Question

A rocket is moving up with a velocity v. If the velocity of this rocket is suddenly tripled, what will be the ratio of two kinetic energies?

Answer 1

Step 1: Given data

1. Initially, the velocity of the rocket is $v$.
2. Finally, the velocity of the rocket is $3v$.

Step 2: Kinetic energy

1. Kinetic energy is the energy possessed by a body when it is in motion.
2. Kinetic energy is defined by the form, $K_E=\frac{1}{2}m{v}^2$, where, m is the mass of the body and v is the velocity of the body.

Step 3: Finding the ratio of kinetic energies

Initially, the kinetic energy of the rocket is $K=\frac{1}{2}m{v}^2$.

Finally, the kinetic energy of the rocket is

$$\begin{gathered} K\text{'}=\frac{1}{2}m{\left(3v\right)}^2=\frac{1}{2}m.9v^2 \\ orK\text{'}=\frac{1}{2}m.9v^2 \end{gathered}$$.

Now,

$$\begin{gathered} \frac{K\text{'}}{K}=\frac{{\displaystyle \frac{1}{2}}m.9v^2}{{\displaystyle \frac{1}{2}}m{v}^2}=9 \\ so,\frac{K\text{'}}{K}=9. \end{gathered}$$

So, the ratio of two kinetic energies is $9:1$.