For a body moving with relativistic speed if the velocity is doubled, then

Question

For a body moving with relativistic speed if the velocity is doubled, then (A) its linear momentum is doubled (B) its linear momentum will be less than double (C) its linear momentum will be more than double (D) its linear momentum remains unchanged

Tardigrade Admin · Accepted Answer

From Einsteins theorem Relativistic momentum =(m0 v/√1-v2 / C2) Putting v=2 v we get ρ=(m0(2 v)/√1-(2 v)2 / c2) =2((m0 v/√1-(2 v)2 / c2)) So the momentum becomes more than the double.

Q. For a body moving with relativistic speed if the velocity is doubled, then

its linear momentum is doubled

its linear momentum will be less than double

its linear momentum will be more than double

its linear momentum remains unchanged

From Einsteins theorem
Relativistic momentum $= \frac{m _{0} v}{1 - v ^{2} / C ^{2}}$
Putting $v = 2 v$
we get $ρ = \frac{m _{0} ( 2 v )}{1 - ( 2 v ) ^{2} / c ^{2}}$
$= 2 (\frac{m _{0} v}{1 - ( 2 v ) ^{2} / c ^{2}})$
So the momentum becomes more than the double.

Q. For a body moving with relativistic speed if the velocity is doubled, then

its linear momentum is doubled

its linear momentum will be less than double

its linear momentum will be more than double

its linear momentum remains unchanged

From Einsteins theorem Relativistic momentum =1−v2/C2​m0​v​ Putting v=2v we get ρ=1−(2v)2/c2​m0​(2v)​ =2(1−(2v)2/c2​m0​v​) So the momentum becomes more than the double.

From Einsteins theorem
Relativistic momentum $= \frac{m _{0} v}{1 - v ^{2} / C ^{2}}$
Putting $v = 2 v$
we get $ρ = \frac{m _{0} ( 2 v )}{1 - ( 2 v ) ^{2} / c ^{2}}$
$= 2 (\frac{m _{0} v}{1 - ( 2 v ) ^{2} / c ^{2}})$
So the momentum becomes more than the double.