Question

Question 2.15 A famous relation in physics relates ‘moving mass’ m to the ‘rest mass’ m0 of a particle in terms of its speed v and the speed of light, c. (This relation first arose as a consequence of special relativity due to Albert Einstein). A boy recalls the relation almost correctly but forgets where to put the constant c. He writes:

Class XI Physics Units And Measurements Page 36

Answer 1

Given the relation,m = m0 / (1-v2)1/2
Dimension of m = [M1]
Dimension of m0 = [M1]
Dimension of v = [L1 T–1]
Dimension of v2 = [ L2 T–2]
Dimension of c = [ L1 T–1]
The given formula will be dimensionallycorrect only when the dimension of L.H.S is the same as that of R.H.S. This is only possible when the factor,
(1-v2)1/2 is dimensionless
i.e., (1 – v2) is dimensionless.
This is only possible if v2 is divided by c2. Hence, the correct relation ism = m0 / (1 - v2/c2)1/2
hope u will understand it

Answer 2

According to the principle of homogeneity of dimensions, powers of M,L,T on either side of the formula must be equal. For this, on RHS, the denominator (1-v^2)^1/2 should be dimensionless. Therefore, instead of (1-v^2)^1/2, we should write (1-v^2/c^2)^1/2

Hence the correct formula would be
m=m0/(1-v2/c^2)^1/2