91Ó°ÊÓ

1. Mass of block,$\mathrm{M}=5.4\mathrm{kg}$
2. Spring constant, $\mathrm{k}=6000\mathrm{N}/\mathrm{m}$
3. Mass of bullet, role="math" localid="1655095781171" $m=9.5g=0.0095\mathit{}kg$
4. Velocity of bullet, $v=630\mathrm{m}/\mathrm{s}$

Using the concept of an elastic collision between the block and the bullet, we can find the velocity of the block after collision with the help of momentum conservation. We find the amplitude of SHM with the help of the given spring and its spring constant.

Formulae:

The momentum of a body, $p=mv$…..(¾±)

The potential energy of a body,$PE=\frac{1}{2}k{x}^2$…..(¾±¾±)

The kinetic energy of the system, $KE=\frac{1}{2}m{v}^2$…..(¾±¾±¾±)

As we know there is elastic collision between bullet and block.

We can write conservation of momentum as:

$$\begin{gathered} \left(M×0\right)+\left(m×V\right)=\left(M+m\right)v^{,} \\ {v}^{,}=\frac{m×v}{M+m} \\ {v}^{,}=\frac{\left(0.0095\right)\left(630\right)}{5.4+0.0095} \\ =1.11\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence, the value of speed of block is 1.11 m/s

Using equation (iii) & given values, we get the total kinetic energy of the system as:

$$\begin{gathered} KE=\frac{1}{2}\left(M+m\right){\mathrm{v}}^{,2} \\ =\frac{1}{2}\left(5.4+0.0095\right){\left(1.10\right)}^2 \\ =3.27\mathrm{J} \end{gathered}$$

At maximum compression, the spring also has some PE which equals to KE

So, we can write

$$\begin{gathered} PE=\frac{1}{2}k{A^2}_{{\mathrm{maximum}}_{}} \\ 3.27=\frac{1}{2}×6000×{A^2}_{\mathrm{maximum}} \\ \frac{6.54}{6000}={A^2}_{\mathrm{maximum}} \\ {A}_{\mathrm{maximum}}=0.033\mathrm{m} \\ =3.3×{10}^{-2}\mathrm{m} \end{gathered}$$

Hence, the value of maximum amplitude is$3.3×{10}^{-2}\mathrm{m}$