Question

Two masses of \[1\,g\] and \[4\,g\] are moving with equal kinetic energies. The ratio of the magnitudes of their linear momenta is
A. \[4:1\]
B. \[\sqrt 2 :1\]
C. \[1:2\]
D. \[1:16\]

Answer 1

Hint:A particle or an object that is in motion has a sort of energy called kinetic energy. When work, which entails the transfer of energy, is done on an object by applying a net force, that object acquires kinetic energy.

Formula Used:
The equation to calculate kinetic energy is as follows:
\[K.E. = \dfrac{1}{2}m{v^2}\]
Here, \[m\] is mass and \[v\] is the velocity of the object.

Complete step by step solution:
The product of an object's mass, m, and its velocity, v, is the vector quantity known as linear momentum. It is represented by the letter "p," which is also used to denote momentum. Please be aware that the body's momentum always points in the same direction as its vector of velocity. Because it is a conserved quantity, a system's overall momentum is always constant. The unit of linear momentum is \[kg\,m/s\]. The linear momentum formula is given by,
\[p = mv\]
Here, \[p\] is linear momentum, \[m\] is mass and \[v\] is the velocity of the object.

The given masses are as follows:
\[{m_1} = 1\,g\]
\[\Rightarrow {m_2} = 4\,g\]
The kinetic energy for object 1 is as follows:
\[K{E_1} = \dfrac{1}{2}{m_1}{v_1}^2 \\
\Rightarrow K{E_1} = \dfrac{1}{2}\left( 1 \right){v_1}^2 \\
\Rightarrow K{E_1} = \dfrac{1}{2}{v_1}^2\,\,\,...\,(1) \\ \]
The kinetic energy for object 2 is as follows:
\[K{E_2} = \dfrac{1}{2}{m_2}{v_2}^2 \\
\Rightarrow K{E_2} = \dfrac{1}{2}\left( 4 \right){v_2}^2 \\
\Rightarrow K{E_2} = 2{v_2}^2\,\,\,...\,(2) \\ \]
The kinetic energy is equals for the given masses,
\[K{E_1} = K{E_2} \\
\Rightarrow \dfrac{1}{2}{v_1}^2 = 2{v_2}^2 \\
\Rightarrow {v_1}^2 = 4{v_2}^2 \\ \]
Further solving,
\[{v_1} = 2{v_2}\,\,\,\,...(3)\]
Linear momentum of object 1 is as follows:
\[{p_1} = {m_1}{v_1}\]
Linear momentum of object 2 is as follows:
\[{p_2} = {m_2}{v_2}\]
Taking the ratio of the linear momentum as follows:
\[\dfrac{{{p_2}}}{{{p_1}}} = \dfrac{{{m_2}{v_2}}}{{{m_1}{v_1}}} \\
\Rightarrow \dfrac{{{p_2}}}{{{p_1}}} = \dfrac{{4 \times {v_2}}}{{1 \times 2{v_2}}} \\
\Rightarrow \dfrac{{{p_2}}}{{{p_1}}} = \dfrac{1}{2} \\ \]
Therefore, the ratio of the magnitude of their linear momentum is \[1:2\] .

Note: A linear motion has linear momentum as its defining feature. In bodies of various masses, the same force exerted over the same period of time results in the same change in linear momentum.