Question

Two bodies of masses m, and m, have equal momenta. Their kinetic energy are in the ratio

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Answer 1

Given:-

• Masses of the two bodies are m and m.

• Their momentum is equal.

To Find:-

• Ratio of their Kinetic Energy.

Solution:-

Here, We know that  $K.E = \dfrac{1}{2} m v^{2}$

So, Lets find Relation Between Kinetic Energy and Momentum :-

$K.E = \dfrac{1}{2} m v^{2}$

$K.E = \dfrac{1}{2} m v^{2} \times \dfrac{m}{m}$                  [ We can multiply as well as divide the equation with same number ]

$K.E = \dfrac{ m^{2} v^{2}}{2m}$

$K.E = \dfrac{ (mv)^{2} }{2m}$  

$K.E = \dfrac{p^{2}}{2m}$                                  [ We know that momentum (p) = mv ]

$K.E = \dfrac{p^{2}}{2m}$  

This is the required relation.

Hence, Now put the values in the above formula

$\dfrac{K.E_1}{K.E_2} = \dfrac{\dfrac{p^{2}}{2m}}{\dfrac{p^{2}}{2m}}$                   [ Since, It is given that mases of both bodies are m and m and momentum is equal ]

So, $\dfrac{K.E_1}{K.E_2} = \dfrac{p^{2}}{p^{2}}$                [ 2m and 2m are equal so it is cancelled ]

Now, $\dfrac{K.E_1}{K.E_2} = \dfrac{1}{1}$             [ momentum of both bodies are equal hence it is cancelled ]

Hence, Ratio of their Kinetic Energy is 1 : 1 .

   

Some Important Terms:-

• Potential Energy = mgh

• Kinetic Energy $(K.E )= \dfrac{1}{2} m v^{2}$