Kinetic energy and momentum in an elastic collision

[Monsterboy]

Homework Statement

Mass m1 = 2kg traveling at v = 3 m/s
Mass m2 = 3kg traveling at v = 2 m/s
After an elastic collision (from opposite directions) what will be the momentum and velocities of each of the bodies ?

Homework Equations

[/B]
Momentum = mass x velocity
$K.E = \frac {1}{2}.m.v^2$

The Attempt at a Solution

[/B]
The momentum of each body is the same in magnitude. as $m1.v1 = m2v2$

But, kinetic energies will be different. Using the formula we get $K.E1 = 9 J$ and $K.E2 = 6 J$

In an elastic collision both kinetic energy and momentum are conserved.
After a perfectly elastic collision (I think)both the bodies will retain the same momentum that will leave velocity as the only variable, does that mean they will retain the same kinetic energies ?

 

[Nugatory]

After a perfectly elastic collision (I think)both the bodies will retain the same momentum

That assumption is not valid.

But before we go there, you have to more precise about the problem. Are the two objects on a head on collision course (one moving to the left, one moving to the right) or are they moving in the same direction (object m2 is moving to the right at 2 m/sec, m1 is behind it but moving faster so that it catches up and strikes m2 from behind with a speed of 1 m/sec)?

Either way, you have two unknowns: the speed of m1 after the collision and the speed of m2 after the collision. Conservation of kinetic energy and conservation of momentum will you give two equations in these unknowns. Write down those equations and the algebra will see you home from there.

 

[PeterO]

Homework Statement

Mass m1 = 2kg traveling at v = 3 m/s
Mass m2 = 3kg traveling at v = 2 m/s
After an elastic collision what will be the momentum and velocities of each of the bodies ?

Homework Equations

[/B]
Momentum = mass x velocity
$K.E = \frac {1}{2}.m.v^2$

The Attempt at a Solution

[/B]
The momentum of each body is the same in magnitude. as $m1.v1 = m2v2$

But, kinetic energies will be different. Using the formula we get $K.E1 = 9 J$ and $K.E2 = 6 J$

In an elastic collision both kinetic energy and momentum are conserved.
After a perfectly elastic collision (I think)both the bodies will retain the same momentum that will leave velocity as the only variable, does that mean they will retain the same kinetic energies ?

Once you have decided which direction the bodies are going (opposite or same) (see NUGATORY's response) You could then calculate the velocity of the Centre of Mass. With an elastic collision, each mass begins moving towards the centre of mass at a vertain speed. After the collision each mass is traveling away from the centre of mass at exactly the same speed. (doing all the calculations NUGATORY speaks of and check that idea).

 

[Monsterboy]

I want to do it for opposite directions. So, momentum are equal and opposite and they cancel out right ?
What about the kinetic energy ? is the net kinetic energy of the two body system $9 - 6 = 3 J$ ?

 

[Nugatory]

I want to do it for opposite directions.

Velocities have signs - moving left is a negative velocity and moving right is a positive velocity. So you'll want to correct your initial problem statement accordingly.

So, momentum are equal and opposite and they cancel out right ?

Not in general, but for this particular choice of masses and speeds it does. But you should actually do the (very easy) calculation using the definition of momentum once you've gotten the signs on the velocities right to see that the total momentum comes out to be zero instead of just intuiting that everything cancels.

What about the kinetic energy ? is the net kinetic energy of the two body system $9 - 6 = 3 J$ ?

No. Just as the total momentum is the sum of the individual momenta after you've gotten the sign of the velocities right, the total kinetic energy is the sum of the individual kinetic energies.

 

[Monsterboy]

I solved the following equations
Where v1 and v2 are assumed to be velocities after collision.

$2v1^2 + 3v2^2 = 30$ from conservation of K.E
$2v1 - 3v2 = 0$ from conservation of momentum.

taking $v1 = \frac{3v2}{2}$ and substituting that in the K.E for $v1$

I get the exact same velocities that were present before the collision.

 

[Nugatory]

I solved the following equations
Where v1 and v2 are assumed to be velocities after collision.

$2v1^2 + 3v2^2 = 30$ from conservation of K.E
$2v1 - 3v2 = 0$ from conservation of momentum.

taking $v1 = \frac{3v2}{2}$ and substituting that in the K.E for $v1$

I get the exact same velocities that were present before the collision.

There are two solutions (remember, the signs on the velocities matter) and you've only found one; the other is the one you're looking for.

However, you should also reread @PeterO's post #3 in this thread. Perhaps by accident, you have chosen a rest frame in which the center of mass is not moving (equivalently, the total momentum is zero) and the solutions take on a particularly simple form in that frame. It would be a good exercise to try working this problem when m1 has an initial speed of 4 m/sec and and m2 has an initial speed of -1m/sec; note that this is exactly the same physical situation, just described from the point of view of someone who is moving to the right at 1 m/sec.

