Einstein’s Special Theory of Relativity :: Discussion


4.For Einstein’s relation, E2 – p2c2 =?
A.moc2
B.mo2c4
C.moc4
D.mo2c6
Answer:  Option  B
Explanation:

We know, E = mc2 and momentum, p = mv
Now, E2 – p2c2 = m2c4 – m2v2c2
Now, we know, m = m01V2c2
Therefore, E2 – p2c2 = m2c4(1-v2/c2)
=c4m2(1-v2/c2)
Using, m2(1-v2/c2) = mo2 we get
E2 – p2c2 = mo2c4.

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Published by:Michael Daani

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