Two bullets of equal mass were fired at the same speed into the opposite sides of a target. The target was formed of two different pieces of metal stuck together. The first was 10 centimeters thick, and the second was 14 centimeters thick. When the bullets hit the target, the first one passed through the first layer and embedded eight centimeters into the second before it stopped, whereas the other bullet passed through the second layer and embedded five centimeters into the first layer before it stopped. Using the work–energy principle, calculate the ratio 𝑅 one divided by 𝑅 two, where 𝑅 one is the resistance of the first metallic layer and 𝑅 two is that of the second.
The work–energy principle, which we’re asked to use to answer this question, relates the net work done to an object to the total change in the kinetic energy of that object. To effectively use the work–energy principle, we’ll also need to define work. As a formula, we write work equals force times distance because work is actually the increase or decrease in energy of an object associated with moving in the same direction or opposite direction to an external force.
We’ll also need to define kinetic energy, which is one-half times the mass of an object times the square of its speed. So, in essence, the work–energy principle tells us that the net work, that is, the total combination of the work from each individual force, is equal to the net change in the kinetic energy, that is, the difference between the final kinetic energy and the initial kinetic energy.
To see how these formulas relate to answering the question, observe that the question tells us about the mass and speed of the bullets, which are directly related to their kinetic energy. Furthermore, 𝑅 one and 𝑅 two, the components of the ratio that we’re trying to find, are exactly the force in force times distance. The distance part of force times distance is given to us in the form of the thicknesses of the two components of the target and also how far each bullet penetrated into the target.
Okay, so we know we have all the information we need to apply the work–energy principle. So let’s draw a diagram of this situation to organize that information. Here we have the target with its two components labeled with their respective widths and resistances and the two bullets with the same mass and speed, one on each side of the target. Note that the ordering of the two layers of the target is arbitrary as long as the 10-centimeter layer has resistance 𝑅 one and the 14-centimeter layer has resistance 𝑅 two.
Now let’s draw what happens when the two bullets embed themselves in the target. The first bullet, which in our diagram is on the right, passes 10 centimeters through the first layer with resistance 𝑅 one and then embeds eight centimeters into the second layer with resistance 𝑅 two. The second bullet, which in our diagram starts on the left, passes 14 centimeters through the second layer with resistance 𝑅 two and then embeds five centimeters into the first layer with resistance 𝑅 one. Both of these bullets come to a stop, which means that their final speed is zero.
Okay, let’s now write down the work–energy principle equation for these two bullets. We’ll start with the work done on the first bullet. The bullet experiences a force of 𝑅 one for the 10 centimeters it is traveling through the first layer. So the work done by the first layer on the first bullet is 10𝑅 one. The work done by the second layer is 𝑅 two times the length that the first bullet penetrates into the second layer, which is eight centimeters. So the total work done on the first bullet is 10𝑅 one from the first layer plus eight 𝑅 two from the second layer.
As for the kinetic energy, we know the final kinetic energy of the bullet is zero because the bullet stops, so its speed is zero. The initial kinetic energy is just one-half 𝑚𝑣 squared using 𝑚 and 𝑣 as we defined before. Note that the right-hand side of this equation, which is a change in energy, is negative even though energy is always either positive or zero. This is okay because a negative change just means that the energy has decreased.
Alright, let’s now do the same for the second bullet, again starting with the work. This bullet travels 14 centimeters against the resistance 𝑅 two of the second layer, followed by five centimeters against the resistance 𝑅 one of the first layer. So in analogy to what we calculated for the first bullet, for the second bullet, the total work is 14𝑅 two plus five 𝑅 one. As for the kinetic energy, the initial kinetic energy of the second bullet is the same as that of the first bullet because they have the same mass and initial speed. And the final kinetic energy is also zero because the second bullet also comes to a stop. So the total change in kinetic energy of the second bullet is also zero minus one-half 𝑚𝑣 squared.
We’re almost ready to answer the question. Observe that our two expressions for the work, 10𝑅 one plus eight 𝑅 two and 14𝑅 two plus five 𝑅 one, are both equal to the same change in kinetic energy, zero minus one-half 𝑚𝑣 squared. Since two expressions that are equal to the same thing are also equal to each other, we have that 10𝑅 one plus eight 𝑅 two equals 14𝑅 two plus five 𝑅 one.
If we gather all the terms with 𝑅 one on one side of the equation and all the terms with 𝑅 two on the other side of the equation, we’ll be ready to answer the question. Let’s erase the diagram to make some space to do this calculation. To collect the terms as desired, we’ll subtract five 𝑅 one and eight 𝑅 two from both sides of the equation. 10𝑅 one minus five 𝑅 one is five 𝑅 one. Eight 𝑅 two minus eight 𝑅 two is zero. 14𝑅 two minus eight 𝑅 two is six 𝑅 two. And five 𝑅 one minus five 𝑅 one is also zero.
Now we have five 𝑅 one equals six 𝑅 two. To make five 𝑅 one the ratio 𝑅 one over 𝑅 two that we’re looking for, we need to divide by five and also by 𝑅 two. But anything we do to the left-hand side we also must do to the right-hand side. So we have five 𝑅 one divided by five 𝑅 two equals six 𝑅 two divided by five 𝑅 two. On the left-hand side, this gives us 𝑅 one divided by 𝑅 two, the ratio that we’re looking for, because five divided by five is just one. On the right-hand side, we get six-fifths because 𝑅 two divided by 𝑅 two is one. So the ratio 𝑅 one to 𝑅 two of the resistance of the first layer of the target to the second layer is six to five or six-fifths.