Answer
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Hint: Mathematically, linear momentum is given by, \[p=mv\] and kinetic energy is given by, \[K.E.=\dfrac{1}{2}m{{v}^{2}}\]. Where,
\[m\]= mass of the body
\[v\]= velocity
\[p\]= linear momentum
\[K.E.\]= kinetic energy
Complete step by step Solution:
Linear momentum: Linear momentum is the quantity of motion of a moving body. Its magnitude is given by the product of mass and velocity of the body at a given time. It is given by, \[p=mv\]
Kinetic energy: Kinetic energy is defined as the energy possessed by a body because of its motion. It is given by, \[K.E.=\dfrac{1}{2}m{{v}^{2}}\]
Where symbols have their usual meanings.
Now, since the momentum is expressed as,
\[p=mv\]
On squaring both sides,
\[\Rightarrow {{p}^{2}}={{\left( mv \right)}^{2}}\]
On dividing by \[2m\] both sides,
\[\Rightarrow \dfrac{{{p}^{2}}}{2m}=\dfrac{1}{2}m{{v}^{2}}\]
As we know that the kinetic energy of a body is given by, \[K.E.=\dfrac{1}{2}m{{v}^{2}}\] so we can write the above equation as,
\[\Rightarrow \dfrac{{{p}^{2}}}{2m}=K.E.\]
\[\Rightarrow {{p}^{2}}=2m\times (K.E.)\]
On taking square root both sides,
\[\Rightarrow p=\sqrt{2m(K.E.)}\]
Hence, the relation between the linear momentum and the kinetic energy is, \[p=\sqrt{2m(K.E.)}\]
Additional Information:
Momentum is directly proportional to the object’s mass and its velocity. Thus, the greater an object’s mass or the greater its velocity, the greater will be its momentum.
Momentum, \[p\] is a vector quantity having the same direction as the velocity \[v\]. The SI unit for momentum is\[kg.m/s\]
Kinetic energy is the form of energy that an object or a particle possesses because of its motion. If work, which transfers energy, is done on an object by applying a net force, the object speeds up and thereby gains kinetic energy.
Kinetic energy, \[K.E.\] is a scalar quantity. The SI unit for kinetic energy is $\text{joule}$.
Note: Students should understand the physical significance of both the physical quantities i.e., linear momentum and kinetic energy with their units. Students need to memorize the basic mathematical expressions of both quantities so that by performing some mathematical operations they can get the required relation.
\[m\]= mass of the body
\[v\]= velocity
\[p\]= linear momentum
\[K.E.\]= kinetic energy
Complete step by step Solution:
Linear momentum: Linear momentum is the quantity of motion of a moving body. Its magnitude is given by the product of mass and velocity of the body at a given time. It is given by, \[p=mv\]
Kinetic energy: Kinetic energy is defined as the energy possessed by a body because of its motion. It is given by, \[K.E.=\dfrac{1}{2}m{{v}^{2}}\]
Where symbols have their usual meanings.
Now, since the momentum is expressed as,
\[p=mv\]
On squaring both sides,
\[\Rightarrow {{p}^{2}}={{\left( mv \right)}^{2}}\]
On dividing by \[2m\] both sides,
\[\Rightarrow \dfrac{{{p}^{2}}}{2m}=\dfrac{1}{2}m{{v}^{2}}\]
As we know that the kinetic energy of a body is given by, \[K.E.=\dfrac{1}{2}m{{v}^{2}}\] so we can write the above equation as,
\[\Rightarrow \dfrac{{{p}^{2}}}{2m}=K.E.\]
\[\Rightarrow {{p}^{2}}=2m\times (K.E.)\]
On taking square root both sides,
\[\Rightarrow p=\sqrt{2m(K.E.)}\]
Hence, the relation between the linear momentum and the kinetic energy is, \[p=\sqrt{2m(K.E.)}\]
Additional Information:
Momentum is directly proportional to the object’s mass and its velocity. Thus, the greater an object’s mass or the greater its velocity, the greater will be its momentum.
Momentum, \[p\] is a vector quantity having the same direction as the velocity \[v\]. The SI unit for momentum is\[kg.m/s\]
Kinetic energy is the form of energy that an object or a particle possesses because of its motion. If work, which transfers energy, is done on an object by applying a net force, the object speeds up and thereby gains kinetic energy.
Kinetic energy, \[K.E.\] is a scalar quantity. The SI unit for kinetic energy is $\text{joule}$.
Note: Students should understand the physical significance of both the physical quantities i.e., linear momentum and kinetic energy with their units. Students need to memorize the basic mathematical expressions of both quantities so that by performing some mathematical operations they can get the required relation.
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