In the movie Monty Python and the Holy Grail a cow of mass 110 kilograms is catapulted from the top of a castle wall 9.1 meters high onto the people down below. The cow is launched from a spring with a constant of 1.1 times 10 to the fourth newton-meters that was initially compressed by 0.50 meters from equilibrium. The gravitational potential energy is set to zero at the base of the castle wall. What is the gravitational potential energy of the cow as it just clears the wall, treating the cow as a point particle? What is the elastic spring energy of the cow before the catapult is released? What is the speed of the cow at the instant that it hits the ground?
This is a multipart problem we’ll work on piece by piece. Let’s start with part one, the gravitational potential energy of the cow as it just clears the castle wall. We’ll make a couple of assumptions as we work through this problem. First, as stated in the problem statement, we’ll treat the cow as a point particle; and second, we’ll treat the acceleration due to gravity, lowercase 𝑔, as exactly 9.8 meters per second squared. Let’s take a moment to write down the critical information that was given to us in the problem statement.
We’re told that the mass of our object, the cow, is 110 kilograms; we’ll write that as lowercase 𝑚. We were told that the wall that the cow is catapulted from has a height of 9.1 meters; we’ll call that lowercase ℎ. The cow was launched using a spring with a spring constant of 1.1 times 10 to the fourth newton-meters; we’ll represent that value as 𝑘. We’re also told how much that spring was compressed due to the cow’s weight, and we’re told the compression is 0.50 meters from equilibrium; we’ll call that 𝑥 sub 𝑐.
In part one of our problem, we’re asked to solve for the gravitational potential energy of the cow as it just clears the castle wall; we’ll call that PE sub 𝑔. In part two, we want to solve for the elastic spring energy of the cow before the catapult is released; we’ll call that PE sub 𝑒, for elastic potential energy. And finally, in the third part, you wanna solve for the speed of the cow just as it hits the ground; we’ll call that 𝑣 sub 𝑐 for the speed of the cow.
Now that we’ve stated our assumptions, recorded the given information, and identified what we want to solve for in each part of this problem, let’s begin with part one and draw in a diagram of the scenario. Now here we have a sketch of our cow, modeled again as a point particle, being launched from a catapult at the top of the castle wall 9.1 meters above ground level. In this diagram, it looks as though the cow is in fact being launched from a height greater than 9.1 meters due to the length of the arm of the catapult. But in our modeling, we’ll assume that the launch height from the catapult is equal to the height of the wall above the ground. 9.1 meters.
So we first turn our attention to part one, solving for the gravitational potential energy of the cow just as it clears the castle wall. We recall the equation for gravitational potential energy. This type of potential energy is equal to the mass of the object involved times the acceleration due to gravity, 𝑔, multiplied by the height of the object. Now interestingly, when it comes to gravitational potential energy, we can set our value of zero anywhere we like. Any height value we choose can be set as the zero point.
In this problem, we’ve chosen the ground level as our zero point. That means when we apply this general equation to our situation, the height we’ll use is 9.1 meters. We then plug in for the mass, 𝑚, and acceleration due to gravity, 𝑔. Those values are given, the mass, as 110 kg, and 𝑔 is a constant of 9.8 meters per second squared. When we plug these numbers into our calculator, we get a gravitational potential energy of 9.8 times ten to the third joules. This is how much energy the cow would have at the moment it leaves the wall simply due to its gravitational potential.
Now that we’ve solved for this component of the cow’s potential energy, let’s move on to part two where we solve for the potential energy due to the elastic energy of the spring in the catapult. Let’s begin by recalling the equation for elastic spring energy or the elastic potential energy due to the compression of a spring. That equation states that the elastic potential energy of a spring is equal to half the spring constant multiplied by 𝑥 squared, where 𝑥 is the displacement of the spring from equilibrium, either stretching or compressing.
When we apply this formula to our particular scenario, we see that we’ve been given each one of the variables in this equation: 𝑘, the constant of our spring, is 1.1 times 10 to the fourth newton-meters; and 𝑥, in our case, is called 𝑥 sub 𝑐, the compression of our spring given a 0.50 meters. Let’s plug those values into this equation to solve for elastic potential energy of the spring. When we plug these values into our calculator, we find that the elastic spring energy of the cow is equal to 1.4 times 10 to the third joules. This energy of the cow is transmitted to it by the compression of the spring in the catapult.
We’ve now solved for both potential energy terms of the cow: the potential energy due to gravity and the potential energy due to the spring. Now we move on to the final part of our problem, solving for the speed of the cow, 𝑣 sub 𝑐, just as it hits the ground. Now it’s interesting that in this problem, we’re not told the angle at which the cow is fired from the catapult. We don’t know whether it was fired horizontally or nearly vertically or anywhere in between, and yet we’re still asked for the speed of the cow as it finally lands on the ground.
We’re able to solve this problem by referring back to the conservation of energy. This law tells us that energy can’t be created or destroyed but that it can be transformed from one form to another. In our case, that means that all of the potential energy we just solved for, gravitational and elastic, is converted to the cow’s kinetic energy when its height above the ground is zero; that is, when it lands. Using this conservation of energy approach, we don’t need to know the specific angle at which the cow was launched from the wall. All we need to know is the initial energy of the cow when it’s launched. This will let us figure out the speed of the cow when it lands. So what is the initial energy of the cow; that is, the energy when it’s launched from the catapult?
At the very moment before the spring of the catapult is released, the cow is stationary; it’s speed is zero, so all of its energy is potential, gravitational potential energy and elastic potential energy. By the conservation of energy, the initial energy of the cow will also be equal to its final energy. That’s what energy conservation implies in a lossless environment: the final energy of the cow — that is, it’s energy you just as it hits the ground — is purely kinetic. That’s because at that point, it has no gravitational potential energy due to the fact that ℎ at the ground is zero.
So we can write E sub final in terms of the cow’s kinetic energy. Recall that an object’s kinetic energy is equal to one-half its mass times its speed squared. We’ll write that in to our equation, substituting 𝑣 sub 𝑐 for the speed, because that’s the speed of the cow that we’re searching for. So what we’ve established is that the sum of the potential energies of the cow, the gravitational and elastic, equals the final kinetic energy of the cow.
We can now rearrange this equation to solve for 𝑣 sub 𝑐, the speed of the cow. To do that, we’ll multiply both sides by two divided by 𝑚, which will effectively cancel those terms out on the right side of our equation, and we’ll then take the square root of both sides, which has the effect of canceling out the squared term with the square root on the right side of our equation.
When we write a simplified and cleaned up version of this expression, we see that 𝑣 sub 𝑐, the speed of the cow when it hits the ground, equals the square root of two over 𝑚 times the quantity of the potential energy due to the gravity plus the potential energy due to the elasticity of the spring. Those potential energies are the terms that we solved for in parts one and two.
When we substitute those values into this equation as well as the mass of the cow, 110 kilograms, and enter these values into our calculator, we find a speed of the cow as it hits the ground of 14 meters per second. That’s how fast the cow must be moving when it lands due to the conservation of energy.