Question

A particle of charge Q and mass m travels through a potential difference V from rest. The final momentum of the particle is:
(A). \[P = \sqrt {\dfrac{{Qm}}{V}} \]
(B). \[P = \sqrt {QV} \]
(C). \[P = \sqrt {2mQV} \]
(D). \[P = \sqrt {\dfrac{{2QV}}{m}} \]

Answer 1

Hint: In this question the formula used is derived from work energy theorem and the another formula comes into account is of momentum -
\[{K_f} - {K_i} = W\]
\[p = m \times \sqrt {\dfrac{{2vq}}{m}} \]
\[p\]= momentum of the body by using this information we can easily approach our answer.

Complete step-by-step answer:
Work Energy Theorem – The fluctuation in kinetic energy of a particle is regarded as the work done on it by the net force. Work coined as to the force and displacement over which it acts. Work is completed by the equivalent force over a particular displacement on the body. The W-E theorem is scalar from the second law for two or three dimensions in a vector form. The information w.r.t directions contained in Newton’s second law is not present in it.
\[{K_f} - {K_i} = W\]

Kinetic Energy – It is a scalar quantity. It is a measure of the work an object can do because of its motion. Sailing ships use KE of wind. KE of a fast-flowing stream has been used for grinding corn.

Momentum (\[p\]) – It is defined as the quantity of motion contained in a body. It is measured as the product of mass of the body and its velocity and has the same direction as that of velocity. It is a vector quantity. It is represented by (\[p\]).
Momentum (\[p\]) = \[mass \times velocity\]
The S.I coined for momentum is kg m/s.

 How KE and linear momentum is connected – mass (m) and velocity (v)
(\[p\]) = \[mass \times velocity\]
KE = \[\dfrac{1}{2}m{v^2} = \dfrac{1}{{2m}}\left( {{m^2}{v^2}} \right)\]
\[KE = \dfrac{{{p^2}}}{{2m}}\]

By work - energy theorem,
Work done = change in kinetic energy
\[v \times q = \dfrac{1}{2}m\left( {{x^2} - {u^2}} \right)\]
\[vq = \dfrac{{{x^2}m}}{2}\]
Velocity = \[x = \sqrt {\dfrac{{2vq}}{m}} \]
Momentum = \[mass \times velocity\]
\[p = m \times \sqrt {\dfrac{{2vq}}{m}} \]
\[\therefore p = \sqrt {2vqm} \]
So the correct option is (C).

Note:
KE depends on the frame of reference. KE is always positive. KE holds even when the magnitude or direction of force changes. So, expression is valid irrespective of how the body acquires the velocity v.
W-E theorem is an integral form of Newton’s second law.