Tag Info

Hot answers tagged

29

Look at it this way: Suppose you are in a train travelling at 10 m/s. Somebody inside the train throws a ball at you in the opposite direction at 10 m/s. You feel the pain belonging to your first experiment. However, somebody looking at this experiment from outside the train would say that the ball is standing still and you are travelling towards the ball ...


13

Yes, unfortunately. Because of the equivalence of inertial reference frames, the the physical laws are the same in both reference frames. However, another possibility, which is non abelian, is that instead of feeling the same amount of pain, you could be feeling the opposite amount of pleasure. It depends if pain (X) are fermions or bosons, that is, if ...


8

Let's take everything out of our scenario other than you and the ball. No baseball stadium, no Earth, no spherical cows, NOTHING in the entire universe but you and the ball. (Nope, not even microwave background radiation) Now the question has changed. Now you need to ask whether it's possible to decide whether you're moving towards the ball or vice versa.


4

No mater the inertial referential: The pain is linked to the energy dissipated by the change of speed of the 2 objects in any inertial referential. Remarque 1: a part of the energy can be absorbed by the ball by deformation or heating. So a soft ball would make less pain. Remarque 2: The pain depends on what is static behind you and what is static behind ...


3

If the kinetic energy of both the ball decreases, how can their velocities be equal?? The front one ( B ),from the beginning, had low KE; if it decreases during the deformation, how can its velocity be equal to the velocity of the rear ball ( A )? In order to get a clear picture, let's consider the extreme case when the velocity of B = 0 Let's ...


2

Think of it this way: Ball A is moving at 10 m/s Ball B is moving at 3 m/s Both balls have the same mass Here we have Ball A collides with Ball B, transferring energy. During this transfer (ignore the deformation for now, since that doesn't seem to be an issue at the moment, so we've got an elastic collision), consider what happens in the exchange of ...


2

Pavel Petrman gave a different perspective, but since impact remains same pain should remain same, given that properties of the situation remain same. To try take a ball in your hand and run towards a weighing scale and hit it and in next case exchange ball and weighing scale and repeat the procedure. Take a note of readings on weighing scale.


2

I have heard that we can maximize the force by minimizing the time of impact in a punch That is not true. if you want to crack a bone hit at the right place with right timing.... but if you want to break bone the other way.. have a good mass and bang on with full speed, no need to pull the hand backwards like karate chop... I got the ...


2

Conservation of momentum works here like everywhere else. When A (with mass $m_A$) throws a ball with mass $m_b$ with velocity $v$, then $v_A=-v_b\frac{m_b}{m_A}$ so that after the ball is thrown, the net momentum is zero; note that the ball will not be moving towards $B$ at velocity $v$ but instead at $v-v_A$ since $A$ started moving backwards... When B ...


2

The simple answer is that in an elastic collision (for objects >> in mass than typical molecules) energy moves from kinetic to potential then back to kinetic as long as the "elastic limits" of the materials are not exceeded. In other words, as long as they act like springs. In non-elastic collision the energy goes mostly from kinetic of the colliding masses ...


2

What is the difference that leads to conservation of kinetic energy in elastic collision ? The difference is only in the properties of the material of a body. If it is elastic (happy ball) it can deform itself (thus absorbing KE) and then recover the original shape, giving back roughly the same amount of KE, which is considered as temporarily stored ...


2

A lot of the answers get distracted by the "feeling pain" part of the question. So let me start by focusing on that - simplifying a very complex psycho-physiological issue to a simple physical statement (this is a physics forum - we leave the other stuff for other sites): During the impact of two bodies, there will be an exchange of momentum. The force ...


2

The only factor is the capacity to return to the original shape when it is deformed by an external force. An elastic object recovers its shape after it has been compressed and deformed, if you crush a plastic bottle and remove the cap it will partially return to his original length, a lead ball is almost completely irreversibly deformed. The Coefficient ...


2

*elastic collision occur between atomic particles? inelastic collision occur between ordinary objects? perfectly inelastic collision occur during shooting? super elastic collision occur during explosion?* as John Rehnnie has explained, an elementary particle as the term implies, is not made of other particles, has no lattices ...


2

You're making a mistake in assuming that there is any left. Heat and sound account for all of it. There is one exception, though. If it crashes into something, and that breaks or bends the object, then the potential energy of the molecules is higher. That's why cars end up smashed after a collision. The molecules of the metal or plastic or whatever have ...


1

Assuming that hands remain completely static, and the object do not breaks, then all of its energy can be considered to be converted into heat and sound (as you have already described).


