Conservation of energy of a bow and arrow

AI Thread Summary
The discussion centers on the conservation of energy in archery, specifically regarding the conversion of elastic potential energy into kinetic energy when an arrow is shot. It highlights that both the arrow and the bow experience motion due to the recoil effect, which raises questions about energy distribution. In scenarios where the archer is on firm ground, the bow's recoil is absorbed, allowing more energy to be transferred to the arrow. Conversely, on frictionless surfaces, the energy is shared between the arrow, bow, and archer. The conversation also draws an analogy to the Mössbauer effect, illustrating energy conservation principles at a nuclear level.
Sam Jelly
Messages
15
Reaction score
1
Homework Statement
Imagine a drawn bow with an arrow, storing elastic potential energy. When we release the string, this elastic potential energy converts to kinetic energy, launching the arrow with a certain velocity. However, the bow itself also recoil due to the equal and opposite reaction force. Wouldn't this recoil give the bow some kinetic energy as well? If both the arrow and the bow are moving, doesn't that violate the law of conservation of energy? Shouldn't all the initial elastic potential energy be solely converted to the arrow's kinetic energy?
Relevant Equations
Total Energy = Constant
I thought the initial elastic potential energy would be converted to the arrow's kinetic energy but it also appears that the bow has some velocity.
 
Physics news on Phys.org
Typically there is an archer holding the bow so that it does not recoil. If it does recoil then yes, the energy will be split between bow and arrow.
 
How much movement of the bow do you see in her shot?



Can you comment about other lossy energy mechanisms in archery shooting? Can you comment about the differences in those losses for a compound bow versus a recurve bow? :smile:
 
Sam Jelly said:
When we release the string, this elastic potential energy converts to kinetic energy, launching the arrow with a certain velocity. However, the bow itself also recoil due to the equal and opposite reaction force. Wouldn't this recoil give the bow some kinetic energy as well? If both the arrow and the bow are moving, doesn't that violate the law of conservation of energy? Shouldn't all the initial elastic potential energy be solely converted to the arrow's kinetic energy?
Let's change the situation a bit and imagine the archer standing on frictionless ice. He shoots the arrow standing up at rest with respect to the ice. The arrow moves forward and the bow + archer move backward to conserve linear momentum. The initial potential energy stored in the stretched bow is converted into kinetic energy of the arrow, bow and archer. It is not necessarily true that all the elastic potential energy must be converted to kinetic energy of the arrow.

However, when the archer stands on firm ground wearing cleated boots and there is no slipping, the bow + archer + Earth take up the recoil which means essentially no recoil . In this case, the arrow will travel farther than the archer-on-ice case because now it is true that all the elastic potential energy is converted into the arrow's kinetic energy.

The Mössbauer effect is an analog of this at the nuclear level. Nuclear transitions from an excited state to the ground state emit gamma rays (essentially) without recoil in which case the emitted gamma ray has essentially the same energy as the difference between the excited and ground state. That in itself may be just a curiosity until one realizes that these extremely monochromatic gamma rays can be used for nuclear gamma ray spectroscopy to probe the environment of nuclei in all sorts of interesting materials
 
Thread 'Minimum mass of a block'
Here we know that if block B is going to move up or just be at the verge of moving up ##Mg \sin \theta ## will act downwards and maximum static friction will act downwards ## \mu Mg \cos \theta ## Now what im confused by is how will we know " how quickly" block B reaches its maximum static friction value without any numbers, the suggested solution says that when block A is at its maximum extension, then block B will start to move up but with a certain set of values couldn't block A reach...
TL;DR Summary: Find Electric field due to charges between 2 parallel infinite planes using Gauss law at any point Here's the diagram. We have a uniform p (rho) density of charges between 2 infinite planes in the cartesian coordinates system. I used a cube of thickness a that spans from z=-a/2 to z=a/2 as a Gaussian surface, each side of the cube has area A. I know that the field depends only on z since there is translational invariance in x and y directions because the planes are...
Thread 'Calculation of Tensile Forces in Piston-Type Water-Lifting Devices at Elevated Locations'
Figure 1 Overall Structure Diagram Figure 2: Top view of the piston when it is cylindrical A circular opening is created at a height of 5 meters above the water surface. Inside this opening is a sleeve-type piston with a cross-sectional area of 1 square meter. The piston is pulled to the right at a constant speed. The pulling force is(Figure 2): F = ρshg = 1000 × 1 × 5 × 10 = 50,000 N. Figure 3: Modifying the structure to incorporate a fixed internal piston When I modify the piston...
Back
Top