Question

A ball of mass $2\mathrm{Kg}$ moving with velocity $3{\mathrm{ms}}^{-1}$, collides with spring of natural length $2\mathrm{m}$ and force constant $144{\mathrm{Nm}}^{-1}$. What will be the length of the compressed spring?

• A. $2\mathrm{m}$
• B. $1.5\mathrm{m}$
• C. $1\mathrm{m}$
• D. $0.5\mathrm{m}$

Answer 1

The correct option is B

$1.5\mathrm{m}$

Step 1: Given data

Mass of a ball $\mathrm{m}=2\mathrm{Kg}$

Velocity of a ball $\mathrm{v}=3{\mathrm{ms}}^{-1}$

Natural length of a spring $=2\mathrm{m}$

Force constant of a spring, $\mathrm{k}=144{\mathrm{Nm}}^{-1}$

Let the displacement of spring be$=\mathrm{X}\mathrm{meter}$

Step 2: Calculating the length of compresses spring

Here, when ball collides with the spring, the kinetic energy of a ball will get converted into the potential energy of spring, that is

$$\begin{gathered} \frac{1}{2}{\mathrm{mv}}^2=\frac{1}{2}{\mathrm{kX}}^2 \\ \Rightarrow {\mathrm{mv}}^2={\mathrm{kX}}^2 \\ \Rightarrow 2\times 3^2=144\times {\mathrm{X}}^2 \\ \Rightarrow 18=144\times {\mathrm{X}}^2 \\ \Rightarrow \mathrm{X}=\frac{1}{2\sqrt{2}}\mathrm{m} \end{gathered}$$

Now the length of compressed spring is given by:

Compressed length$=\mathrm{natural}\mathrm{length}-\mathrm{X}$

$$\begin{gathered} =2-\frac{1}{2\sqrt{2}} \\ =\frac{4\sqrt{2}-1}{2\sqrt{2}} \\ =1.5\mathrm{m} \end{gathered}$$

Hence, option(B) is the correct answer.