Consider an object that at one time has energy E1 and momentum p1 and at a later time has energy E2 and momentum p2. Use the relativistic energy-momentum equation E2=p2c2+m2c4 to find the value of E22−E21.

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Question:

Consider an object that at one time has energy E1 and momentum p1 and at a later time has energy E2 and momentum p2. Use the relativistic energy-momentum equation E2=p2c2+m2c4 to find the value of E22−E21.

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Complete QuestionConsider an object that at one time has energy E_1 and momentum p 1 and at a later time has energy E_2 and momentum  p_2 . Use the relativistic energy-momentum equation E 2 = p 2 c 2 + m 2 c 4  to find the value of E 2 2 − E 2 1 . Express your answer in terms of p_1 , p_2 , m and cAnswer: The value of  [tex]E_2^2 -E_1^2[/tex] [tex]=(p_2^2 -p_1^2)c^2[/tex]Explanation:   The objective of the Question is to obtain                  [tex]E_2^2 -E_1^2[/tex]Energy momentum equation is mathematically represented as            [tex]E^2 = p_1 c^2 +m^2 c^2[/tex]   Where c is velocity and m is mass From the question we are told that          [tex]E_2^2 = p_2c^2 +m^2c^2[/tex]Therefore         [tex]E^2_1 = p_1c^2 +m^2c^2[/tex] Now             [tex]E_2^2 -E_1^2[/tex] [tex]=(p_2^2 -p_1^2)c^2[/tex]

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