Answer
The ratio of the pellet's kinetic energy to the bullet's kinetic energy is $~~255$
Work Step by Step
In part (a), we found that the kinetic energy of the pellet in the reference frame of the satellite is $~~4.6\times 10^5~J$
We can find the kinetic energy of a bullet:
$K = \frac{1}{2}mv^2$
$K = \frac{1}{2}(0.0040~kg)(950~m/s)^2$
$K = 1805~J$
We can find the ratio of the pellet's kinetic energy to the bullet's kinetic energy:
$\frac{4.6\times 10^5~J}{1805~J} = 255$
The ratio of the pellet's kinetic energy to the bullet's kinetic energy is $~~255$
