A proton moves with a speed of `7.45xx10^(5) m//s` dirctly towards a free proton originally at rest. Find the distance of closest approach for the two protons.
Given : `(1)/(4pi in_(0))=9xx10^(9)(N-m^(2))/C^(2), m_(p)=1.67xx10^(-27) kg` and `e=1.6xx10^(-19) C`

Explanation

To find the distance of closest approach for two protons, we can use the principles of conservation of momentum and conservation of energy. Here’s a step-by-step solution:

### Step 1: Define the initial conditions
Let:
- \( m_p \) = mass of a proton = \( 1.67 \times 10^{-27} \) kg
- \( e \) = charge of a proton = \( 1.6 \times 10^{-19} \) C
- \( u \) = initial speed of the moving proton = \( 7.45 \times 10^5 \) m/s
- \( K \) = electrostatic constant = \( \frac{1}{4 \pi \epsilon_0} = 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \)
...