
The kinetic energy of a body has been increased by$21\%$. What is the percentage increase in the magnitude in the linear momentum of the body?
$\begin{align}
& A.10\% \\
& B.25\% \\
& C.15\% \\
& D.40\% \\
\end{align}$
Answer
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Hint:The kinetic energy will be conserved here. So conservation should be taken care of. From this the relation between the velocities in the two cases should be found out. And then the relation between momentums is found out and hence the percentage difference.
Complete answer:
First of all let us discuss the kinetic energies and the linear momentum. The kinetic energy of an object is the energy a body possesses because of its motion.
Kinetic energy is given by the equation
$K.E=\dfrac{1}{2}m{{v}^{2}}$
Where m is the mass of the body or the object, v is the velocity with which it is moving and the unit of kinetic energy is given in the units of$J$. As it is an energy, the quantity is a scalar quantity also. Linear momentum is described as the product of a system's mass multiplied by its velocity. Linear momentum is expressed as
$P=mv$
Momentum is directly proportional to the mass of the object and also its velocity. Thus if the object's mass is greater or the velocity is greater, then the momentum will also be greater.
Kinetic energy in the first case will be equal to that of the second case.
$\begin{align}
& {{E}_{K2}}=\dfrac{121}{100}{{E}_{K1}} \\
& \dfrac{1}{2}m{{v}_{2}}^{2}=\dfrac{121}{100}\dfrac{1}{2}m{{v}_{1}}^{2} \\
\end{align}$
Hence the velocity relation will be,
${{v}_{2}}=\dfrac{11}{10}{{v}_{1}}$
Multiplying this with mass will give momentum,
$\begin{align}
& m{{v}_{2}}=\dfrac{11}{10}m{{v}_{1}} \\
& {{P}_{2}}=\dfrac{11}{10}{{P}_{1}} \\
\end{align}$
Therefore percentage change in momentum will be
$\dfrac{{{P}_{2}}-{{P}_{1}}}{{{P}_{1}}}\times 100=\dfrac{1}{10}\times 100=10\%$
Hence the percentage increase in momentum will be$10\%$.
Note:
kinetic energy is explained as the work required to accelerate a body of a given mass from rest to its particular velocity. During its acceleration, the body have gained energy the body will maintain this kinetic energy unless its speed changes.
Complete answer:
First of all let us discuss the kinetic energies and the linear momentum. The kinetic energy of an object is the energy a body possesses because of its motion.
Kinetic energy is given by the equation
$K.E=\dfrac{1}{2}m{{v}^{2}}$
Where m is the mass of the body or the object, v is the velocity with which it is moving and the unit of kinetic energy is given in the units of$J$. As it is an energy, the quantity is a scalar quantity also. Linear momentum is described as the product of a system's mass multiplied by its velocity. Linear momentum is expressed as
$P=mv$
Momentum is directly proportional to the mass of the object and also its velocity. Thus if the object's mass is greater or the velocity is greater, then the momentum will also be greater.
Kinetic energy in the first case will be equal to that of the second case.
$\begin{align}
& {{E}_{K2}}=\dfrac{121}{100}{{E}_{K1}} \\
& \dfrac{1}{2}m{{v}_{2}}^{2}=\dfrac{121}{100}\dfrac{1}{2}m{{v}_{1}}^{2} \\
\end{align}$
Hence the velocity relation will be,
${{v}_{2}}=\dfrac{11}{10}{{v}_{1}}$
Multiplying this with mass will give momentum,
$\begin{align}
& m{{v}_{2}}=\dfrac{11}{10}m{{v}_{1}} \\
& {{P}_{2}}=\dfrac{11}{10}{{P}_{1}} \\
\end{align}$
Therefore percentage change in momentum will be
$\dfrac{{{P}_{2}}-{{P}_{1}}}{{{P}_{1}}}\times 100=\dfrac{1}{10}\times 100=10\%$
Hence the percentage increase in momentum will be$10\%$.
Note:
kinetic energy is explained as the work required to accelerate a body of a given mass from rest to its particular velocity. During its acceleration, the body have gained energy the body will maintain this kinetic energy unless its speed changes.
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