
The kinetic energy of a body is increased by 21%. What is increased in the momentum of the body.
A river of salty water is flowing with a velocity of 10 m/sec. If the density of the water is 1.2 gm/$c{m^3}$. Then the kinetic energy of each cubic metre of water is
Answer
589.8k+ views
Hint: We will first see the definition of kinetic energy. Then we will see the formula to find the kinetic energy of a body having mass ‘m’ in two cases. In the first case when mass ‘m’ and velocity ‘v’ is given and in the second case when mass and momentum are given. Finally we will calculate the kinetic energy of the given body.
Complete answer:
Let us first see the definition of kinetic energy.
This is the energy associated with the motion of the body. It is defined as the energy possessed by a body of mass ‘m’ moving with velocity ‘v’. The kinetic energy is given as:
K.E. = $\dfrac{1}{2}m{v^2}$ . Its unit is joule.
Momentum of a body having mass ‘m’ and velocity ‘v’ is given as:
P = mv. Its unit is m/sec.
Kinetic energy in terms of momentum of the body and mass of the body is given as:
K.E. = $\dfrac{{{p^2}}}{{2m}}$ .
$ \Rightarrow p = \sqrt {2m \times K.E} $
Let the initial K.E. be ‘y’ joule.
Therefore, initial momentum is given as:
${p_i} = \sqrt {2my} $ .
Now the final K.E. = y + $\dfrac{{21y}}{{100}}$ =$\dfrac{{121y}}{{100}}$
$\therefore $ Final momentum = ${p_f} = \sqrt {\dfrac{{2 \times 121}}{{100}}my} = \dfrac{{11}}{{10}}\sqrt {2my} $
Change in momentum = ${p_f} - {p_i} = \dfrac{{11}}{{10}}\sqrt {2my} - \sqrt {2my} = \dfrac{1}{{10}}\sqrt {2my} $
Percentage change in momentum =\[\dfrac{{{p_f} - {p_i}}}{{{p_i}}} \times 100\% = \dfrac{{\dfrac{1}{{10}}\sqrt {2my} }}{{\sqrt {2my} }} \times 100\% = 10\% \]
Now it is given that the mass of 1$c{m^3}$ of is 1.2 gm.
Therefore, the mass of 1${m^3}$ of water = $\dfrac{{1.2 \times {{10}^6}}}{{{{10}^3}}}$ kg =$1200$kg.
Kinetic energy of 1 cubic metre of water = $\dfrac{1}{2} \times $ mass of water $ \times $${\left( {{\text{Velocity of water}}} \right)^2}$
=$\dfrac{1}{2} \times 1200 \times 10 \times 10$ = 60KJ.
Note: In this question, you should remember the important relationship between kinetic energy and momentum of the body. Unlike potential energy, the kinetic energy does not depend on the final and initial position of the body. It depends on the velocity of the body. A smaller body with a high velocity has a more impact because of its high energy, example is a bullet from a gun.
Complete answer:
Let us first see the definition of kinetic energy.
This is the energy associated with the motion of the body. It is defined as the energy possessed by a body of mass ‘m’ moving with velocity ‘v’. The kinetic energy is given as:
K.E. = $\dfrac{1}{2}m{v^2}$ . Its unit is joule.
Momentum of a body having mass ‘m’ and velocity ‘v’ is given as:
P = mv. Its unit is m/sec.
Kinetic energy in terms of momentum of the body and mass of the body is given as:
K.E. = $\dfrac{{{p^2}}}{{2m}}$ .
$ \Rightarrow p = \sqrt {2m \times K.E} $
Let the initial K.E. be ‘y’ joule.
Therefore, initial momentum is given as:
${p_i} = \sqrt {2my} $ .
Now the final K.E. = y + $\dfrac{{21y}}{{100}}$ =$\dfrac{{121y}}{{100}}$
$\therefore $ Final momentum = ${p_f} = \sqrt {\dfrac{{2 \times 121}}{{100}}my} = \dfrac{{11}}{{10}}\sqrt {2my} $
Change in momentum = ${p_f} - {p_i} = \dfrac{{11}}{{10}}\sqrt {2my} - \sqrt {2my} = \dfrac{1}{{10}}\sqrt {2my} $
Percentage change in momentum =\[\dfrac{{{p_f} - {p_i}}}{{{p_i}}} \times 100\% = \dfrac{{\dfrac{1}{{10}}\sqrt {2my} }}{{\sqrt {2my} }} \times 100\% = 10\% \]
Now it is given that the mass of 1$c{m^3}$ of is 1.2 gm.
Therefore, the mass of 1${m^3}$ of water = $\dfrac{{1.2 \times {{10}^6}}}{{{{10}^3}}}$ kg =$1200$kg.
Kinetic energy of 1 cubic metre of water = $\dfrac{1}{2} \times $ mass of water $ \times $${\left( {{\text{Velocity of water}}} \right)^2}$
=$\dfrac{1}{2} \times 1200 \times 10 \times 10$ = 60KJ.
Note: In this question, you should remember the important relationship between kinetic energy and momentum of the body. Unlike potential energy, the kinetic energy does not depend on the final and initial position of the body. It depends on the velocity of the body. A smaller body with a high velocity has a more impact because of its high energy, example is a bullet from a gun.
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