What is Elastic collision: Definition and 531 Discussions
An elastic collision is an encounter between two bodies in which the total kinetic energy of the two bodies remains the same. In an ideal, perfectly elastic collision, there is no net conversion of kinetic energy into other forms such as heat, noise, or potential energy.
During the collision of small objects, kinetic energy is first converted to potential energy associated with a repulsive or attractive force between the particles (when the particles move against this force, i.e. the angle between the force and the relative velocity is obtuse), then this potential energy is converted back to kinetic energy (when the particles move with this force, i.e. the angle between the force and the relative velocity is acute).
Collisions of atoms are elastic, for example Rutherford backscattering.
A useful special case of elastic collision is when the two bodies have equal mass, in which case they will simply exchange their momenta.
The molecules—as distinct from atoms—of a gas or liquid rarely experience perfectly elastic collisions because kinetic energy is exchanged between the molecules’ translational motion and their internal degrees of freedom with each collision. At any instant, half the collisions are, to a varying extent, inelastic collisions (the pair possesses less kinetic energy in their translational motions after the collision than before), and half could be described as “super-elastic” (possessing more kinetic energy after the collision than before). Averaged across the entire sample, molecular collisions can be regarded as essentially elastic as long as Planck's law forbids black-body photons to carry away energy from the system.
In the case of macroscopic bodies, perfectly elastic collisions are an ideal never fully realized, but approximated by the interactions of objects such as billiard balls.
When considering energies, possible rotational energy before and/or after a collision may also play a role.
So i started off breaking up the problem into two sequences, right before the collision and after the collision has happened. I need to find the first ball's speed immediately before the collision which is no problem. PEi = KEf > mghi = (1/2)mvf (vf being the velocity right before the collision...
Given that we're working with an elastic collision we want to populate the following system:
##k_{i} = k_{f}##
##p_{i} = p_{f}##
Solve for kinetic energy just before and after the collision:
##k_{i} = \frac{1}{2}mv_{i}^{2}##
##k_{f} = \frac{1}{2}mv_{f}^{2} + \frac{1}{2}I_{P}^{sys}...
Assuming there are four robot inside a stationary spacecraft (mass 2 ton) , robot A has an electric gun that can shoot iron balls weighing 1kg at a speed of 8m/s. These balls collide with red balls at a 45-degree angle in a two-dimensional elastic collision. Each ball has a mass of 1kg. The red...
The portion from the text : I copy and paste the portion from the text. After discussing a situation where the result should be symmetric, he discusses one where we should not expect the answer to be symmetric. In this case, he means that were we to substitute the values of the masses, that is...
With given information in the problem I can't decide which direction balls will go after collision. I assumed they will go in opposite directions. I know that is not a full solution, but I can't come up with anything else.
From conservation of energy we have...
Hi,
Here is the problem
What is required to answer this question is two assumptions. Firstly, the component of the momentum normal to the centre line is the same before and after. Therefore, secondly, A must recoil entirely in the horizontal plane. This is the only way to answer this question...
I found that 1/2m1v1i^2+1/2m2v2i^2=1/2m1v1f^2+1/2m2v2f^2
=>0.5*200*55^2+0.5*46*0^2=0.5*40^2*200+0.5*46*0*vf^2=>vf=78.713 m/s.
The true answer is 65.2 m/s and is solved using m1v1i+mvv2i=m1v1f+m2v2f. Are these equations not interchangeable? Why can I not use the equation I used?
Equation 1 is equating the kinetic energies of the objects before and after the elastic collision. Equation 2 is equating the momentums of the objects after the elastic collision. They can be used interchangeably as long as the collision is elastic.
Am I right in my conclusion?
For this problem,
The solution is,
However, is the reason why they don't include electrical potential energy because the time interval for which we are applying conservation of energy over is very small so the change in electric potential energy is negligible?
Also, when they said, "electrons...
Using principle of conservation of momentum:
m×u=m×v1 + M×v2
Where m=mass of moving particle in the beginning
u=Initial velocity of particle m
v1= final velocity of particle m
v2=velocity of object M
m×u-(mv1)=Mv2
(mu-mv1)÷M=v2
My answer is this (mu-mv1)÷M
However, it is nowhere close to...
I'm a little confused as to what the answer could be. This was one of my homework questions that I got wrong as I chose 0.5 v as the answer. Would someone be able to tell me what the correct answer would be?
-Solved for vf using equation 3 to get 20.0m/s (speed before explosion) then solved for the distance to reach the explosion using equation 4, to get 20.0m, which felt wrong having the same numbers but that may just be coincidence.
-Found the distance travelled of the lighter piece using 530m -...
Suppose I have two spheres in 3 dimensions of equal mass. In cartesian coordinates, sphere A is traveling with velocity uAi, and sphere B travels with vBi. They will collide elastically.
I want to find the final velocities after the collision, ie uAf and vBf.
Am I correct in saying that...
From conservation of momentum:
m1u1 + m2u2 = m1v1 + m2
u1 - u2 = v1 + v2 (u2 is negative because the object moves to the left)
From conservation of KE, I got answer (C)
So there are two correct answers, (B) and (C)?
