
Two bodies with kinetic energies in the ratio $4:1$ are moving with equal linear momentum. The ratio of their masses is?
A. $1:1$
B. $1:2$
C. $4:1$
D. $1:4$
Answer
566.4k+ views
Hint: kinetic energy is the energy of a body due to its motion. The kinetic energy is given by the formula \[K=\dfrac{m{{v}^{2}}}{2}\], where m is the mass of the body and v is its velocity. It is clear that if the body is at rest, then its velocity will be zero and thus zero kinetic energy. Also, momentum of a body is given by the relation, $p=mv$.
Complete step by step answer:
The relation between the momentum and kinetic energy is given by \[K=\dfrac{{{p}^{2}}}{2m}\]
Now as per the question, \[\dfrac{{{K}_{1}}}{{{K}_{2}}}=\dfrac{4}{1}\]and linear momentum is same for both.
\[\dfrac{{{K}_{1}}}{{{K}_{2}}}=\dfrac{4}{1}\\
\Rightarrow\dfrac{{{K}_{1}}}{{{K}_{2}}} =\dfrac{{{p}_{1}}^{2}\times
2{{m}_{2}}}{2{{m}_{1}}\times p_{2}^{2}}\]
$\Rightarrow \dfrac{{{p}_{1}}^{2}\times 2{{m}_{2}}}{2{{m}_{1}}\times p_{2}^{2}}=\dfrac{4}{1} \\
\therefore \dfrac{{{m}_{2}}}{{{m}_{1}}}=\dfrac{4}{1} \\$
So, the correct option is C.
Additional Information:
The work-energy theorem states that the net work done by the forces on an object equals the change in its kinetic energy. The law of conservation of momentum states that the momentum of an isolated system remains constant. mechanical energy can be summed up as the sum of potential and kinetic energy stored or possessed by the body which is either at rest, either in motion
Note: If no external force acts on the system, then the total momentum of the system must remain conserved. Here we make use of the relationship between kinetic energy and momentum to arrive at the solution. Kinetic energy is a scalar quantity while momentum both linear and angular are vector quantities. Momentum is a vector quantity because it is the product of mass and velocity and velocity is a vector quantity.
Complete step by step answer:
The relation between the momentum and kinetic energy is given by \[K=\dfrac{{{p}^{2}}}{2m}\]
Now as per the question, \[\dfrac{{{K}_{1}}}{{{K}_{2}}}=\dfrac{4}{1}\]and linear momentum is same for both.
\[\dfrac{{{K}_{1}}}{{{K}_{2}}}=\dfrac{4}{1}\\
\Rightarrow\dfrac{{{K}_{1}}}{{{K}_{2}}} =\dfrac{{{p}_{1}}^{2}\times
2{{m}_{2}}}{2{{m}_{1}}\times p_{2}^{2}}\]
$\Rightarrow \dfrac{{{p}_{1}}^{2}\times 2{{m}_{2}}}{2{{m}_{1}}\times p_{2}^{2}}=\dfrac{4}{1} \\
\therefore \dfrac{{{m}_{2}}}{{{m}_{1}}}=\dfrac{4}{1} \\$
So, the correct option is C.
Additional Information:
The work-energy theorem states that the net work done by the forces on an object equals the change in its kinetic energy. The law of conservation of momentum states that the momentum of an isolated system remains constant. mechanical energy can be summed up as the sum of potential and kinetic energy stored or possessed by the body which is either at rest, either in motion
Note: If no external force acts on the system, then the total momentum of the system must remain conserved. Here we make use of the relationship between kinetic energy and momentum to arrive at the solution. Kinetic energy is a scalar quantity while momentum both linear and angular are vector quantities. Momentum is a vector quantity because it is the product of mass and velocity and velocity is a vector quantity.
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