In the Hunger Games movie, Katniss Everdeen fires a 0.0200-kilogram arrow from ground level to pierce an apple up on a stage. The spring constant of the bow is 330 newtons per meter, and she pulls the arrow back a distance of 0.550 meters. The apple on the stage is 5.00 meters higher than the launching point of the arrow. At what speed does the arrow leave the bow? At what speed does the arrow strike the apple?
As we work through this problem, we’ll make a few assumptions. First, we’ll assume that all of the spring energy of the bow is transferred to the arrow, none is lost to friction or otherwise. Second, we’ll assume that 𝑔, the acceleration due to gravity, is in exact 9.8 meters per second squared. Let’s highlight some of the vital information given to us in this problem statement. First, we’re told the mass of the arrow, 0.0200 kilograms. We’re also told the spring constant of the bow, that’s 330 newtons per meter. We’re given that the arrow is pulled back a distance of 0.550 meters and that the target, the apple on the stage, is 5.00 meters higher than the launching point of the arrow.
There are two parts to this problem. First, we wanna solve for the speed at which the arrow leaves the bow. We’ll call that 𝑣 sub 𝑏. And then in part two, we want to solve for the speed at which the arrow strikes the apple. We’ll call that 𝑣 sub 𝑎. We’ve assigned these variables based on the letters 𝑏 and 𝑎 to represent bow and apple. Don’t let the fact that 𝑎 comes before 𝑏 in the alphabet confuse about which one is which.
Now let’s move on to drawing a diagram of the situation. So here we have our diagram of our archer firing an arrow from what we’ll call ground level up on to a stage where an apple sits. And we’re told that the distance in height between the launching point of the arrow and the height of the apple on stage is 5.00 meters. In this problem, we’ll represent that height with the letter lowercase ℎ. We’ll let 𝑚 represent the mass of the arrow 0.0200 kilograms. 𝑘 represent the spring constant of the bow 330 newton meters. And 𝑥 to represent the distance the arrow is pulled back in the bow, 0.550 meters. And what we’re solving for is 𝑣 sub 𝑏, the speed of the arrow as it leaves the bow, as well as 𝑣 sub 𝑎. 𝑣 sub 𝑏 and 𝑣 sub 𝑎, the speed of the arrow when it’s released from the bow and the speed of the arrow when it strikes the apple respectively are what we’re looking to solve for.
We’ll approach this problem by using the conservation of energy to our advantage. Remember that law as it’s written in equation form. The law of conservation of energy tells us that in a given system, the initial energy of that system is equal to the final energy of the system. That is, energy is neither created nor destroyed. We can write this expression in an even simpler form by letting 𝑖 stand for initial and 𝑓 for final so that 𝐸 sub 𝑖 equals 𝐸 sub 𝑓. Now it’s very important as we move through this problem to keep clear about which moments in time we speak of when we say initial and final. So here at the outset, let’s define three moments in time over the course of the action described here.
The first moment in time is the instant right before the arrow is released from the bow. That’s the instant when the arrow is pulled back farthest in the bow. We’ll call that time one and abbreviate it 𝑡 sub one. Our next time 𝑡 sub two will be the instant in time when the arrow is released from the bow. This is the time when the arrow and the bow string lose contact with one another after the arrow has been shot. And finally, we’ll define a time 𝑡 sub three. This will be the time at which the arrow strikes the apple after it is flown through the air and up on stage.
So we have these three snapshots in time: 𝑡 one, when the arrow is pulled back as far as it ever goes. 𝑡 two, when the arrow is released from the bowstring. And 𝑡 three, when the arrow finishes its course and strikes the apple. As we solve for 𝑣 sub 𝑏, the speed of the arrow when it leaves the bow, we’ll focus on 𝑡 one and 𝑡 two as our initial and final states. So what is the initial energy of our system that is the energy at 𝑡 one? We’ll solve for that in a moment, but we know whatever that value is, that it’s equal to the final energy which is the energy of our system at time equals 𝑡 two. Now that we’re clearer on that, it’s our task to figure out what are the energy terms at these two time values, 𝑡 one and 𝑡 two, that make up our energy balance equation.
Consider the scenario at 𝑡 one, when the arrow is maximally pulled back in the bow. At that moment, the arrow is not in motion, so it has no kinetic energy. It’s also at the lowest point of its height throughout its trajectory and therefore, it’s gravitational potential energy can be considered zero. That leaves the spring or elastic energy of the bow on the arrow. What is that energy? Well, recall the equation for spring elastic energy. Spring elastic energy is a potential energy which equals one-half the spring constant multiplied by 𝑥, the distance the spring is displaced from equilibrium, squared. This is the only source of energy at time equals 𝑡 one. So we’ll rewrite the left-hand side of our equation using this spring elastic energy term.
