# Glancing elastic collision, and center of mass

• WrongMan
In summary, the problem can be solved using only the information provided if two of the seven unknown quantities are known.
WrongMan

## Homework Statement

So this is just something i came up with to check if i am understanding everything allright, it all comes down to collisions in regard to the center of mass

So two "masses" experience a glancing collision
M1:
vi - 3m/s (refered to as v1i) ;
m - 2Kg (m1)
M2
vi - 0 (v2i) ;
m - 6Kg (m2)

## Homework Equations

what i did was figure out the speed of the center of mass (Vcm) using:
Eq(1) (v1i*m1 +v2i*m2)/(m1+m2)
and then since the Vcm is constant before and after the collision (right?) i figured out v1f and v2f because:
eq(2) v2f=2vcm-v2i (and same for v1f)

this is were i get stuck, using the system:
Sys(1)
-v1f*m1*sin(theta) + v2f*m2*sin(theta2)=0
m1*v1i=v1f*m1*cosin(theta) + v2f*m2*cosin(theta2)

## The Attempt at a Solution

so using eq 1 and 2 i get Vcm=3/4 and v1f=-3/2 and v2f=3/2
and i can't solve the system, all the other glancing colision problems i have seen give one of the angles.
plus the masses are different so the angle bettwen theta and theta2 isn't 90º, i don't know what to do.
I am probably doing something wrong as everything i have seen says i can't solve this kind of problem if i am not given one of the 4 unknown quantities (the two angles and the two final velocities) so probably my mistake is regarding the two final velocities, i don't know.

It seems to me that there is a solution where the Ms continue to move in the same direction (with the small M going in the opposite sense of course). Therefore, if some other specific solution is required, there needs to be some piece of information which precludes the simple one - such as a given final speed which does not match or an angle of contact. Otherwise there will be many other solutions depending on the angle.
Since the other extreme case is grazing incidence, with M1 continuing undeviated and M2 remaining stationary, I think any angle of deviation for M1 is possible.

Edit: I don't think you have done anything wrong. If you have shown all the given data, nothing you can work out can add anything.

WrongMan
WrongMan said:
and then since the Vcm is constant before and after the collision (right?) i figured out v1f and v2f because:
eq(2) v2f=2vcm-v2i (and same for v1f)
This is true in one dimension only, where "glancing collision" does not make sense. The magnitude of the velocity stays the same, but the direction can change.

WrongMan
mfb said:
This is true in one dimension only, where "glancing collision" does not make sense. The magnitude of the velocity stays the same, but the direction can change.
Ohh i see, thanks, so there is no way to solve 2-d exercices using the centre of mass?
so, new questions:
1. there is no way of solving a problem like this using only the information provided, correct?
2. provided 1 is correct, imagine say v1f is a given value by the exercice, i should figure out v2f using conservation of kinectic energy, what should i do next? i still don't think i have enough information to solve the system...

i've set v1f to be 3/2, so v2f is +/- sqrt(2), how do i figure out the angles? i don't know any of the components of the two velocity vectors, only the magnitude...

Im failing to see how to relate the angles of the system.
The book clearly states as long as no more than 2 of the (seven) quantities of the equations in the system are unknown it is solvable, so there is a way of solving the system i just can't figure out how

WrongMan said:
Ohh i see, thanks, so there is no way to solve 2-d exercices using the centre of mass?
so, new questions:
1. there is no way of solving a problem like this using only the information provided, correct?
2. provided 1 is correct, imagine say v1f is a given value by the exercice, i should figure out v2f using conservation of kinectic energy, what should i do next? i still don't think i have enough information to solve the system...

i've set v1f to be 3/2, so v2f is +/- sqrt(2), how do i figure out the angles? i don't know any of the components of the two velocity vectors, only the magnitude...

Im failing to see how to relate the angles of the system.
The book clearly states as long as no more than 2 of the (seven) quantities of the equations in the system are unknown it is solvable, so there is a way of solving the system i just can't figure out how
You can use centre of mass, but whatever method you use you need to know the angle of impact. That is, the line that the initial relative velocity makes to the normal where the surfaces contact. (This is ignoring friction.). If that normal does not pass through the mass centres then it gets nastier, since torques will be involved and the objects will acquire spin. So assume the objects are spheres.

Whether it is a matter of solving for two unknowns, or three, or four, depends on how you count them. You could argue that there are four unknowns: the final speeds and angles. Counting that way, you will need four equations. You have conservation of momentum overall in the direction normal to the surface. It is only conserved 'overall' because there is an unknown impulse between the surfaces. In the direction parallel to the surfaces, with no friction, there is no impulse, so momentum is conserved separately for each object. That gets you to three equations. The fourth is conservation of energy, but that requires you to know the coefficient of restitution. Note, the elasticity operates only normal to the surfaces, so only apply it to those components of velocity.

WrongMan
thanks @haruspex, alllow me to clarify, this is a simple case (i think)
this is the sort of collision I am talking about:

So, imagine i know, v1i (v2i is 0) and i know the masses (which are different), and I am also given v1f

with this info, knowing its elastic, i can use the conservation of kinectic energy equation to figure out v2f.

all that is left to find out is theta and phi,
with:
m1*v1f*cos(theta)+m2*v2f*cos(phi)=m1*v1i
and
-m1*v1f*sin(theta)+m2*v2f*sin(phi)=0
because there are 2 unknowns and 2 equations. I am supposed to find out theta and phi but i don't know how (algebra failure?)

