Question

A block of mass$\mathit{M}\mathit{=}\mathit{5}\mathit{.}\mathit{4}\mathbf{kg}$, at rest on a horizontal frictionless table, is attached to a rigid support by a spring of constant$\mathit{k}\mathbf{=}\mathbf{6000}\mathbf{}\mathit{N}\mathbf{/}\mathit{m}$. A bullet of mass$\mathit{m}\mathit{=}\mathit{9}\mathit{.}\mathit{5}\mathit{}\mathit{g}$and velocity$\overrightarrow{\mathbf{v}}$of magnitud$\mathbf{630}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s}$strikes and is embedded in the block (SeeFigure). Assuming the compression of the spring is negligible until the bullet is embedded.

(a) Determine the speed of the block immediately after the collision and

(b) Determine the amplitude of the resulting simple harmonic motion.

Short answer

(a) Speed of the block immediately after the collision is$1.10\mathrm{m}/\mathrm{s}$

(b) Amplitude of the resulting SHM is$3.3\times {10}^{-2}\mathrm{m}$

Step-by-step solution

The given data

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, $v=630\mathrm{m}/\mathrm{s}$

Understanding the concept of simple harmonic motion

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, $p=mv$…..(i)

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

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

(a) Calculation of speed of block after collision

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

We can write conservation of momentum as:

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

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

(b) Calculation of amplitude in SHM

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

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

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