Now we’d like to figure out what are the energy terms of the system at time equals 𝑡 two, when the arrow is released from the bow? At time equals 𝑡 two, we no longer have an elastic spring source of energy. And if we assume that the elevation or altitude of the arrow is not changed appreciably between being pulled back and released, then the height of the arrow can also be considered zero. This means the arrow’s effective gravitational potential energy is zero. The one form of energy we do find in our system at time equals 𝑡 two is kinetic energy. It’s at this moment when the arrow has its maximum speed. Recalling the equation for kinetic energy, 𝖪𝖤 is equal to one-half the mass of our object multiplied by its speed squared, we can rewrite the right-hand side of our equation now with this term for our arrow. We’ll use the mass of our arrow 𝑚 and for 𝑣, we’ll use 𝑣 sub 𝑏 which is the speed of the arrow just as it’s released from the bow.
From this equation, we want to solve for 𝑣 sub 𝑏. Let’s rearrange this equation to do that. First, we can cancel out the one-half terms common to both sides of our equation. Next, we can divide both sides by 𝑚, the mass of the arrow. Doing this means that the 𝑚s on the right-hand side of our equation cancel out. Finally, we can take the square root of both sides of our equation so that the squared term above 𝑣 sub 𝑏 and the square root cancel. Now believe it or not, we have 𝑣 sub 𝑏 all by itself on one side of our equation. Let’s rewrite a cleaned up and simplified version of this equation.
Our equation simplifies to 𝑣 sub 𝑏, the speed of the arrow as it leaves the bow, is equal to 𝑥, the displacement from equilibrium of the string in the bow, multiplied by the square root of the spring constant divided by 𝑚, the mass of our arrow. All three of these values are given to us in the problem statement. So let’s insert them now into this expression. Let’s now enter these values into our calculator. And when we do that, we find a speed of the arrow as it leaves the bow of 70.6 meters per second. That’s how fast the arrow was moving at 𝑡 sub two when it’s released from the bow.
That’s part one. Now let’s move on to part two where we want to solve for the speed of the arrow when it hits the apple up on stage. To do that, we’ll again rely on the fact that in our scenario, energy is conserved. This time, the two moments in time we refer to when we say initial and final are not 𝑡 sub one and 𝑡 sub two like before. Now our initial time is at 𝑡 sub two when the arrow was first released. And our final time is 𝑡 sub three when the arrow strikes the apple. So let’s again consider, like before, what energy terms are present at these moments in time?
Recall from our analysis of the situation at time equals 𝑡 sub two, that the only energy term present was the kinetic energy of the arrow. That’s still the case, so let’s write that term in for the left side of our equation. One-half the mass of the arrow multiplied by its speed, as it leaves the bow, squared is equal to the energy of the system at 𝑡 sub three, when the arrow hits the apple.
At that moment in time, what energy terms are present in our system? We know that the arrow will continue to be in motion at some new speed which we’ve called 𝑣 sub 𝑎. So we’ll have a kinetic energy term in 𝐸𝑡 sub three. We will also have a gravitational potential energy term because looking back at our diagram, we can see that at that final moment, 𝑡 sub three, the arrow will have gained 5.00 meters of elevation. So there are two terms in 𝐸𝑡 sub three, a kinetic energy term and a gravitational potential energy term. Recall that gravitational potential energy is equal to the mass of the object multiplied by the acceleration due to gravity 𝑔 multiplied by ℎ, the height of the object above zero. So we write in our kinetic and potential energy terms for 𝐸𝑡 sub three. And we see in this form of the equation that one-half the mass of our arrow multiplied by its speed as it leaves the bow squared is equal to half its mass times its speed as it hits the apple squared plus its mass times gravity times ℎ, the height of the arrow.
We want to solve this equation for 𝑣 sub 𝑎, the speed of the arrow as it hits the apple. We can simplify this equation by dividing the entire equation by the mass 𝑚, cancelling it out everywhere it appears. We can then multiply both sides of our equation by two, which will cancel out the one-half terms in our kinetic energy terms and leave us with a factor of two on our potential energy term. When we clean this equation up a bit, we see that the speed of the arrow as it leaves the bow squared is equal to the speed of the arrow as it strikes the apple squared plus two times the acceleration due to gravity times ℎ, the height of the arrow above zero. We then subtract two times 𝑔 times ℎ from both sides which cancels that term out on the right side of our equation. And we then take the square root of both sides which cancels out the square and square root terms on the right.
Now we have an expression that tells us that the speed of the arrow when it hits the apple, 𝑣 sub 𝑎, equals the square root of the speed of the apple when it leaves the bow squared minus two times 𝑔 times ℎ. We have solved for 𝑣 sub 𝑏, we are given ℎ, and 𝑔 is a constant value. So let’s plug all three of those into this square root. With these values inserted as shown, we can now enter these numbers into our calculator to solve for the speed of the arrow just as it hits the apple, and we find a value of 69.9 meters per second.
So the speed of the arrow as it left the bow is 70.6 meters per second, and the speed of the arrow as it hits the apple is 69.9 meters per second.