Collision Definition and 1000 Threads

While attempting this question , velocity of ##B## wrt ##A## ,##u'_x=\frac{u_x-v}{1-u_xv/c^2}## where ##u_x=-0.6c,v=0.8c## comes out to be ##-0.945c## (approaching).. The distance between ##A## and ##B## seen by ##A## at ## t=0## is ##d=\sqrt(1-.8^2)4.2×10^8## comes out to be ##252*10^6m##...

I honestly have no clue where to start, any help would be great.

m1v1i + m2v2i = (m1+m2)vf (1.67 × 10^-27)v1i = (1.67 × 10^-27 + 1.67 × 10^-27) vf (1.67 × 10^-27/3.34 × 10^-27)v1i = (3.34 × 10^-27/3.34 × 10^-27) vf (1.67 × 10^-27/3.34 × 10^-27)v1i = vf (0.5)(v1i) = vf not sure what to do from here nor if I'm in the correct path ?

I read (in "The View From The Center") that Jupiter protects the Earth from collision with large space rocks, asteroids, etc. What I can't get out of my mind is that could it also cause collisions. A large rock (initially not heading for Earth) could be put on a different path by Jupiter's...

So far I found the answer for a and b, but when I attempted to do the other ones I was completely lost. A.) P= MV M = 25g = .025kg V = 18 .025 * 18 = .45kg*m/s B.) KE= 1/2 mv^2 1/2 (.025)(18)^2 4.05 J

Attempt: I know that the conservation of momentum and energy also applies. Solution: Correct answer is C. But I can't understand how any of the two conservation laws lead to the answer C. The exam report to this question did not even mention anything about this - so, I guess it should be...

How is it that momentum is being preserved in a non elastic collision? for example let's say that two balls are colliding head-on, not elastically and heat is produced, does that not reduce the momentum of the system?

Change in KE = Change in thermal energy 0.5 * (6)* vblock^2 = 0.4 * 6 * 9.81* 0.1 vblock = 0.885 By Conservation of Momentum, (0.05)(854) = (0.05)*vbu + (6)(0.885)I am not sure whether Change in KE = Change in thermal energy is true coz there should be a change in internal energy of the block...

I've read some stuff on pileup.In one paper, it says the number of inelastic event approximately equals pileup event in detector.I don't quitely understand it.Can someone explain?Thanks

In inelastic collision their is loss of energy and according to my current knowledge energy can be transferred by either work out heat...Now The problem is that we use law of conservation of momentum in problems related with inelastic and if energy is transferred from our system (two masses)...

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I know that if the collision was not elastic, some of the kinetic energy of the incident neutron wound be used up in some other process. But, I can't understand how I can figure out exactly how much. Even if I can calculate it, I don't know how to find the condition for the collision to go from...

I tried to solve this question using conservation of momentum. The momentum of the system is mc(vc)=mcvc'+mbvf'. But after that I have no idea I want to use the conservation of kinetic energy but the question doesn't say it's elastic collision, I need to find the velocity of the mb after the...

In collisions that are inelastic or partially elastic, how can we predict how much of the energy lost to the surroundings becomes heat, and how much becomes sound? What determines that fraction?

Conservation of Energy: 9GeV + E = 5.3GeV + 5.3GeV Therefore E = 1.6GeV for the threshold energy. How would I find the velocity of B0 mesons so that I can calculate their mean distance? Then it would just be distance = velocity of b0 * mean proper lifetime Right?

Two identical point-sized particles with the same Y-coordinate were traveling along the X and Z axes respectively. Given that gravity is acting parallel to the Y-axis, will the particles when they eventually collide, continue traveling along the same linear path due to work done by either being 0?

So I am making the assumption that the resulting particle Z is emitted at rest. For part a I believe that since the two positron beams are symmetric they would each provide half of the energy to create the Z particle so the KE of each positron would be 91.187GeV/2, I am ignoring the rest energy...

I really want to know which answer is correct. I don’t really know if I should include velocities to the left as negative velocities in the equation. Is it -1 or 4.33? Please help! Thanks!

My initial thought was to use the conservation of energy law since there're no external forces acting on the system bullet + rod. The rod is in rest, the bullet is moving. Then after the collision, the bullet and the rod are rotating around the pivot together, so the kinetic energy of the bullet...

A simple model often used to explain solar system gravitational slingshots is to consider a mass moving to the right with initial velocity v1i and a much larger mass moving to the left with initial velocity v2i. After the collision, the first mass is moving to the left with velocity v1f and the...

Homework Statement: an object has a total mass of 75kg and is stationary, it then ejects as a smaller object to the left with a velocity of(+ve direction)7 ms-1(v2) with a mass of m2 , the original object then recoils backwards to the right (-ve direction) at a velocity of -2 ms-1 and a mass of...

My teacher wants me to know how to solve for missing values in a 1D collision when v2 does NOT equal 0. Could someone do me a huge favour and make me a practice question to solve for a missing value when v2 does not equal 0? Or even point one out to me online? And then let me try it out and...

initial total KE= (1/2)(0.6kg)(8m/s)^2 = 19.2J (0.6kg)(8m/s) = (0.6kg+1.8kg)(vf) vf= 2m/s final KE= (1/2)(0.6kg+1.8kg)(2m/s)^2 = 4.8J I tried to use linear speed=angular speed * radius : thus 2m/s= angular speed * (3.3m/2) angular speed= 1.2 rad/s Apparently that is wrong.

Homework Statement: Verify the claim of Section 7.2 that the electrons of a metal collide with the surface at a rate of about 10^30 per second per square centimeter. Do this by estimating the collision frequency of electrons in a 1.00-cm cube of copper metal with one face of the cube surface...