 

[Monsterboy]

You could then calculate the velocity of the Centre of Mass.

Sorry, I don't know how to do this.

 

[Monsterboy]

There are two solutions (remember, the signs on the velocities matter) and you've only found one; the other is the one you're looking for.

Two solutions, one for collision while in the same direction and other for opposite directions ?

 

[Nugatory]

Two solutions, one for collision while in the same direction and other for opposite directions ?

Find them both. One of them is for the case in which they collide, one of them is for the case in which they miss one another. Which one is which?
(This might be a bit clearer if you try an example in which one of them starts at rest and the other one hits it).

 

[PeterO]

Sorry, I don't know how to do this.

The velocity of the centre of mass is simply the total momentum divided by total mass (taking account of positive and negative velocities, and therefore momentum).
In this case, the total momentum is zero.

 

[PeterO]

I solved the following equations
Where v1 and v2 are assumed to be velocities after collision.

$2v1^2 + 3v2^2 = 30$ from conservation of K.E
$2v1 - 3v2 = 0$ from conservation of momentum.

taking $v1 = \frac{3v2}{2}$ and substituting that in the K.E for $v1$

I get the exact same velocities that were present before the collision.

Note: the mathematics equations just describe the system of bodies when the KE and Momentum have those values. That is the case both before and after the collision. It sounds like you have calculated the "before " answer (which is not news) you also need to calculate the "after" answer.
These problems always have the "before" and "after" situation - you 'calculate' both, recognise one as the (given) 'before' conditions, so the other answer is the 'after' conditions.

 

[PeterO]

From your original post:

In an elastic collision both kinetic energy and momentum are conserved.
After a perfectly elastic collision (I think)both the bodies will retain the same momentum that will leave velocity as the only variable, does that mean they will retain the same kinetic energies ?

These statements are a little bit "loose with the truth" or misleading.

At all times, during any collision, the TOTAL momentum remains the same - though the momentum of each body (contributing to the total) may be changing.
For an elastic collision the TOTAL KE before the collision is the same as the TOTAL KE after collision, however the KE of each body involved may be different. It is common for one body to slow down while the other speeds up during a collision. The TOTAL KE usually reduces during the collision, then increases back to what it was.
There are even some examples when the KE of each of the bodies just happens to be the same before and after a collision, but those examples should be accepted as a coincidence rather than the rule.

 

[PeterO]

I want to do it for opposite directions. So, momentum are equal and opposite and they cancel out right ?

The momenta of the bodies are equal in magnitude and opposite in direction, but I would prefer that they sum to zero rather than that they cancel out.
Consider 10 tonne truck traveling at 20 miles per hour in one direction, about to collide with a 2 tonne car traveling at 100 miles per hour in the opposite direction. To say that their momentums cancel out almost makes it sound like there is not going to be much happen when they collide. To say the momenta sum to zero predicts that when the mayhem is over, both wrecks will be stationary, and (approximately) at the point where they collided.

 

[Monsterboy]

Find them both. One of them is for the case in which they collide, one of them is for the case in which they miss one another. Which one is which?
(This might be a bit clearer if you try an example in which one of them starts at rest and the other one hits it).

Considering motion in the same direction.

$2v1 + 3v2 = 12$
$2v1^2 + 3v2^2 = 30$

I get $v1 = 1.8 m/s$
and $v2 = 2.8m/s$ are these velocities after collision ?

this looks weird because I expected the body with more mass to move with lesser velocity after collision...for some reason.

 

[PeterO]

Considering motion in the same direction.

$2v1 + 3v2 = 12$
$2v1^2 + 3v2^2 = 30$

I get $v1 = 1.8 m/s$
and $v2 = 2.8m/s$ are these velocities after collision ?

this looks weird because I expected the body with more mass to move with lesser velocity after collision...for some reason.

Just a comment on solving this sort of problem.
While it is common to use $v1 & v2$ as the velocities, I prefer to use (in this case) $v2 & v3$ where $v2$ is the velocity of the 2kg mass and $v3$ is the velocity of the 3 kg mass. That way when I reach the answers, there is no doubt about which velocity related to which mass. I will even use $v3 & v5$ if that is what the masses were.

 

[PeterO]

Considering motion in the same direction.

$2v1 + 3v2 = 12$
$2v1^2 + 3v2^2 = 30$

I get $v1 = 1.8 m/s$
and $v2 = 2.8m/s$ are these velocities after collision ?

this looks weird because I expected the body with more mass to move with lesser velocity after collision...for some reason.

