Collisions, added mass, speed increase, and energy transfer via quantum entanglement

This paper (preprint) is the author’s analytical research of elastic and inelastic collisions (impacts) of two bodies in a linear motion. An experiment (inelastic collision) described in the paper is designed to understand how an additional mass can physically (not in physical formulas) make a final speed increase of one of two colliding bodies. A first body (big block), which is linearly moving at some constant speed, hits a second body (small block) which is motionless. Accelerations of and final speeds of the blocks, and the impact time are measured. Then, some additional mass is assembled to the big block so that a length of the big block can stay the same. Next, the big block (with an added mass) hits the small block keeping the initial speeds of the blocks the same. Accelerations of and final speeds of the blocks, and the impact time are measured. In according to the classical mechanics formulas, a final speed of the small block gets higher if a total mass of the big block is increased, even though the initial speeds of the blocks do not change. However, it is unclear how a mass, which is added to the big block, may physically make an increase of the small block final speed if the initial speeds of and lengths of the colliding blocks do not change. A speed of shock waves is the same in the two blocks if they are made of the same material. Based on the physics of collision, if a length of the small block is less than a length of the big block, then the small block must begin to move when its shock wave returns to the impact area before a shock wave in the big block returns to the impact area. In other words, the small block must begin to move at the same time in the collision with or without an added mass, so the initial speeds of and lengths of the two blocks do not change. Therefore, the physical impact time must be the same for the collision with or without an added mass. Hence, the blocks cannot squeeze each other faster or longer during the physical impact when

sort of quantum entanglement, anisotropy, relativity, that is, without any fields, or the energy transfer between fields happens via some sort of quantum entanglement anisotropically. Though, such an assumption requires a supplementary experimental research.

