ln your job in a police lab, you must design an apparatus to measure the muzzle velocities of bullets fired from handguns. Your solution is to attach a 2.00-kg wood block that rests on a horizontal surface to a light horizontal spring. The other end of the spring is attached to a wall. Initially the spring is at its equilibrium length. A bullet is fired horizontally into the block and remains embedded in it. After the bullet strikes the block, the block compresses the spring a maximum distance d. You have measured that the coefficient of kinetic friction between the block and the horizontal surface is 0.38. The table lists some firearms that you will test.

A grain is a unit mass equal to 64.80 mg. (a) Of bullets A through E, which will produce the maximum compression of the spring? The minimum? (b) You want the maximum compression of the spring to be 0.25 m. What must be the force of constant of the spring? (c) For the bullet that produces the minimum spring compression, what is the compression d if the spring has the force constant calculated in part (b)

The given data is listed below as-

• The mass of the wooden block is, M =2.00 kg
• The frictional coefficient between surface and block is $\mu =0.38$
• Mass equivalent of grain: $1\hspace{0.33em}\mathrm{grain}=64.80\hspace{0.33em}\mathrm{mg}$

The kinetic energy of a particle equals the amount of work required to accelerate the particle from rest to speed V. Therefore, kinetic energy on the particle is given by-

$\mathbf{K}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{mV}}^{\mathbf{2}}$

The kinetic energy is a scalar and it is always positive or zero.

Let mass of bullet be and speed be v.

Let M be mass of the block.

If the speed of block right after the collision is V , by conservation of momentum we get the relation-

Therefore,

$mv=(m+M)V$

$V=\frac{m}{(m+M)}v$ ………….(1)

After the collision the kinetic energy of block and bullet is given by

$K=\frac{1}{2}(m+M)V^2$

Put value of V from equation (1) in above equation

$K=\frac{1}{2}\frac{m^2}{m+M}{v}^2$ ………(2)

Since, mv=p , the momentum of bullet, equation (2) can be written as:

$K=\frac{p^2}{2(m+M)}$

Non-conservative forces comprise of friction and elastic potential energy.

$K=U_{e1}+W_{fric}$

$K=\frac{1}{2}k{d}^2+\mu (M+m)gd$ ……..(3)

Using the conversion $1\hspace{0.33em}\mathrm{ft}=0.3048\hspace{0.33em}\mathrm{m}$ and equation (2), the kinetic energy for various bullets is given as-

$$\begin{gathered} \mathrm{A})K_{\mathit{A}}=\frac{1}{2}\frac{{\left(\left(80\times 64.8\times {10}^{-8}\right)\mathrm{kg}\right)}^2}{\left(80\times 64.8\times {10}^{-6}\right)\mathrm{kg}+2\mathrm{kg}}((1667\times 0.3048)\mathrm{m}/\mathrm{s}{)}^2 \\ \mathrm{B})K_{\mathit{B}}=\frac{1}{2}\frac{{\left(\left(125\times 64.8\times {10}^{-6}\right)\mathrm{kg}\right)}^2}{\left(125\times 64.8\times {10}^{-6}\right)\mathrm{kg}+2\mathrm{kg}}((945\times 0.3048)\mathrm{m}/\mathrm{s}{)}^2 \\ \mathrm{C})K_{\mathit{C}}=\frac{1}{2}\frac{{\left(\left(240\times 64.8\times {10}^{-6}\right)\mathrm{kg}\right)}^2}{\left(240\times 64.8\times {10}^{-6}\right)\mathrm{kg}+2\mathrm{kg}}((851\times 0.3048)\mathrm{m}/\mathrm{s}{)}^2 \\ \mathrm{D})K_{\mathit{D}}\mathit{=}\frac{1}{2}\frac{{\left(\left(200\times 64.8\times {10}^{-8}\right)\mathrm{kg}\right)}^2}{\left(200\times 64.8\times {10}^{-6}\right)\mathrm{kg}+2\mathrm{kg}}((819\times 0.3048)\mathrm{m}/\mathrm{s}{)}^2 \\ \mathrm{E})K_{\mathit{E}}=\frac{1}{2}\frac{{\left(\left(140\times 64.8\times {10}^{-6}\right)\mathrm{kg}\right)}^2}{\left(140\times 64.8\times {10}^{-6}\right)\mathrm{kg}+2\mathrm{kg}}((1335\times 0.3048)\mathrm{m}/\mathrm{s}{)}^2 \\ =1.391\mathrm{J} \end{gathered}$$

Thus, it can be noted from above that $K_c$is the largest and hence the bullet C will compress the spring the most whereas the bullet B will compress the spring the least.

The bullet C will cause the maximum compression d. For Bullet C, the equation of spring constant will be

$k=\frac{2}{d^2}[K_C-\mu (M+m_C\left)gd\right]$

For, d =0.25 m , the value of k will be

$$\begin{gathered} k=\frac{2}{{\left(0.25\right)}^2}\left[4.037J-0.38\left(\right(240\times 64.8\times {10}^{-6})kg+2kg)9.8m/s\times 0.25m\right] \\ =69.14N/m \end{gathered}$$

Thus, the force of constant of the spring for Bullet C is 69.14 N/m

The bullet B will have the minimal compression.

Again use equation (3)

$$\begin{gathered} \frac{1}{2}k{d}^2+\mu \left(M+m_B\right)gd-K_B=0 \\ \frac{1}{2}\left(\begin{array}{c}69.14N/m{\left(\frac{d}{1m}\right)}^2+0.38\left(\right(125\times 64.8\times {10}^{-6})kg+2kg)(9.8m/s^2)\left(\frac{d}{1m}\right)\\ -1.335J=0\end{array}\right) \\ d=0.117m \end{gathered}$$

The above equation has a positive solution d =11.7 cm

Thus, the compression d of the spring for bullet B is 11.7 cm.