Question

You have been asked to design a “ballistic spring system” to measure the speed of bullets. A spring whose spring constant is k is suspended from the ceiling. A block of mass M hangs from the spring. A bullet of mass m is fired vertically upward into the bottom of the block and stops in the block. The spring’s maximum compression d is measured.

a. Find an expression for the bullet’s speed $v_B$ in terms of m, M, k, and d.

b. What was the speed of a data-custom-editor="chemistry" $10\mathrm{g}$bullet if the block’s mass is data-custom-editor="chemistry" $2.0\mathrm{kg}$and if the spring, with data-custom-editor="chemistry" $\mathrm{k}=50\mathrm{N}/\mathrm{m}$, was compressed by data-custom-editor="chemistry" $45\mathrm{cm}$?

Short answer

a. The velocity is ${\mathrm{v}}_{\mathrm{B}}=\sqrt{\frac{2(\mathrm{m}+\mathrm{M})}{{\mathrm{m}}^2}\left(\left(\mathrm{m}+\mathrm{M}\right)\mathrm{gd}+\frac{1}{2}\mathrm{k}(\frac{\mathrm{Mg}}{\mathrm{k}}-{\mathrm{d}}^2)\right)}$.

b. The bullet speed is$600\mathrm{m}.$

Step-by-step solution

Part (a) Step 1: Given information

We have given,

Speed constant = k

Block mass = M

Bullet mass = m

Compressed length =d

We have to find the speed of the bullet.

Simplify

From the conservation of momentum which tells us that the momentum before collision and after collision of the system will be same.

Then,

Momentum before Collison = momentum after collision

$$\begin{gathered} {\mathrm{mv}}_{\mathrm{B}}=(\mathrm{M}+\mathrm{m}){\mathrm{v}}_{\mathrm{f}} \\ {\mathrm{v}}_{\mathrm{f}}=\frac{{\mathrm{mv}}_{\mathrm{B}}}{(\mathrm{M}+\mathrm{m})}........................\left(1\right) \end{gathered}$$

Now we are going to use the conservation of energy in the system the tells that the energy will always be conserved.

Here we are going to consider the potential energy of the spring before and after the Collison due to spring and also the potential energy of masses due to height and the kinetic energy of the system.

Then, initially the block with spring system will be in static. such that the force due to gravity is balance by the spring force.

$$\begin{gathered} \left|{\mathrm{F}}_{\mathrm{s}}\right|=\left|{\mathrm{F}}_{\mathrm{G}}\right| \\ \mathrm{ky}=\mathrm{mg} \\ \mathrm{y}=\frac{\mathrm{mg}}{\mathrm{k}}...................\left(2\right) \end{gathered}$$

After the Collison the system of masses will be compressed the system and the height of the masses will be changed then, Potential energy of the spring will be,

${\mathrm{U}}_{\mathrm{f}}=\frac{1}{2}\mathrm{k}{(\mathrm{y}-\mathrm{d})}^2..................\left(3\right)$

The change in the kinetic energy of the system will be,

$∆\mathrm{K}=\frac{1}{2}(\mathrm{m}+\mathrm{M}){\mathrm{v}}_{\mathrm{f}}^2$

put the value from equation (1)

$$\begin{gathered} ∆\mathrm{K}=\frac{1}{2}(\mathrm{m}+\mathrm{M}){\left(\frac{\mathrm{m}}{(\mathrm{m}+\mathrm{M})}\right)}^2{\mathrm{v}}_{\mathrm{B}}^2 \\ ∆\mathrm{K}=\frac{1}{2}\left(\frac{{\mathrm{m}}^2}{\mathrm{M}+\mathrm{m}}\right){\mathrm{v}}_{\mathrm{B}}^2..............\left(4\right) \end{gathered}$$

Now we will consider the change in height of the masses,

${\mathrm{U}}_{\mathrm{m}}=(\mathrm{m}+\mathrm{M})\mathrm{gd}..................\left(5\right)$

Then using equations (2),(3),(4),(5), we can write,

$$\begin{gathered} ∆\mathrm{K}={\mathrm{U}}_{\mathrm{m}}+{\mathrm{U}}_{\mathrm{f}} \\ \frac{1}{2}\left(\frac{{\mathrm{m}}^2}{\mathrm{M}+\mathrm{m}}\right){\mathrm{v}}_{\mathrm{B}}^2=(\mathrm{m}+\mathrm{M})\mathrm{gd}+\frac{1}{2}\mathrm{k}{(\frac{\mathrm{mg}}{\mathrm{k}}-\mathrm{d})}^2 \\ {\mathrm{v}}_{\mathrm{B}}=\sqrt{\frac{2(\mathrm{m}+\mathrm{M})}{{\mathrm{m}}^2}\left(\left(\mathrm{m}+\mathrm{M}\right)\mathrm{gd}+\frac{1}{2}\mathrm{k}(\frac{\mathrm{Mg}}{\mathrm{k}}-{\mathrm{d}}^2)\right)} \end{gathered}$$

Part (b) Step 1: Given information

We have given,

Block mass = $2\mathrm{kg}$

Bullet mass = $10\mathrm{g}$

Compressed length =$40\mathrm{cm}$

spring constant =$50\mathrm{N}/\mathrm{m}$

We have to find the speed of the bullet.

Simplify 

The speed of the bullet is,

$$\begin{gathered} {\mathrm{v}}_{\mathrm{B}}=\sqrt{\frac{2(\mathrm{m}+\mathrm{M})}{{\mathrm{m}}^2}\left(\left(\mathrm{m}+\mathrm{M}\right)\mathrm{gd}+\frac{1}{2}\mathrm{k}(\frac{\mathrm{Mg}}{\mathrm{k}}-{\mathrm{d}}^2)\right)} \\ {\mathrm{v}}_{\mathrm{B}}=\sqrt{\frac{2(0.01\mathrm{kg}+2\mathrm{kg})}{{(0.01\mathrm{kg})}^2}\left(\left(0.01\mathrm{kg}+2\mathrm{kg}\right)(9.8\mathrm{m}/{\mathrm{s}}^2(0.4\mathrm{m})+\frac{1}{2}(50\mathrm{N}/\mathrm{m}\left)\right(\frac{(2\mathrm{kg})(9.8\mathrm{g}\mathrm{m}/{\mathrm{s}}^2)}{(50\mathrm{N}/\mathrm{m})}-{(0.4\mathrm{m})}^2)\right)} \\ {\mathrm{v}}_{\mathrm{B}}=600\mathrm{m}/\mathrm{s} \end{gathered}$$