Question

A curling stone travelling at 10m/s undergoes a two-dimensional elastic collision with an identical stationary stone. The target stone has a speed of 8.0m/s after impact. What is the speed of the incident stone after impact?

Do I solve this problem using the conservation of momentum law p=mv? Where the momentum of the initial stone is equal to the sum of the momentum of the two stones after collision?

Answer 1

total momentum before = total momentum after

yep, the conservation of momentum

Answer 2

m1v1 + m2v2 = m1v3 + m2v4

Answer 3

I tried using the momentum to solve it but I didn't get the right answer.
10m/s*"M=M*8.0m/s+M*v ; where M is some arbitrary mass of the curling stone and v is the velocity of the second stone.
solving for v, I got 2.0m/s.

Answer 4

it says that its a elastic collision right?
that means the kinetic energy is also conserved

Answer 5

what is the mass of the stone?

Answer 6

the question said it was an elastic collision so then I tried using the conservation of kinetic energy E=.5mv^2 and I got a different answer. I'm confused why they came out to be different...

Answer 7

The mass of the stone is not given, so I used 1kg for simplicity for both stones as they were identical

Answer 8

Since both stones have equal masses and the second stone is initially stationary, this should be quite easy. The first stone is traveling at 10m/s before the collision and the second stone is traveling at 8m/s after the collision. The first stone had to be reduced to a speed of 2 m/s.

Answer 9

Unless I'm just not awake yet :P

Answer 10

I got 2.0m/s for the second stone as well, and I thought it was right, but it's incorrect :(
Using kinetic energy where the kinetic energy of the initial moving stone is equal to the kinetic energy of the two stones after collision, I get that the second stone has a velocity of 6.0m/s, this knowing the the first stone after collision has a velocity of 8.0m/s.

Answer 11

With equivalent masses, the equation should be reduced to simply:
\[v_{1i}+v_{2i}=v_{1f}+v_{2f}\]

Answer 12

oh

Answer 13

For a non-head-on elastic collision between equal masses the angle between the velocities after the collision will always be 90 degrees...

Answer 14

Shane B, the equation you wrote is essentially what I used initially to solve the problem and I got 2.0m/s for the second stone. The problem is that this isn't the right answer :O

Answer 15

whats the right answer?

Answer 16

I understand it wasn't an accepted answer...trying to figure out what we're missing here.

Answer 17

The question that I posted is the exact question on my test that I'm studying (past test already written, just studying for exam)

Answer 18

It was a multiple choice question:

a) 2.0m/s
b)4.0m/s
c) 5.0m/s
d) 6.0m/s

Using momentum, I got A, which is wrong :(
Using kinetic energy, I got d

Answer 19

so D is the right answer?

Answer 20

I'm not sure, my answer was wrong and I didn't ask what the correct answer was. :(

Answer 21

I just worked out "d" as a solution a few minutes ago

Answer 22

I appreciate you guys helping me though :)

Answer 23

I think it would be A in this case, because through out your school work, you would have been using the conservation of momentum to figure out the velocities. And as for the kinetic energies..

Answer 24

yeah the total kinetic energy before = 50 and after = 34, using 2ms-1 which would prove that its a inelastic collision but nope its elastic. This is a tricky one, im still thinking:/

Answer 25

I know, I feel like it should be A since momentum is always conserved... Unfortunately it doesn't give the right answer... It also bothers me why using kinetic energy, it gives a different solution for velocity of the second stone.

Answer 26

We're missing something simple...since KE and momentum should be conserved in a perfectly elastic collision.

Answer 27

indeed

Answer 28

@eashmore : Figure out where we went wrong!

Answer 29

Let me take a gander.

Answer 30

lolya XD

Answer 31

I think we broke physics.

Answer 32

Shane... you read my mind....

Answer 33

The multiple choice options include both the conservation of momentum and conservation of energy answers. They differ, which bothers me because conservation of momentum should always hold when no external forces are present. Because the prompt explicitly states that the collision is elastic, a special case where conservation of energy holds, I'm inclined to choose that answer.

Answer 34

However, both should hold.

Answer 35

I really expected the answers to be the same..no matter which method was used.

Answer 36

I really thought they would both hold, but I tried the conservation of kinetic energy with another question where I solved velocity using momentum (correct solution for velocity), and kinetic energy was not conserved.....O.o

Answer 37

I tried this out of curiosity, to my surprise, it was not conserved D:

Answer 38

I wish my physics exam wasn't this week or I'd pose this strange perdicament to my teacher and stump her XD

Answer 39

I have to wonder if one of the velocities were supposed to be a component velocity...

