Answer: [tex]\sqrt{2}[/tex]
The linear momentum [tex]p[/tex] is given by the following equation:
[tex]p=m.v[/tex] (1)
Where [tex]m[/tex] is the mass and [tex]v[/tex] the velocity.
On the other hand, the kinetic energy [tex]K[/tex] is given by:
[tex]K=\frac{p^{2}}{2m}[/tex] (2)
Which is the same as:
[tex]K=\frac{1}{2}m.v^{2}[/tex]
Now, if we double the kinetic energy, equation (2) changes to:
[tex]2K=2\frac{p^{2}}{2m}[/tex]
[tex]2K=\frac{p^{2}}{m}[/tex] (3)
So, if we want to obtain the kinetic energy as shown in (3), the only option that works is increasing momentum by a factor of [tex]\sqrt{2}[/tex] or [tex]2^{1/2}[/tex]:
Applying this in (2):
[tex]K=\frac{(\sqrt{2}p)^{2}}{2m}[/tex]
[tex]K=\frac{(2p)^{2}}{2m}[/tex]
[tex]K=\frac{p^{2}}{m}[/tex]>>>As we can see, this equation is the same as equation (3)
Therefore, the correct answer is B