So after not being able to solve this problem I did some researching online. I was looking around and came across this video, where they give the following equations for solving for the final velocities of both balls: v1f=((m1-m2)/(m1+m2))*v1i v2f=(2m1/(m1+m2))*v1i I plugged in my numbers and...

x(before) y(before) x(after) y(after) puck (0.36)(13.8)=4.968 0 (0.36)(10.41)cos(α) (0.36)(10.41)sin(α) brick 0 0 (1.35)(1.34)cos(β) -(1.35)(1.34)sin(β) total 4.968 N*s 0 4.968 N*s 0 thus: (0.36)(10.41)sin(α) =(1.35)(1.34)sin(β) β= sin^(-1)[3.747sin(α)/1.809]= sin^(-1)[2.071sin(α)]...

From what I understand, a black hole can exist that was formed entirely from antimatter. If this antimatter black hole were to collide or merge with a black hole that has the exact same mass, but was made of regular matter, what would happen? Obviously the matter and antimatter would...

At the time of asteroid impact 65 million years ago wiping out the dinosaurs and 50 % of all life on Earth - presuming all but the smallest mammals or sea creatures suggests that humans most likely wouldn't be here today if not for that event because the tiny mammals which then existed would not...

If the initial kinetic energy is equal to the final kinetic energy where two objects that collide stick together, this collision is elastic or inelastic?

I have an example from my machinery handbook which shows the kinetic energy of a pile driver and shows it will drive a pile some distance at a force until the energy is expended. The formula is (Average force of blow times distance)= Weight of object (driver) times distance it falls) Just...

I recently heard in the news that the small asteroid Apophis will be so close to Earth it will be within the Earth's satellite orbits. This asteroid will likely not hit earth, even if it did it wouldn't have an extinction level impact. But there will come a day when we will have to face off with...

Hey everyone I'm struggling on the last part of this assignment. I need to find the total work done by the block and the bullet, when the collision happens. The informations is: mblock=0.3 kg mbullet=0.01 kg vg=700 m/s Height=0.72m The final speed after the collision is vf=22.6 m/s and the...

I need to develop some kind of simplified formula that will allow me to estimate pitch and volume of a sound of two colliding objects in a simulation. It doesn't have to be exact, mostly it is enough that it follows the intuition - large object produce lower sounds, large and fast object produce...

Okay, say we have two balls(equal mass and size), 1 and 2. #1 has kinetic energy and #2 is at a standstill, they collide. Ignore all friction, heat, sound losses etc.. Now I know that 1 exerts a force on 2, hence doing work, which in turn uses its kinetic energy up. Therefore, 2 speeds up...

How can I calculate the ACCELERATION of a stationary steel ball after being hit by a moving steel ball. I know how to get the final velocity but how long does it take to accelerate to that velocity from zero? Does it depend on the elasticity of the materials? I guess we need to know long did...

Can someone explain me (even in poor worda) the relationship between collision and relaxation? I mean, a structure is relaxed when the particles (or stars) inside have frequent collisions? Or not?An elliptical galaxy is a relaxed structure. Can i deacribe the internal stars with a...

I am stuck, please help. I did the conservation part but then i can't move on.

I've come up with the following causes: - air resistance - parallax - during the collision, some of the kinetic energy gets converted into thermal energy. - invisible deformations But I'm not sure which would be the biggest effect on the total momentum change. Are there any other reasons that...

My approach: Let us take two orthogonal axes: x, parallel to the racket's plane and y, perpendicular to it. For the ball to not spin, the components of initial velocities of the racket and the ball along x-axis must be same. Also, as the line of collision is along the normal to the racket's...

I calculated that the velocity of the head prior to the collision is sqrt(2gh), as all of the gravitational potential energy is converted to kinetic energy. And I believe the velocity at point A after the collision is given by the formula vf=e•sqrt(2gh), with e representing the coefficient of...

This seems like a fairly simple problem but I got the same answer for both so I'm unsure if I am doing it right. I found the accelerations as such a1 = (fk)/(ma) = (98/20) = 4.9 m/s2 a2 = (fk)/(ma) = (73.5/15) = 4.9 m/s2 Am I doing something wrong or is this problem just that simple?

I was wondering why in the video the moment of inertia for the clay ball (upon collision) was simply 1ml^2. That is the constant for a hollow cylinder. The problem specifies that the object is a ball, so the cylinder classification makes no sense, and also I'm pretty sure clay is rather dense...

a) Let m be the vehicle's mass, M the truck's mass, vt the truck's speed, vc the car's speed, vf the final speed, θ the angle both vehicles make with the horizontal axis (west-east direction) after the collision. Conservation of linear momentum: In the x direction: M vt = (m + M) vf cos(θ) In...

Considering the quantum mechanical model for an atom, what exactly happens when two atoms (say, two Ca2+ ions in a Brownian motion) collide with each other? As I know, this collision is not like a regular elastic or inelastic collision between two macroscopic objects. Is it mainly due to the...

They collide when their positions are the same, so I plugged the information for the boat into equation 1 to get an expression for d which is (2i, +j)t^2 Then I used equation 4 to get an expression for d for the branch, which is (-4i, +j)t I would need to take into account the different...

Good Morning May I ask about spatial collisions of bodies? In undergraduate dynamics, we study that when two particles college, we have two final unknowns: the final velocity of each particle. We first use the conservation of linear momentum. However, we supplement the analysis with the...

The speed of the sphere after the impact will be the same since the collision is elastic and the kinetic energy remains the same. So the change of momentum will be given by the cosine law right? What bothers me is the second question about the force that acts on the sphere (which can be given by...