Question

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

Answer 1

Initial observations

Let the initial mass be m₁

Let the initial velocity be v₁

The initial kinetic energy is K₁

$\mathsf{K_1=\frac{1}{2}m_1v_1^2}$

Final observations

The final mass remains the same .

Final mass = initial mass = m₁

Final velocity is triple the initial velocity

v₂ = 3 v₁

The final kinetic energy is K₂

$\mathsf{K_2=\frac{1}{2}m_1v_2^2}$

$\mathsf{\implies K_2=\frac{1}{2}m_1(3v_1)^2}$

RATIO :

$\mathsf{\frac{K_1}{K_2}}\\\implies\mathsf{\frac{\frac{1}{2}m_1(v_1)^2}{\frac{1}{2}m_1(3v_1)^2}}$

$\implies \mathsf{\frac{K_1}{K_2}=\frac{1}{3^2}}$

$\implies \frac{1}{9}$

ANSWER :

The ratio of kinetic energy would be $\boxed{\frac{1}{9}}$

Hope it helps :)

______________________________________________________________