The distance they covered was 17 Km. Not sure that their displacement was though sorry.
The distance r between the pions when the criteria are satisfied is 1.45 × 10⁻¹⁶ m
If we consider the potential energy (U) between the pions, then (U) can be expressed as:
![\mathbf{U = \dfrac{kq^2}{r} ---- (1)}](https://tex.z-dn.net/?f=%5Cmathbf%7BU%20%3D%20%5Cdfrac%7Bkq%5E2%7D%7Br%7D%20----%20%281%29%7D)
Given that at some instance, the potential energy becomes negligible compared to the final K.E.
As such the conservation of the total energy in the system can be given as:
Again, if we consider the ratio of the potential energy to the kinetic energy to be about 0.01, then:
![\mathbf{\dfrac{U}{K}= 0.01} \\ \\ \mathbf{U = 0.01 K----(2)}](https://tex.z-dn.net/?f=%5Cmathbf%7B%5Cdfrac%7BU%7D%7BK%7D%3D%200.01%7D%20%5C%5C%20%5C%5C%20%5Cmathbf%7BU%20%3D%200.01%20K----%282%29%7D)
∴
Equating both equations (1) and (2) together, we have:
![\mathbf{\dfrac{kq^2}{r} = 0.01 K}](https://tex.z-dn.net/?f=%5Cmathbf%7B%5Cdfrac%7Bkq%5E2%7D%7Br%7D%20%3D%200.01%20K%7D)
![\mathbf{\dfrac{kq^2}{r} = 0.01 \Bigg [ m_{o \pi}c^2 \Big [\dfrac{1}{\sqrt{1 - \dfrac{v_{\pi}^2}{c^2} }} \Big ] \Bigg] }](https://tex.z-dn.net/?f=%5Cmathbf%7B%5Cdfrac%7Bkq%5E2%7D%7Br%7D%20%3D%200.01%20%5CBigg%20%5B%20m_%7Bo%20%5Cpi%7Dc%5E2%20%5CBig%20%5B%5Cdfrac%7B1%7D%7B%5Csqrt%7B1%20-%20%5Cdfrac%7Bv_%7B%5Cpi%7D%5E2%7D%7Bc%5E2%7D%20%7D%7D%20%5CBig%20%5D%20%5CBigg%5D%20%7D)
![\mathbf{r =\dfrac{kq^2}{ 0.01 \Bigg [ m_{o \pi}c^2 \Big [\dfrac{1}{\sqrt{1 - \dfrac{v_{\pi}^2}{c^2} }} \Big ] \Bigg] }}](https://tex.z-dn.net/?f=%5Cmathbf%7Br%20%3D%5Cdfrac%7Bkq%5E2%7D%7B%200.01%20%5CBigg%20%5B%20m_%7Bo%20%5Cpi%7Dc%5E2%20%5CBig%20%5B%5Cdfrac%7B1%7D%7B%5Csqrt%7B1%20-%20%5Cdfrac%7Bv_%7B%5Cpi%7D%5E2%7D%7Bc%5E2%7D%20%7D%7D%20%5CBig%20%5D%20%5CBigg%5D%20%7D%7D)
where:
- r = distance
- k = Columb's constant
- q = charge on a proton
- m_o = rest mass of each pion in the previous question
- c = velocity of light
= calculated velocity of proton in the previous question
Replacing their values in the above equation, the distance (r) between the pions is calculated as:
![\mathbf{r =\dfrac{(9\times 10^9 \ N.m^2/C^2) (1.6022 \times 10^{-19} \ C)^2}{ 0.01 \Bigg [ (2.5 \times 10^{-28\ kg } )\times (3\times 10^8 \ m/s)^2 \Big [\dfrac{1}{\sqrt{1 - \dfrac{(2.97 \times 10^8 \ m/s)^2}{(3 \times 10^8 \ m/s)^2} }} \Big ] \Bigg] }}](https://tex.z-dn.net/?f=%5Cmathbf%7Br%20%3D%5Cdfrac%7B%289%5Ctimes%2010%5E9%20%5C%20N.m%5E2%2FC%5E2%29%20%281.6022%20%5Ctimes%2010%5E%7B-19%7D%20%5C%20C%29%5E2%7D%7B%200.01%20%5CBigg%20%5B%20%282.5%20%5Ctimes%2010%5E%7B-28%5C%20kg%20%7D%20%29%5Ctimes%20%283%5Ctimes%2010%5E8%20%5C%20m%2Fs%29%5E2%20%5CBig%20%5B%5Cdfrac%7B1%7D%7B%5Csqrt%7B1%20-%20%5Cdfrac%7B%282.97%20%5Ctimes%2010%5E8%20%5C%20m%2Fs%29%5E2%7D%7B%283%20%5Ctimes%2010%5E8%20%5C%20m%2Fs%29%5E2%7D%20%7D%7D%20%5CBig%20%5D%20%5CBigg%5D%20%7D%7D)
distance (r) = 1.45 × 10⁻¹⁶ m
Therefore, we can conclude that the distance r between the pions when the criteria are satisfied is 1.45 × 10⁻¹⁶ m
Learn more about electric potential energy here:
brainly.com/question/21808222?referrer=searchResults
Given:
u = 8.5 cm
f = -14.0 cm
v = image distance
Using the mirror formula 1/u + 1/v = 1/f
1/8.5 + 1/v = 1/-14.0
Rewrite to solve for v:
v = (8.5 * -14.0) / (8.5 - (-14.0))
v = -119 / 22.5
v = -5.29 cm ( round answer as needed.)