Question

A proton moves with speed of 7.45 x 10⁵ ms^-1 directly towards a free proton originally at rest. Find the distance of the closest approach for the two protons (m_p = 1.66 x 10^-27 kg).

Answer 1

Let initial speed of proton = v

Let distance of the closest approach = r

If v₁ be the speed of two protons after attaining distance r, then

From law of conservation of momentum

mv + 0 = mv₁ + mv₂

or v₁ = \(\frac{v}{2}\) .............(i)

From law of conservation of energy