Collision Definition and 1000 Threads

Let's assume I have a ball moving at a constant velocity and it collides with a spring and the spring compresses n cm. If I know how much mass the ball has and the spring constant D, how would I calculate the Force? I mean since F = dp/dt I would have to know the time in which the stopping...

Problem : The statement is given above. Here I attach the image of the problem to the right. Attempt : (1) Condition for collision : For the two particles to collide, there should be no relative velocity perpendicular to their "separation vector" ##\boldsymbol{s}##. Hence, we must have ...

Solutions are given in the book, but I could not understand some part of them. For problem 2.2, denote the 4-momentum of the photon by ##\mathbf P_\gamma##, that of the particle by ##\mathbf P## and the values after scattering by primes. Then by the conservation of momentum, we have ##...

See working attached. My problem is that I have come up with 3 equations but 4 unknowns (mq, up, vp, vq). Is this problem solvable? The answers say: a) 0.5 m/s b) 0.65 m/s and c) 0.34 m/s Many thanks

Hi, I'm studying how distance affects the impact of magnet collision. Would like to have some idea before conducting experiment. Imagine 1. holding a magnet at different distance near a fixed metal 2. release the magnet; the magnet attract and collide to the fixed metal Will the distance affect...

For car 1, the parametric equations are x = 1 + 0.8t and y=t. For car 2, the parametric equations are x=0.6s and y=2+s. (Let t and s represent time). Solving the system of equations, when the x values are equated are the y values are equated, I get s = -13 and t = -11. I assume that the 2 cars...

2 balls (Ball 1 and Ball 2) collide fully elastically and their relative velocity stays the same as but in sign opposite to that before the collision. Is there any sort of reference frame in which Ball 2 is always fixed (at rest) so that one can look at their relative velocity always in that...

Consider the system of the mass and uniform disc. Since no external forces act on the system, the angular momentum will be conserved. For elastic collision, the kinetic energy of the system stays constant.Measuring angular momentum from the hinge: ##\vec L_i = Rmv_0 \space\hat i + I \omega_0...

I think it has to do with conversation of momentum and projectile motion.

[Mentor Note: thread split off from a different thread] https://www.physicsforums.com/threads/heavier-objects-fall-faster.1002022/ Since seeing this thread yesterday, I have been trying to derive the time equation for the collision of two masses due to Newtonian gravity. Unfortunately, this...

https://en.wikipedia.org/wiki/Elastic_collision By the angle θ they mean some angle before or during the collision, or after the collision?

https://www.plasmaphysics.org.uk/collision2d.htm This is the only one I found, but when I plug in the numbers of his example I get a wrong result. Do you know any others who solved it i.e. considering the angle of impact? Angle of impact I name the angle that is shaped between the initial dx...

The way I learned to solve this was to switch to a frame of reference where one object is stationary. given: m1 =0.6kg v1 = 5.0m/s [W], m2 = 0.8kg v2 = 2.0 m/s [E] Setting v2 to rest by adding 2.0 m/s W to each object New velocities are v1 = 7.0 m/s [E] and v2 = 0.0m/s Then using the...

https://en.wikipedia.org/wiki/Elastic_collision μα+mβ=μx+my, μα^2+mβ^2=μx^2+my^2 I want x in relation of all variables except y, therefore I need to replace-eliminate y: μα+mβ=μx+my =>y=(μα+mβ-μx)/m μα^2+mβ^2=μx^2+my^2=>y=((μα^2+mβ^2-μx^2)/m)^0.5 and it is eliminated if I equate these two parts...

I calculated as attached and got it right. However, I just wonder why we can't use conservation of energy as the question has already specified 'frictionless', meaning no energy loss and energy distributed to the rotation only.

I was thinking about ballistic pendulums and the symmetry they exhibit. In the simplest case, you have one ball that begins at a certain height and collides with another ball at rest. You can calculate via conservation of momentum and energy the new velocities and max vertical displacements...

1. Hello, so the difficulty I am having with this problem is that is seems relatively straightforward. I have tried to solving it by assuming that this is a collision in which momentum is conserved. Therefore, I found the total momentum before the collision and used this to resolve it must be...

When reconstructing traffic accidents to obtain the collition velocities it is needed to obtain the "crush energy" which is the amount of kinetic energy transformed into permanent damage to the cars involved. From that kinetic energy we get the Energy Barrier Speed which is used with the...

I am guessing you have to find the speed at which the lighter ball moves and then to use the formula for kinetic energy, which I've tried but I'm not getting it quite right. We got m1=5kg v1=3,5kg m2=2.5kg v2=0 v1prim=? v2prim=? Ek=? How do you get the speed of the lighter ball after collision...

I honeslty don't quite know how to start. It seems like the Hooke's coefficent k is independent of the answer to this problem. I would also appreciate any clue of expressing the condition when "balls will collide again". The fact that all balls can keep moving make this rather difficult. It...

So I've managed to confuse myself on this problem :) Since the problem says we can assume ##m_p << m_b##, I'm assuming that the velocity of the bowling ball will be unchanged, such that ##\vec v_{b,i} = \vec v_{b,f} = -v_{b,0} \hat i## I started out using the energy-momentum principle, ##(\vec...

So I know from a previous part of the problem that the kinetic energy right before the collision is 94.556. The non conserved work would also equal the change in kinetic energy + change in potential energy. What I don't understand is how the potential or kinetic energy would change during the...

