Question

a boy recalls the relation for relativistic mass (m) in terms of rest mass mó velocity of particle V but forget to put the constant C (velocity of light) he writes m= mó/1 -v² correct the equation by putting the missing C​

Answer 1

Solution :

According to the principle of homogeneity of dimensions, powers of M, L, T on either side of the forumal

must be equal. For this on RHS, the denominator (1−υ2)1/2 should be dimensionless. Therefore instead

of (1−υ2)1/2, we should write (1−υ2/c2)1/2

Hence, the correct formula would be m=m0(1−υ2/c2)1/2

Answer 2

The incorrect relation between the moving mass (m), rest mass (m0), speed of the moving mass (v) and the speed of light (c) is,

m=m0(1−v2)1/2

Rearrange the above equation further,

(1−v2)1/2=m0m1−v2=m0m−−−√

Since the RHS term (m0m−−−√) is dimension less, so the relation to be dimensionally correct, the LHS term (1−v2) must be dimensionless.

To make the LHS term dimensionless replace the term (1−v2) by (1−v2c2) .

Hence, the correct relation is given as,

m=m0(1−v2c2)1/2=m0(1−v2/c2)−1/2