And doing the Centre of mass consideration.
The total momentum is 12 units, the total mass is 5kg, so the velocity of the centre of mass is 2.4 m/s
That means the 2 kg mass (travelling at 3 m/s) is approaching the CofM at 0.6 m/s while the CofM is approaching the 3 kg mass at 0.4 m/s
After the collision the 2kg mass is receding from the CofM at 0.6 m/s (2.4 - 0.6 = 1.8m/s), while the 3 kg mass will be moving away from the CofM at 0.4 m/s (2.4 + 0.4 = 2.8 m/s)
This "shortcut"/alternative method ONLY works for elastic collisions.

 

[Monsterboy]

I get v1=1.8m/sv1=1.8m/sv1 = 1.8 m/s
and v2=2.8m/sv2=2.8m/sv2 = 2.8m/s are these velocities after collision ?

Does this same result apply for head on collision ?

 

[PeterO]

Does this same result apply for head on collision ?

I think the answer to your question is NO.

The mathematics (solution of the two sumiltaneous equations) yields a pair of answers: (3,2) and (1.8,2.8) One of those answers represents the velocities before the collision and the other represents the velocities after. (3,2) is clearly the before situation, so (1.8, 2.8) is the situation after collision.
Because they are traveling in the same direction, all the velocities were positive (they could also all be negative - you just change your perspective. after all , a positive, rightward velocity is exactly the same as a negative, leftward velocity. (face Right and run forwards (normally) vs face Left and "run backwards"; you are still traveling the same way.
(Note also: because all the velocities were in the same direction - be could just concern ourselves with the magnitudes of those velocities (the speed), but to start looking at just speed for every problem can lead to errors - we must always start by looking at velocities..

If you consider the head on collision, one of the velocities before-hand has to be negative - either will do. So the initial state is (3,-2) or (-3,2) depending which direction you chose to be positive.

Once you apply the maths (solve the simultaneous equations) you get two answers (3, -2) and (-3, 2) one of those represents before, the other after So is (3,-2) is before - (-3,2) is after but if (-3,2) is before the (3,-2) is after.
If that seems confusing, let's say positive velocity means going East, while Negative is going West. The before/after pair (3,-2) - (-3,2) means (3E, 2W) changing to (3W, 2E) each mass changes direction.
Note:
The other answer (3,2) becoming (1.8, 2.8) means (3E, 2E) becoming (1.8E, 2.8E). No direction change, just one mass slowed down (from 3 to 1.8) while the other sped up (from 2 to 2.8)
Using the "centre of mass calculation" for the head on collision: the velocity of the Centre of Mass is zero - leading to an almost trivial solution.
Each mass is approaching the centre of mass at its assigned speed. After collision, each body will be rebounding from the CofM at exactly that same speed. SO the SPEED of each body will be the same before and after, but the VELOCITY will be very much different; 3 m/s vs -3m/s and -2 m/s vs 2 m/s. ( or 3E become 3W and 2W become 3E)

 

[Monsterboy]

So the magnitudes of momentum and magnitudes of velocities remain the same before and after collision in case of head on collision.

Thanks !

 

[neilparker62]

I have been doing the rounds of these type of questions and pointing out a very simple method of dealing with them.

The colliding bodies in an elastic collision both experience the same change in momentum Δp = 2μΔv where μ is the reduced mass [m1*m2/(m1+m2)] of the colliding objects and Δv is their relative velocity. If the bodies are traveling in opposite directions, simply subtract Δp from both before pre-collision momentums to obtain post collision momentums. Divide by respective masses to obtain respective velocities.

If the bodies are traveling in the same direction subtract Δp from the faster moving one and add Δp to the slower one.

 

[neilparker62]

Alternate method as described above:

Equal and opposite impulse during elastic collision given by 2 x 1.2 x 5 = 12 Ns (head on collision) or 2 x 1.2 x 1 = 2.4 Ns ('catch-up' collision).

1.2 is the reduced mass of the colliding pair = 2 x 3 / (2 + 3).

For the head on collision this results in both masses reversing their velocities since they started with momentums 6Ns and -6Ns respectively and end with momentums (6 -12) = -6Ns and (-6+12) = 6Ns respectively.

 

[neilparker62]

So the magnitudes of momentum and magnitudes of velocities remain the same before and after collision in case of head on collision.

Yes - the conditions for this are that the colliding bodies have equal (but opposite) momentum vectors and the collision is elastic. Then the collision impulse experienced will be (+-) 2p where p is the magnitude of the initial momentum of each body. As a result both momentum vectors simply reverse direction.

Elastic collision with initial parameters m1 = 2kg v1 = 3m/s , m2 = 6kg, v2 = -1m/s would work the same way. Assume 1-dimensional collision.