{u} \\ \therefore v_{0} = g t_{u} = 9.8 \times 0.5 = 4.9 \frac{m}{s} \\ \therefore p = Mv_{0} = 0.5 \times 4.9 = 2.45 \frac{kgm}{s}$