1

The force from the ball on the wall is exactly equal and opposite to the force of the wall on the ball. Both forces however are perpendicular to the wall (and must be assuming the wall is frictionless) and not necessarily perpendicular to the ball's initial direction of motion. Being perpendicular to the wall the force on the ball has absolutely no effect on ...


1

Assuming a direct hit - so traveling through about 50 km of atmosphere - at 0.1 c that would take about 2 ms if it didn't get slowed down too much by the atmosphere. What about drag force? Let's assume a radius $r$, density $\rho$, mass $m = \frac43 \pi r^3 \rho$. If it is a sphere, it experiences a drag force $F=\frac12 \rho_a v^2 C_d A$. Putting $\rho_a=1 ...


1

Consider the following results: From the definition of scalar product of four vectors, $$ \tag{1}(p_1 p_2)^2 \equiv (p_{1\mu}p_2^\mu )^2 = (E_1E_2 - \textbf{p}_1 \cdot \textbf{p}_2 )^2.$$ The usual dispersion relations: $$ \tag{2} E_i = \sqrt{ | \textbf{p}_i |^2 + m_i^2}.$$ The velocity $\textbf{v}_i$ in terms of momentum and energy: $$ \tag{3} ...


1

You can be sure that at least one reasearch team may help answer your question. Try to mail them. It seems that one can make such model. But still, as we can see it's a bit probabilistic field. http://trb.metapress.com/content/v4t5712601175275/ ...


1

Two laws govern collisions: conservation of momentum and conservation of energy. Momentum is the product of mass and velocity, so we can write $$\sum m_i\cdot \vec{v_i} = const$$ Conservation or energy is a little bit trickier, since energy can be converted from one type to another. In an elastic collision, the kinetic energy is conserved, so $$\sum ...


1

First of all you should note that Newton's law says when $F$ acts on a mass $m$, then that mass will move with acceleration $a$. Here, we should apply the laws of collision and by using the conservation of momentum, find out what your velocity will be after the collision. Before collision we have: $p_{tot}=mv$ and after collision $p_{tot}'=mv'+MV$ where $M$ ...


1

The difference in force to stop the trains you are talking about here is the difference in force is needed to bring the train to a stop within a particular distance Let me tell you what I mean. When you try to stop the train, you'll obviously be dragged in front of the train. Say the dragging causes an uniform force (due to the friction from the ground) ...


1

If I throw a small rock (m = 1kg) at a big rock (100kg) the small rock rebounds. Let's say my weight is 80kg, if I would jump into a big rock instead of bouncing back I would move in the same direction as a big rock. The big rock is heavier but it is not reflecting me. Why is that? There are two conditions which are to be met if a body A ...


1

In answer to the main question, matter does, in fact, "pass" through other matter. Starting from the macro scale (stars , galaxies), down to the micro scale (atoms), it happens all the time. The "free" movement of matter starts to get impeded, as the atoms start making latices (solids, crystals). But even at this scale, as Rutherford demonstrated, matter ...


1

Would A move backwards at the same speed? no.. Momentum will be conserved,not speed and since A has greater mass than the ball so his speed will be less.. if A throws the ball with speed 'u1',then his speed(in the backward direction) will be m1u1/M1 where m1 is the mass of the ball and M1 is the mass of the person A Now, what would happen when B catches the ...


1

This follows on from my answer to your previous question: Factors on which Coefficient of restitution depend. The coefficient of restitution of a collision depends on the available degrees of freedom for energy to be lost. If you take your example of the collision of atomic particles, let's say two electrons, then there isn't anywhere for the initial ...


1

Although I think the question is related more to algorithm (as a branch of programming), but since it incorporates aspects of physics, I will give it a go. OP, know that there is not one answer to your problem. It can be accomplished in several ways. Here I would introduce you to some important variables in your scenario: The kinetic energy of the body in ...


1

by conservation of linear momentum the center of mass of the system will keep moving, thus the two ball are together they will keep moving in the same direction than the faster ball, at the same speed than the center of mass, that is, slower than the initial speed of the faster ball. Assuming a totally elastic collision, after the balls push back each other, ...


1

Let's say that the ball strikes the floor with the linear momentum $-p_1$ (you see I take the direction of the momentum in consideration). Some part of the momentum is lost (some energy is lost to the floor) s.t. the ball rises again, but at the floor level it has a linear momentum $+p_2$. So, what the floor "gets" in this scenario? It GETS a linear momentum ...



Only top voted, non community-wiki answers of a minimum length are eligible