Question

A body of mass $5kg$ is moving with a velocity of $10m{s}^{-1}$. what will be the ratio of its initial kinetic energy and final kinetic energy, if the mass of the body is doubled and the velocity is halved?

Answer 1

Step 1: Given data and assumptions

Given data:

Mass of the body $\left(m\right)=5kg$

Initial velocity $\left(v_1\right)=10m{s}^{-1}$

Note: As the mass is doubled and the velocity is halved:

The final mass of the body $$\begin{gathered} \left(m_1\right)=2m=2\times 5kg \\ {m}_1=10kg \end{gathered}$$

The final velocity of the body will be $\left(v_2\right)=\frac{v_1}{2}=\frac{10m{s}^{-1}}{2}=5m{s}^{-1}$

Assumptions:

Let the initial kinetic energy of the body be $\text{'}{K}_1\text{'}$

And the final kinetic energy of the body be $\text{'}{K}_2\text{'}$

Step 2: Formula used

According to the formula of kinetic energy:

$K=\frac{1}{2}m{v}^2.......\left(a\right)$

Step 3: Calculating the initial kinetic energy

On, putting all the initial values in equation (a), we get

$$\begin{gathered} K_1=\frac{1}{2}\times 5kg\times {\left(10m{s}^{-1}\right)}^2 \\ {K}_1=\frac{1}{2}\times 5\times 100=250J \\ {K}_1=250J.........\left(b\right) \end{gathered}$$

Step 4: Calculating the final kinetic energy

$$\begin{gathered} K_2=\frac{1}{2}\times 10kg\times {\left(5m{s}^{-1}\right)}^2 \\ {K}_2=\frac{1}{2}\times 10\times 25=125J \\ {K}_2=125J.......\left(c\right) \end{gathered}$$

Step 5: Finding the ratio of kinetic energy

On dividing the equation (b) and equation (c)

$$\begin{gathered} \frac{K_1}{K_2}=\frac{250J}{125J}=4 \\ \frac{K_1}{K_2}=\frac{2}{1} \end{gathered}$$

Hence, the ratio of kinetic energy be $2:1$