Since in an elastic collision, both momentum and energy is conserved, P(initial)=P(final) m1(3v)=m1v+m2v m2/m1=2 Which was the given answer but if we use conservation of energy, K.E(initial)=K.E(final) 1/2*m1*(3v)^2=1/2*m2*v^2+1/2*m1*v^2 m2/m1=8 Why do we get two different answers and why...

Hi, I have the following problem: A homogeneous disc with M = 1.78 kg and R = 0.547 m is lying down at rest on a perfectly polished surface. The disc is kept in place by an axis O although it can turn freely around it. A particle with m = 0.311 kg and v = 103 m/s, normal to the disc's surface at...

I don't know if I did this correctly. ##\int Fdx = \int dE## ##F \Delta x = \Delta E## ##Fx = Mc^2 - (mc^2 + mc^2)## ##M = \frac{Fx + 2mc^2}{c^2}## ##M## is the mass of the resulting particle. ##2mc^2## is the total energy before the collision. The issue is I'm assuming that the resulting...

Xcomp: [-(1500kg)(100km/h)(sin(30))+(3500kg)(70km/h)(cos(90))]/5000kg = -15km/h ycomp: [(1500kg)(100km/h)(cos(30))+(350(70km/h)(sin(90))]/5000kg = 74.98km/h

This problem got me kinda confused since I cannot really understand the question... who tells me how the energy dissipated in this case? Has it all transformed into heat to cause the coalesce of the two particles, or ar the two particles now merged together still traveling with a certain amount...

My homework consists of trying to create a simple model for a collision. But I have trouble understanding a specific part of the assignment, namely what g-force is. I'm guessing that I'm allowed to make assumptions. But without understanding adequately the definition for g-force, I don't think...

What came to my mind for this question is: Consider one of the cars. The velocity and mass of this car are V and M respectively. And the velocity and mass of the piece attached to the car are m, v respectively. Before the collision, the velocity of this piece relative to this car is zero. So its...

When something approaches black hole time dilation slows the event down from our frame of reference such that nothing seems to cross the event horizon. How is it then we can observe two black holes colliding? From our frame of reference wouldn’t it seem the event never happens?

1.p=mv Before the collision: p hydrogen = 1.7x10^-27 * 500 =8.5*10^-25 kg ms^-1 p carbon = 2.0x10^-26 * 0 = 0 kg ms^-1 p total before = 8.5*10^-25 kg ms^-1 The sum of momentum prior to the collision is equal to the total momentum after a collision, momentum is constant, therefore; p before = p...

Question 1: Since Force=Change in momentum = ∆P / ∆t = mv / t Momentum of the water coming out=mv mv=ρVv=ρAv Force d/dt (ρAv2t)=ρAv^2 Force = 1000*40*30ms^2 Force = 3.6*10^7 N Question 2: This is where I am confused because I understand in an elastic collision total kinetic energy and...

Consider a slope with mass, that can move in the horizontal plane without friction. A ball is dropped and hits the slope with restitution coefficient e. How to calculate the final velocities? How can I solve something like this? Note that it's not a simple 2D collision, it has a restriction...

The following is from the 2018 AQA AS Further Maths/Mechanics Paper 2 exam: 'Two smooth spheres A and B of equal radius are free to move on a smooth horizontal surface. The masses of A and B are m and 4m respectively. The coefficient of restitution between the spheres is e. The spheres are...

Let's say you have two masses on either side of a spring. Mass 1 is connected to the end of a spring. The spring itself has no mass. Mass 2 is free in space. So you have: [M1]-[spring] [M2]So it's more descriptive, I'll name the variables like you might in programming. Let's define...

I am trying to understand the science of stone skipping. First, if the stone is thrown vertically would it bounce at least once for any possible velocity or does it go down only? What sort of collision is it when the stone touches the water and how to look at conservation of momentum in this...

IS my solution right? Comparing with the other solutions, the answer just exchange the signals, i don't know why, THats what ifound. And here is the three equations: {i use the point which occurs the collision} Lo = Lf >> 0 = Iw + M*Vcm(block) Eg = ct> mvo² = mvf² + MVcm² + Iw² I = ml²/3...

By using PV=nRT formula, I have found the volume of the vessel. As far as I have learned to calculate the number of collision in a unit volume. So, it is being difficult for me to find the right way to solve. I searched on the internet and have got this...

Since the mass of both vehicles is the same, it's possible to calculate Ki which happens to be 512,5J and from there, multiply it by 0,25 since 75% of the energy is gone and I end up with 128,125J. Now my problem is that for the velocity, I have: 25,0 + -20,0 = V1 + V2 which is 5,00 = V1 + V2...

First of all, we can apply the third kepler of law, and call a by the major axis i.e, the distance between the particles. Replacing μ = m1m2/(m1+m2) Now, the particle is distanced by a and is stopped, and, in a reference r1 and r2 are the position of the particles, and r = r1-r2 their distance...

Hi, I'm looking for help making a graph/model for evaluating the "bounce" of a mass behind a spring that collides with a wall. The setup would include one simple spring mass system that is attached to a wall, and another wall which is closer to the mass than the spring's free length. The mass is...

Hi guys, I'm studying my first-year physics in college, and I'm having to write a report of some proton-proton collisions that were registered in the LHC of CERN years ago. The main goal is to identify different bosons (W and Z) that are decaying into other elemental particles. I've been asked...

A disk is dropped on a platform rotating at a constant angular speed ##\omega_i## as shown below. The question asks whether the final kinetic energy of the platform is conserved. I understand the angular momentum is always conserved provided that the net torque is 0, so I wrote the following...

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