Introduction
Elastic and inelastic collisions of two bodies (solid blocks) are analyzed in this paper. To better understand what is going on during elastic and inelastic collisions, let's first analyze an elastic collision of two identical particles.
Assume there are two identical particles (charges) which have the same masses and fields (field lines are depicted) [1], see Figure #1.
Assume that a particle p2 is not exerted by any forces before the collision and is motionless. A particle p1 reaches p2 at a uniform speed 1 meter per second. At some point, the field of p1 begins to compress the field of p2 [2] so that the initial speeds of both particles begin to change: the p1 speed decreases and the p2 speed increases. As a result, the fields of both particles are compressed at some distance x , see Figure   #2. Qeios Now, let's consider a more complicated process when two different groups [3] of identical particles of steel collide with one another along one axis (uniaxially [4]). The fields of all the particles are identical too.
Assume there are not any friction, air resistance, other forces exerted on the groups. The left group of the particles is bigger than the right group of the particles. A length L 1 of the left group is more than a length L 2 of the right group. The left group begins to move toward the right group at a uniform speed 1 meter per second at time t = 0 , see Figure #3. When the collision (impact) starts, the fields of the left group surface particles begin to compress the fields of the right group surface particles along one axis at the left group initial speed, see Figure #4. Assume that the collision between the groups starts at time t 1 . The shock wave propagates at the same speed in both groups because they represent the same material (steel) [5]. The impact should end when the shock waves return to the impact area [6,7]. However, the shock wave reflected from the right end of the right group returns to the impact area earlier than the reflected shock wave in the left group returns from its left end because L 2 is less than L 1 . Hence, the impact should end when the shock wave in the right group returns to the impact area. The groups both begin to deform at the impact area along one axis.
The deformation is assumed to be elastic, that is, the groups restore their initial shapes without strains after the impact [5,6,7].
The linear momentum of any mass is defined by a mass m multiplied by its velocity v [8]: Next, increase a width of the left group but keep the same length so that the centers of the mass of the groups are in line, see Figure #5. Therefore, a total mass of the left group becomes larger but a mass of the right group stays unchanged.
The left group hits the right group at the same initial speed. Likely the deformation in the collision (impact) before increasing the left group, the deformation of the groups happens along one axis (uniaxially).
Because a total mass of the left group increases, a force exerted on each group becomes larger than that force in the collision before increasing the left group. The force F (magnitude of the force) is calculated based on the following formula via impulse: where m group is a total mass of one of the groups, v group is a final speed (after collision) of a group, i group is an initial speed (before collision) of a group, t collision is the collision time [9]. The same force F is exerted by the groups on each other in according to Newton's third law [10]. The deformation delta L of each group should be larger after increasing the left group because the force F becomes larger based on the following equation: where L is a length of a group before the deformation, A is the collision contact area between the groups (impact area), E is Young's modulus (elastic modulus) for steel. The equation (3) can be derived from the formulas for epsilon (engineering strain), sigma (applied or engineering stress), and E (Young's modulus or elastic modulus) [11]. The length L of each group and the impact area A do not change in the collisions.
Notice that the deformation of the left group is always less than the deformation of the right group if L 2 is less than L 1 based on the equation (3). It should be mentioned that the fields of the particles are compressed during the process of the elastic deformation. When the pressure in the contact (impact) area reduces to zero, all the fields of the particles restore their initial state.
The deformation of both groups gets larger because of the larger force F based on the equation (3). If the initial speeds of the groups do not not change, then the larger deformation can physically happen only if the impact time gets longer.
If the impact time increases after the left group gets bigger, it induces the following point. When the collision ends at some time t 2 (before the left group being increased), the right group begins to move at some speed v 2 . Therefore, the right group gets enough kinetic energy to begin to move at t 2 . After the left group gets increased and if the impact time gets longer, then the right group should begin to move at some time t 2 ' . The latter is supposed to be longer than t 2 . However, it is unclear why the right group should not start to move at t 2 if the right group gets enough kinetic energy to begin to move at a speed v 2 at time t 2 . Notice that the fields of the particles at the impact area compress each other at the same speed in both collisions and a speed of the shock wave propagation does not change. Moreover, if a length of the right group is less than a length of the left group, then the right group must begin its own movement when the shock wave (in the right group) returns to the impact area. Hence, nothing prevents the right group to begin to move at t 2 in the collision with the enlarged left group.
Next, let's first analyze an elastic collision (ideal impact) of two masses that consist of the particles, of course. Assume that there are two different masses (two solid blocks) that are made of steel. These blocks freely move on wheels along a smooth straight metal floor or rails in vacuum (friction is negligible and no air resistance). A big block (Block #1) is 50 kg and a small block (Block #2) is just 1 kg. These two blocks collide with each other in an elastic collision (Situation #1), see Figure #6.
Notice that the centers of mass of each block must be in line during the collision to avoid rotation and vibration. The big block initial speed (before the impact) is 1 meter per second. The small block initial speed (before the impact) is zero. The elastic collision means that no any kinetic energy is lost, that is, turns into heat, sound waves during the collision [12]. If the momentum of the small block is zero before the impact, then the momentum of the big block before the impact must be equal to a sum of the momenta of the two blocks after the impact. The appropriate formula defines the conservation of the linear momentum when two different bodies (blocks) interact in the elastic collision, that is, when no any kinetic energy is lost in the collision: where m 1 is a mass of the big block (Block #1), i 1 is the big block initial speed 1 meter per second (before the impact), m 2 is a mass of the small block (Block #2), i 2 is the small block initial speed 0 meter per second (before the impact), v 1 is the big block final speed (after the impact), v 2 is the small block final speed (after the impact) [13].
If the small block initial speed i 2 is zero, then the equation (4) becomes: hence: Therefore, the formula for the kinetic energy is [14]: (7) hence, the formulas for v 1 and v 2 are [15]: Next, figure out the final speeds of the blocks (after the impact) if the big block initial speed is 1 meter per second: Therefore, the final speed of the small block is almost 1.96 meters per second. This speed is almost two times more than the big block final speed.
Next, consider Situation #2 when an additional mass 150 kilograms is assembled to the big block (Block #1), see Figure #7.  (8) and (9). The additional mass may be tightly assembled wherever: on the top or to the bottom, or behind Block #1, or other positioning. However, the additional mass must be assembled so that the impact contact area can stay unchanged.
The final speed increase of Block #2 after adding the additional mass is explained by a larger force exerted by Block #1 (with the additional mass) on Block #2 during the impact. This force may be calculated via the impulse based on the formula (2): where t elastic is the elastic impact time in seconds. Next, figure out a force F 1,1  During the inelastic collisions, some kinetic energy turns into the shock wave (elastic deformation) and heat. It means that the equation (7) changes, that is, the kinetic energy of Block #1 before the impact of the blocks should be more than the total kinetic energy of the blocks after the impact. Moreover, the final becomes v 2s . Then, the kinetic energy inequality is: therefore, the momentum of Block #1 before the impact should be more than a sum of the momenta of the two blocks after the impact: Though, it is necessary to remember that a total momentum of the system is conserved. To calculate the final speeds of the blocks in the inelastic collisions, the coefficient of restitution C R is used [17,18]: Therefore, a final speed of Block #1 is calculated based on this formula: and a final speed of Block #2 after the inelastic collision is [18,19]: Consequently, if C R is known before making the experiment (with and without the additional mass), the final speeds of the blocks are defined based on the formulas (28) and (29).
Next, first analyze the inelastic collision in Situation #1 (without the additional mass). Block #1 (50kg) collides with Block #2 (1kg). The impact starts at time t 1 and ends at time t 2 when Block #2 begins to move at its final speed v 2s , see Figure #8.  (2) and (29) appropriately. Based on the formulas (2) and (3) To clearly observe whether or not the deformation happens to be larger after the additional mass is assembled, Block #1 or Block #2, which is made of a material with a lower Young's modulus E , can be used for the experimental collisions. However, one of the blocks should be made of steel. In this way, the deformation of another block should be observable.