Question

Two bodies A and B having mass M and m respectively possess the same kinetic energy. Given that $M > m$. If ${\rho _A}$ and ${\rho _B}$ be their momentum, then which of the following statements is true?
A. ${\rho _A} = {\rho _B}$
B. ${\rho _A} > {\rho _B}$
C. ${\rho _A} < {\rho _B}$
D. It cannot be predicted

Answer 1

Hint: When the kinetic energy of a body remains constant then the momentum of the body depends on the mass of the body by the relation
$\rho = \sqrt {2mk} $
Where
$\rho $= momentum of the body
$m$ = mass of the body
$k$ = kinetic energy of the body
If the kinetic energy of the body remains constant then on increasing the mass of the body, momentum of the body also increases because mass is directly proportional to the square root of the mass of the body.

Complete step-by-step answer:
We know that
Kinetic energy of a body given by, $k = \dfrac{1}{2}m{v^2}$
Momentum of a body given by, $\rho = mv$
On multiplying and dividing $m$on left side we get
$k = \dfrac{1}{{2m}}{\left( {mv} \right)^2} \Leftrightarrow k = \dfrac{1}{{2m}}{\rho ^2}$
On further solving we get
$\rho = \sqrt {2mk} $
 Therefore the momentum of body A, given by
${\rho _A} = \sqrt {2Mk} \cdot \cdot \cdot \cdot \cdot \cdot \left( 1 \right)$
Therefore the momentum of body B, given by
${\rho _B} = \sqrt {2mk} \cdot \cdot \cdot \cdot \cdot \cdot \left( 2 \right)$
From (1) and (2) we get
$\dfrac{{{\rho _A}}}{{{\rho _B}}} = \sqrt {\dfrac{M}{m}} $
Since $M > m$ therefore ${\rho _A} > {\rho _B}$
Hence the momentum of body A is greater than body B
Hence the correct option is B.

Note: In the above question if the velocity varies then momentum of the bodies should not be compared by the above relation. Then in order to get the correct relation between them we should put the kinetic energy in the terms of velocity and their corresponding masses. Generally the mass of the body does not vary but if the velocity of the particle becomes equal to velocity of light then their mass varies, in that situation these equations are not valid.