Answer 40

I was thinking about component velocities too, but I don't think it was a component velocity otherwise there wouldn't be a solution that would work in the multiple choice using kinetic energy or momentumm

Answer 41

Yea...you'd need more info to solve it.

Answer 42

Logically, the conservation of momentum answer makes the most sense to me.

Answer 43

@derrick902 Kinetic energy does have to be conserved. Only energy itself. If the coefficient of restitution is not equal to 1, the collision will lose energy as heat within the molecules of the colliding objects or as strain energy if the material becomes deformed.

I'm inclined to say we are missing some information here because it says this is a 2-D collision, yet we lack angles.

In a perfectly elastic, 1-D collision, all the energy of moving block should be transferred to the stationary block.

Answer 44

It just doesn't make sense that an object moving at 10m/s hits a stationary object of the same mass...and both objects move off at a combined velocity higher than the original velocity.

Yes, I saw that two eashmore...it does say 2D but doesn't describe it further.

Answer 45

err *two = too

Answer 46

Easemore, I think you meant Kinetic energy does not have to be conserved, but I follow what you are saying. I reproduced the question I posted exactly as it appears on my test so all the necessary information should be there.

Answer 47

I stated kinetic energy explicitly.

Answer 48

you didn't write "not" XD

Answer 49

Ah yes! Whoops. haha.

I would use the COM answer. It's always conserved regardless of the coefficient of restitution.

Answer 50

I'd love to see your teacher work this one out....I think there has to be something mis-worded or missing in the problem.

Answer 51

Indeed. We can speculate all day.

I would move on to another question.

I'm logging off. Sorry I couldn't be of more help.

Answer 52

No big deal easemore, thanks for the help

Answer 53

But I do have one last comment. Let's assume that the question was worded correctly, and that it was a perfectly elastic collision. My issue is that the resultant velocity of the second stone is different if solved using the idea of conservation of kinetic energy versus the conservation of momentum. Without working it out, I would've assumed that they would yield the same result, but tis not the case D: . I've never had a question that would contradict otherwise until today, and apparently neither have you guys XD

Answer 54

Yea...I agree. This is going to bug me until I figure out why the answers are different. If you get an answer that explains all this, please post it back here.

Answer 55

did you guys take the directions right?

Answer 56

what is the answer?

Answer 57

The question doesn't provide directions...and he only knows that he's been told that 2m/s was incorrect.

Answer 58

Yes, I will definitely post the reasoning behind this madness if I ever find out

Answer 59

I guess we can wait for someone here to bless us with their godly insight in the meantime

Answer 60

i think i got it, the question says that it was a elastic collision.

if we just work out the speed using the conservation of momentum, it gives us the speed of the stone during a in inelastic collision and we know that because using 2ms-1 we work out that the kinetic energy is not conserved. The conservation of momentum does not necessarily have to equal to the conservation of the kinetic energy

but that is not the case, since it is a elastic collision, all of the kinetic energy is conserved. So the speed from the energy equation is correct.

Answer 61

I think momentum and kinetic energy are both at play and should both work in an elastic collision such as this. They both seem to make sense and I'm almost certain you can solve velocities of an elastic system using conservation of momentum.

Answer 62

I have found the reason why we were wrong on this question.

The question was worded perfectly correct and it wasn't because the collision was elastic or inelastic. The reason why we couldn't use the conservation of momentum was because the collision was not one dimensional (ie left to right). We know from the conservation of momentum that the total momentum of the initial objects is equal to the sum of the momentum of the objects after collision, but this only applies to 1 dimensional collisions. In 2 dimensional collisions, the total x components and total y components must be conserved before and after collision. But what we did was take the vector momentum of the first stone plus the second stone (which was stationary) to be equal to the momentum of the two stones after collision. But momentum doesn't work this way. While it is true that the x component and y component of the momenta would be equal before and after collision, the vector momentum of the two stones would not simply be the sum. We essentially used 1D rules in a 2D problem. However, the conservation of kinetic energy holds very nicely here and (correct me if I'm wrong) anywhere there is an elastic collision.

And that my friends, is why the velocity was different when solved using momentum vs kinetic energy.

I think we all knew something was amiss in the back of our minds but we just overlooked this simple fact.

Answer 63

That makes sense. I can't believe I didn't note the distinction that momentum was a vector while kinetic energy was a scaler quantity.

Glad you got it figured out.

Answer 64

I missed it too...but thanks for posting the info back here.

Answer 65

np :)