Thanks
I started by writing the equation v1i + v1f = 2v and then drawing a triangle with v1i, v1f, and 2v as the three sides. Then I used the Law of Cosines to solve for cos theta but this did not lead to a solution. Could I have a hint on how to begin? Thank you!
I have a problem in understanding angular momentum equation (mrv), especially the part where radius is involved.
imagine an elastic collision occurred between sphere of mass (M) attached to a string forming a circle of radius (R) and moving with velocity (V) and another stationary sphere having...
I was able to solve for the velocity of MB and got my answer as 4.47m/s.
The main issue right now for me is how to get the angles. I'm really confused on what most people have been posting as we didn't get a groundwork on this topic and so most of the basics I had them self taught.
So far I...
Standard formula for final velocities ##v_1##, ##v_2## in elastic collision with masses ##m_1##, ##m_2## and initial velocities ##u_1##, ##u_2## is given by $$v_1 = \frac{m_1-m_2}{m_1+m_2}u_1+\frac{2m_2}{m_1+m_2}u_2$$$$v_2 = \frac{2m_1}{m_1+m_2}u_1+\frac{m_2-m_1}{m_1+m_2}u_2$$.
By rearranging...
1 = elephant
2 = fly
So I am trying to find v'2 which is the final velocity of the fly. I have v1 the initial velocity of the elephant 2.1m/s. So I plug it into the equation and have v'2=(2m1/(m1+m2))*2.1m/s. We are not given the masses so I just know m1>m2 but I don't understand how that will...
I attempted to do mvf-mvi to find the impulse, but had trouble figuring out what to use as v (where does the angle of 3degrees come in?), and thought that there had to be more to the problem considering the other details I was given. I then attempted to maybe calculate the kinetic energy lost...
My approach so far is to use F = ma.
The forces acting on the block in the horizonital direction are friction and the force of the spring. Choosing the direction towards the spring as the positive axis.
Therefore: F = ma
-Fr - kx = ma
Solving for a = (-Fr - kx)/m
If I plug in values I end up...
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...
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...
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...
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...
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...
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...
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...
After simplifying the equations, I got:
m1(v1-v1') = m2v2' (momentum) and
m1(v1-v1')(v1+v1') = m2v2'^2 (kinetic energy)
From there, I'm not sure what to do. I referred to a textbook and it said to divide the energy equation by the momentum equation (the simplified versions) and then do a...
Hello
Can someone please tell me what is the use of poisson's ration in determinig stress cos what I know in this case we should have stress=E*strain and so now use for poison
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?
I'm interested in calculating effective gravity for a point-mass in a spinning gyro or swinging pendulum bob on a rotating planet undergoing any translational velocities and/or accelerations.
I want to investigate the theoretical effects of high-energy mechanical oscillation on orbital...
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!
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...
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...
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...
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...
How small should ##\Delta T## be in a collision to be considered elastic? In elastic collisions ##\Delta T =0##, but as far as I know, just atomic collisions are considered perfectly elastic. Then, which criterias are used to considere a collision between two objects elastic?
Homework Statement
In perfectly elastic collision between two atoms, it is always true to say that
a. the initial speed of one atom will be the same as the final speed of the other atom
b. the relative speed of approach between two atoms equals their relative speed of separation
c. the total...
Homework Statement
A ball A is rotating on a table with an angular velocity ω about its vertical axis. An identical ball B collides with the ball A elastically. After collision the ball A starts sliding over the table. The coefficient of friction is µ. Find:
1) the angle α between the angular...
Homework Statement
An atomic nucleous of mass m traveling with speed v collides elastically with a target particle of mass 3.0m (initially at rest) and is scattered at 45o
(a). What are the final speeds of the two particles?
Advice: eliminate the target particle's recoil angle by manipulating...
Just thought I'd post a couple of formulas which I have found useful when assisting (or should I say attempting to assist!) with collisions problems in the "Homework" forums. These formulas work on the basic premise that a collision is essentially a "Newton 3 event" in which equal and opposite...
Homework Statement :
Two blocks with masses m1=3kg and m2=4.5kg are moving on a platform with velocities v1 and v2, respectively. The platform is inclined with θ=30ο and is frictionless. Mass m2 has a very stiff spring with constant k=3000kg/s2 attached as shown in the figure. The two blocks...
Homework Statement
A 4-kg block moving at 7m/s makes a head-on collision with a stationary block of mass 3kg. Find the velocities of the two blocks after the collision.
m1=4kg, vi1=7m/s
m2=3kg, vi2 = 0m/s
Trying to find vf1 and vf2
2. Homework Equations
Using the conservation of momentum...
Homework Statement
A proton strikes a stationary alpha particle (4He nucleus) head-on. Assuming the collision is completely elastic, what fraction of the proton’s kinetic energy is transferred to the alpha particle?
Homework Equations
Pi = Pf
Ki = Kf
The Attempt at a Solution
Tried finding...