Collision of charged particles

Consider the mass of two protons as m, the initial velocity of moving proton as u 1 and the final velocities of two protons be v 1 and v 2 respectively, the angle they make with horizontal as \theta and \phi respectively. Now as the collision is elastic, so Kinetic Energy and Momentum is conserved. Conserving K.E. : \frac 12mu 1^2 = \frac 12mv 1^2 + \frac 12mv 2^2
\Rightarrow u 1^2 = v 1^2 + v 2^2 \Rightarrow u 1^2 - v 1^2 = v 2^2 ......(1) Conserving Momentum : in x-direction mu 1 = mv 1\cos \theta + mv 2\cos \phi \Rightarrow u 1 = v 1\cos \theta + v 2\cos \phi \Rightarrow u 1 - v 1\cos \theta = v 2\cos \phi ........(2) in y-direction 0 = mv 1\sin \theta - mv 2\sin \phi \Rightarrow v 1\sin \theta = v 2\sin \phi .......(3) Squaring and Adding (2) & (3) ( u 1 - v 1\cos \theta)^2 + (v 1\sin \theta)^2 = (v 2\cos \phi)^2 + (v 2\sin \phi)^2 \Rightarrow u 1^2 - 2u 1v 1\cos \theta + v 1^2 = v 2^2 putting value of v 2^2 u 1^2 - 2u 1v 1\cos \theta + v 1^2 = u 1^2 - v 1^2 \Rightarrow v 1 = u 1\cos \theta substituting this value of v 1 in (1) \Rightarrow u 1^2 - u 1^2\cos^2 \theta = v 2^2 \Rightarrow v 2 = u 1\sin \theta substituting value of v 1 and v 2 in (3) u 1(\cos \theta)(\sin \theta) = u 1(\sin \theta)(\sin \phi) \Rightarrow \cos \theta = \sin \phi \Rightarrow \sin(90-theta) = \sin \phi \Rightarrow 90 - \theta = \phi or \theta + \phi = 90

Note that since the two proton system is isolated (there are no external forces) we can apply conservation of linear momentum. Let us denote the initial velocity of the moving proton by $\vec{v}_{0}$ . Then $m \vec{v}_{0}= m \vec{v}_{1} +m \vec{v}_{2}.$ where $v_{1}$ and $v_{2}$ are the final velocities of the protons. In addition, we have from conservation of energy that $m\frac{v_{0}^{2} }{2}= m\frac{v_{1}^{2} }{2}+ m\frac{v_{2}^{2} }{2}.$ We do not need to take into account the electrostatic interaction energy because the protons far away from each other in both the initial and the final configuration. From the first equation we have that $\vec{v}_{0}= \vec{v}_{1} + \vec{v}_{2}$ which implies $v_{0}^{2}= \vec{v}_{0}\cdot\vec{v}_{0}= (\vec{v}_{1}+\vec{v}_{2})\cdot (\vec{v}_{1}+\vec{v}_{2})= v_{1}^{2}+v_{2}^{2}+ 2 \vec{v}_{1}\cdot \vec{v}_{2} .$ The dot ( $\cdot$ ) between the vectors denotes scalar product. Note that the above equation is compatible with conservation of energy if and only if $\vec{v}_{1}\cdot \vec{v}_{2}=0$
which means that either the angle between the vectors is $90^{\circ}$ or one of the final velocities is zero. Therefore if the protons interact, and the collision is not collinear, the angle between the velocities must be $90^{\circ}$ . If the collision is collinear (which is not likely), the initially moving proton stops and transfers its velocity to the other proton. Note, that this result holds for any elastic collision between objects of equal mass.

one proton is at rest and another proton is moving.therefore they can't move at 180 degrees change each others velocity.If they move at 90 degrees,they don't change each others velocity