What is Collision: Definition and 1000 Discussions
In physics, a collision is any event in which two or more bodies exert forces on each other in a relatively short time. Although the most common use of the word collision refers to incidents in which two or more objects collide with great force, the scientific use of the term implies nothing about the magnitude of the force.
Some examples of physical interactions that scientists would consider collisions are the following:
When an insect lands on a plant's leaf, its legs are said to collide with the leaf.
When a cat strides across a lawn, each contact that its paws make with the ground is considered a collision, as well as each brush of its fur against a blade of grass.
When a boxer throws a punch, their fist is said to collide with the opponent's body.
When an astronomical object merges with a black hole, they are considered to collide.Some colloquial uses of the word collision are the following:
A traffic collision involves at least one automobile.
A mid-air collision occurs between airplanes.
A ship collision accurately involves at least two moving maritime vessels hitting each other; the related term, allision, describes when a moving ship strikes a stationary object (often, but not always, another ship).
In physics, collisions can be classified by the change in the total kinetic energy of the system before and after the collision:
If most or all of the total kinetic energy is lost (dissipated as heat, sound, etc. or absorbed by the objects themselves), the collision is said to be inelastic; such collisions involve objects coming to a full stop. An example of such a collision is a car crash, as cars crumple inward when crashing, rather than bouncing off of each other. This is by design, for the safety of the occupants and bystanders should a crash occur - the frame of the car absorbs the energy of the crash instead.
If most of the kinetic energy is conserved (i.e. the objects continue moving afterwards), the collision is said to be elastic. An example of this is a baseball bat hitting a baseball - the kinetic energy of the bat is transferred to the ball, greatly increasing the ball's velocity. The sound of the bat hitting the ball represents the loss of energy.
And if all of the total kinetic energy is conserved (i.e. no energy is released as sound, heat, etc.), the collision is said to be perfectly elastic. Such a system is an idealization and cannot occur in reality, due to the second law of thermodynamics.
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...
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 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)...
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...