
A body of mass ${M_1}$ collides elastically with another body of mass ${M_2}$ at rest. There is maximum transfer of energy when
A. ${M_1} > {M_2}$
B. ${M_1} < {M_2}$
C. ${M_1} = {M_2}$
D. Same for all values of ${M_1}$ and ${M_2}$
Answer
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Hint:In order to solve this question, we will apply the concept of elastic collision in which initial kinetic energy of the system is always equal to the final kinetic energy of the system and using this we will determine the case of maximum energy transfer.
Formula used:
Kinetic energy of a body is calculated as,
$K.E = \dfrac{1}{2}m{v^2}$
where m is the mass of the body and v is the velocity of the body and for elastic collision $K.{E_i} = K.{E_f}$ which means initial kinetic energy equals to the final kinetic energy.
Complete step by step solution:
According to the question, we have given that initially body of mass ${M_1}$ and suppose it have a velocity of v then its kinetic energy will be $K.E = \dfrac{1}{2}{M_1}{v^2}$ collides with another body of mass ${M_2}$ which was at rest so its kinetic energy will be zero so total initial kinetic energy of the system before collision is,
$K.{E_i} = \dfrac{1}{2}{M_1}{v^2} \to (i)$
Now after the collision the maximum energy which can be transferred to another body of mass ${M_2}$ is possible if whole energy of body ${M_1}$ which is $K.E = \dfrac{1}{2}{M_1}{v^2}$ transferred to ${M_2}$ so that body of mass ${M_1}$ became at rest after collision and transfer its whole energy to body of mass ${M_2}$ so after the collision ${M_1}$ became at rest and final kinetic energy of the system will be kinetic energy of ${M_2}$ which is written as,
$K.E = \dfrac{1}{2}{M_2}{v^2} \\
\Rightarrow \dfrac{1}{2}{M_2}{v^2} = \dfrac{1}{2}{M_1}{v^2} \\
\therefore {M_1} = {M_2} \\ $
Hence, the correct answer is option C.
Note: It should be remembered that, in elastic collision there is no loss in energy transfer but in elastic collision final kinetic energy is always less than initial kinetic energy as some of the energy loss in other forms of energy such as heat or sound energy.
Formula used:
Kinetic energy of a body is calculated as,
$K.E = \dfrac{1}{2}m{v^2}$
where m is the mass of the body and v is the velocity of the body and for elastic collision $K.{E_i} = K.{E_f}$ which means initial kinetic energy equals to the final kinetic energy.
Complete step by step solution:
According to the question, we have given that initially body of mass ${M_1}$ and suppose it have a velocity of v then its kinetic energy will be $K.E = \dfrac{1}{2}{M_1}{v^2}$ collides with another body of mass ${M_2}$ which was at rest so its kinetic energy will be zero so total initial kinetic energy of the system before collision is,
$K.{E_i} = \dfrac{1}{2}{M_1}{v^2} \to (i)$
Now after the collision the maximum energy which can be transferred to another body of mass ${M_2}$ is possible if whole energy of body ${M_1}$ which is $K.E = \dfrac{1}{2}{M_1}{v^2}$ transferred to ${M_2}$ so that body of mass ${M_1}$ became at rest after collision and transfer its whole energy to body of mass ${M_2}$ so after the collision ${M_1}$ became at rest and final kinetic energy of the system will be kinetic energy of ${M_2}$ which is written as,
$K.E = \dfrac{1}{2}{M_2}{v^2} \\
\Rightarrow \dfrac{1}{2}{M_2}{v^2} = \dfrac{1}{2}{M_1}{v^2} \\
\therefore {M_1} = {M_2} \\ $
Hence, the correct answer is option C.
Note: It should be remembered that, in elastic collision there is no loss in energy transfer but in elastic collision final kinetic energy is always less than initial kinetic energy as some of the energy loss in other forms of energy such as heat or sound energy.
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