WrongMan said:
with this info, knowing its elastic, i can use the conservation of kinectic energy equation to figure out v2f.

all that is left to find out is theta and phi,
with:
m1*v1f*cos(theta)+m2*v2f*cos(phi)=m1*v1i
and
-m1*v1f*sin(theta)+m2*v2f*sin(phi)=0
because there are 2 unknowns and 2 equations. I am supposed to find out theta and phi but i don't know how (algebra failure?)
What is the relationship between the sin of an angle and its cosine?

WrongMan
haruspex said:
What is the relationship between the sin of an angle and its cosine?
tan(x)=sin(x)/cos(x) i think that's what you mean
I'm not seeing how that will help me but is that it? i got to write it downand try to solve it, but i think I've tried going with the tangent and it didnt seem to help, but il try again tommorow
its almost 4 am, got to get some sleep, il get back on it tommorow
thank you :)

WrongMan said:
tan(x)=sin(x)/cos(x) i think that's what you mean
I'm not seeing how that will help me but is that it? i got to write it downand try to solve it, but i think I've tried going with the tangent and it didnt seem to help, but il try again tommorow
its almost 4 am, got to get some sleep, il get back on it tommorow
thank you :)
No, a relationship involving only the sine and the cosine and constants.

WrongMan
haruspex said:
No, a relationship involving only the sine and the cosine and constants.
sine (x)=cosin (90-x) ?? don't thinks that's it as it doesn't seem to help much
im running through all the identitied and relatioships i can think of, i can't get there :/

WrongMan said:
sine (x)=cosin (90-x) ?? don't thinks that's it as it doesn't seem to help much
im running through all the identitied and relatioships i can think of, i can't get there :/
Pythagoras!

WrongMan
haruspex said:
Pythagoras!
the basic? sin(x)= opp/hyp ? Ok I am doing this and i think its working.
i get something like
m1*v1fx+m2*v2fx=m1*v1i
-m1*v1fy+m2*v2fy=0

and then i write v1fy in respect to v2fy (and v1fx to v2fx), and use the pythagoras hypotnuse formula of the both triangles as a system and solve (since i know the hypotenuses, and i wrote v1x and v1y "as" b*v2x and a*v2y)
Am i correct?

everything seemed to work out.
One last question, i got both positive angles, and both positive values for Vy, i imagine this is because it doesn't matter which particle goes up or down, i just have to pick which one i set..
but the angles are both positive also? that confused me a bit, is it also because "i get to choose" which particle goes up and which goes down?
the angles i might have messed up/confused somehow with the signs in the equations, but I am pretty sure i didnt mess up with the V signs

The way you defined the signs, everything should be positive.

WrongMan
WrongMan said:
the basic? sin(x)= opp/hyp ?
I meant sin2+cos2=1, which is equivalent to the Pythagorean formula. Looks like you got there anyway.

WrongMan
ah i get the values i have now, I am evaluati g each triangle "outside" the reference frame, after figuring out the trianglesnl i have to apply them in a way that makes sense to the problem.

haruspex said:
I meant sin2+cos2=1, which is equivalent to the Pythagorean formula. Looks like you got there anyway.

hm i thought about that but it seemed to me i wouldn't be able to "line up" same angle cosines and sines with a sum, il give it a go on paper when i get home.

Thanks for all the help

Sorry, I'm really lost at this point. Could you please break it down a bit to a simpler explanation.

Mysterio said:
Sorry, I'm really lost at this point. Could you please break it down a bit to a simpler explanation.
I assume you are referring to post #5, but I'll explain in terms of the diagram in post #6.

In post #5 I referred to the line of centres of the balls at the moment of impact, and the angle this makes to the initial velocity of the impacting ball. In the diagram, that is ##\phi##. This is because the impulse between the balls acts normally to the surfaces at the point of contact, so its line of action passes through the centres of the spheres.

Try to draw your own version of the diagram, but instead of showing it with the impacting ball striking the other full on (which is misleading), show it with the impacting ball striking it a little higher up (on a parallel line to the one shown) , so that at the instant of impact the line through their centres is the line along which the impacted ball departs.

Next, write the equations for:
- conservation of momentum in the x direction
- conservation of momentum in the y direction.

If you are assuming perfect elasticity you can also write the equation of conservation of KE.
Otherwise, you have to assume some coefficient of restitution, ##\rho##. I'll take you through that later if you wish.

## What is a glancing elastic collision?

A glancing elastic collision is a type of collision between two objects where both objects retain their original shapes and kinetic energy after the collision. This type of collision is characterized by the objects bouncing off of each other at an angle.

## How is the center of mass related to glancing elastic collisions?

The center of mass is a point in a system where the mass of the system is concentrated. In glancing elastic collisions, the center of mass of the system remains unchanged, meaning that the total momentum of the system is conserved.

## What factors affect the outcome of a glancing elastic collision?

The outcome of a glancing elastic collision is affected by the masses and velocities of the objects involved, as well as the angle at which they collide. Additionally, the elasticity of the objects and the surface on which they are colliding can also affect the outcome.

## How does a glancing elastic collision differ from an inelastic collision?

In an inelastic collision, the objects involved stick together after the collision and their kinetic energy is not conserved. In a glancing elastic collision, the objects bounce off of each other and their kinetic energy is conserved.

## What are some real-life examples of glancing elastic collisions?

Some examples of glancing elastic collisions include a game of pool, a tennis match, or a game of billiards. These involve objects colliding at angles and bouncing off of each other without losing their shape or kinetic energy.

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