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
Correct Answer: C
Solution :
From Einsteins theorem Relativistic momentum\[=\frac{{{m}_{0}}v}{\sqrt{1-{{v}^{2}}/{{C}^{2}}}}\] Putting\[v=2v\] we get\[\rho =\frac{{{m}_{0}}(2v)}{\sqrt{1-{{(2v)}^{2}}/{{c}^{2}}}}\] \[=2\left( \frac{{{m}_{0}}v}{\sqrt{1-{{(2v)}^{2}}/{{c}^{2}}}} \right)\] So the momentum becomes more than the double.
You need to login to perform this action.
You will be redirected in
3 sec
