Question

When the speed of the particle is doubled, the ratio of its kinetic energy to its momentum
A. Remains the same
B. Gets doubled
C. Becomes half
D. Becomes four times

Answer 1

This can be solved by just keeping the formulas of kinetic energy and momentum in a ratio for a particle with mass m and velocity v. See how the ratio varies with the changes in the velocity that we are asked in the question.

Formula used: Kinetic energy
$K.E.=(1/2)mv^2$
Momentum is just
p= mv

Complete step by step answer:
When a particle of mass m is traveling with a velocity v, we know that its
Kinetic energy is given by $(1/2)mv^2$
and momentum is mv
Therefore the ratio r, of the two is just:
$r= \dfrac{(1/2)mv^2}{mv} = \dfrac{v}{2}$
Now, if the velocity of the particle is doubled v becomes 2v and the ratio becomes
$r' = \dfrac{2v}{2} =2r$
which is twice the previous ratio so we can conclude that the ratio of the kinetic energy to momentum doubles when the speed of the particle is doubled.

So, the correct answer is “Option B”.

Additional Information: Kinetic energy is possessed by an object when it undergoes any kind of motion. Here we only used the translational kinetic energy but when objects undergo rotational motion there is also a term of rotational kinetic energy. The translational + the rotational term create the total kinetic energy in such a scenario.

Note: In such types of questions it is better to use a primed variable when the system parameters are changed. Here, we denoted the ratio by r and the ratio after velocity doubling by r' to avoid confusion. Also so r' can be written in terms of the previous p thus producing a clear result.