Question

The kinetic energy of a rocket is increased by a factor of eight after its engines are fired, whereas its total mass is reduced by half through the burning of fuel. By what factor is the magnitude of its momentum changed?

Answer 1

The momentum increases by a factor of 2

Explanation:

We can solve this problem by rewriting the momentum of the rocket in terms of the kinetic energy and the mass.

The kinetic energy of the rocket is:

$K=\frac{1}{2}mv^2$ (1)

where

m is the mass

v is the velocity

The momentum of the rocket is

$p=mv$ (2)

From eq.(1) we get

$v=\sqrt{\frac{2K}{m}}$

and substituting into (2),

$p=\sqrt{2mK}$

Now in this problem we have:

- The kinetic energy of the rocket is increased by a factor 8:

$K' = 8K$

- The mass is reduced by half:

$m'=\frac{m}{2}$

Substituting, we find the new momentum:

$p'=\sqrt{2(\frac{m}{2}(8K)}=\sqrt{4(2mK)}=2\sqrt{2mK}=2p$

So, the momentum increases by a factor of 2.

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