Elastic Collision: Understanding the Role of Positive and Negative Signs

[Iris JT]

Hi guys. There is a question that confusing me for a long time. If an elastic collision occur, do I need to consider the positive and negative sign?? According to my teacher, an object moves in the opposite direction we need to consider the sign. Is it works in this situation?

 

[Kajal Sengupta]

Yes, because in collision you consider the conservation of momentum and momentum ( mass x velocity ) is a vector quantity because velocity is a vector quantity. Thus velocity has a direction and to represent a body moving in opposite directions you have to consider the sign of the velocity.

 

[Iris JT]

Yes, because in collision you consider the conservation of momentum and momentum ( mass x velocity ) is a vector quantity because velocity is a vector quantity. Thus velocity has a direction and to represent a body moving in opposite directions you have to consider the sign of the velocity.

Is this concept works on perfectly elastic collision?? Because there is one question regarding to the collision of the balls that have perfectly elastic collision, and my teacher told me not to consider the sign.. How can it be??

 

[sophiecentaur]

Is this concept works on perfectly elastic collision?? Because there is one question regarding to the collision of the balls that have perfectly elastic collision, and my teacher told me not to consider the sign.. How can it be??

I think you would need to ask your teacher about that. You may have posed a question or taken an answer out of context.
But if you are dealing with Momentum, it is a vector quantity so its sign is very important in any calculation, whether you are dealing with elastic or inelastic collisions. Momentum is always conserved, whatever.

 

[Kajal Sengupta]

See, If you are calculating the magnitude of momentum of a body then the sign is not important but if you are say calculating change in momentum then the sign has to be considered because it has to be done according to laws of vectors. It is slightly difficult to give a clear answer unless we know the context in which your teacher ignored the sign. The rules are the same for elastic and inelastic collision as far as sign is concerned.

 

[Doc Al]

Because there is one question regarding to the collision of the balls that have perfectly elastic collision, and my teacher told me not to consider the sign.. How can it be??

How about you tell us exactly what that question is?

 

[Iris JT]

Alright I think I should write out the question here.
A neutron, of mass 1.67×10^-27 kg, moving with a velocity of 2.0×10^4 ms-1, makes a head-on collision with a boron nucleus,of mass 17.0×10^-27 kg, originally at rest. Find the velocity of the boron nucleus if the collision is perfectly elastic.
This is the question, and the answer given is 3580 ms-1.

 

[sophiecentaur]

Alright I think I should write out the question here.
A neutron, of mass 1.67×10^-27 kg, moving with a velocity of 2.0×10^4 ms-1, makes a head-on collision with a boron nucleus,of mass 17.0×10^-27 kg, originally at rest. Find the velocity of the boron nucleus if the collision is perfectly elastic.
This is the question, and the answer given is 3580 ms-1.

In that question, the total Momentum is greater than zero. The relative velocity of separation will be equal and opposite to the relative approach velocity. You can write out a Momentum equation for before and after and combine that with an equation for the difference between velocities to solve for the final velocity of the boron nucleus. Whether or not a negative sign turns up somewhere will depend on the values of the masses. I can see no reason to go into this, looking for 'no negative velocities'.
Are you having a problem with how to solve the question? What I have written above should help but I can't be bothered to do the arithmetic because I would expect you to be able to do that (if you've been set the question appropriately.

 

[Kajal Sengupta]

To find the velocity you need to consider two equations. One the conservation of momentum and other conservation of kinetic energy ( elastic collision) . Your original dilemma was the sign of velocity. Out of these equations the sign would not matter for conservation of kinetic energy because it involves square of velocities . In conservation of momentum the equation will be : mass of neutron x velocity of neutron + mass of boron x velocity of boron before collision ( this term will be zero as the boron nucleus is at rest before collision) = mass of neutron x velocity of neutron after collision ( v1, say) + mass of boron x velocity of boron after collision ( v2, say). The direction of motion of neutron and boron is not known after collision so we are not in a position to assign any sign to the velocities. Remember the opposite signs of velocities that we usually get is after we solve momentum conservation equation. Hence it is perfectly all right if we do not consider the signs in this question.

 

[Iris JT]

In that question, the total Momentum is greater than zero. The relative velocity of separation will be equal and opposite to the relative approach velocity. You can write out a Momentum equation for before and after and combine that with an equation for the difference between velocities to solve for the final velocity of the boron nucleus. Whether or not a negative sign turns up somewhere will depend on the values of the masses. I can see no reason to go into this, looking for 'no negative velocities'.
Are you having a problem with how to solve the question? What I have written above should help but I can't be bothered to do the arithmetic because I would expect you to be able to do that (if you've been set the question appropriately.

Yes I'm having a problem on solving the question. But I think my minds go more clearer after listening to your explanation.Thanks a lot!

 

[Iris JT]

To find the velocity you need to consider two equations. One the conservation of momentum and other conservation of kinetic energy ( elastic collision) . Your original dilemma was the sign of velocity. Out of these equations the sign would not matter for conservation of kinetic energy because it involves square of velocities . In conservation of momentum the equation will be : mass of neutron x velocity of neutron + mass of boron x velocity of boron before collision ( this term will be zero as the boron nucleus is at rest before collision) = mass of neutron x velocity of neutron after collision ( v1, say) + mass of boron x velocity of boron after collision ( v2, say). The direction of motion of neutron and boron is not known after collision so we are not in a position to assign any sign to the velocities. Remember the opposite signs of velocities that we usually get is after we solve momentum conservation equation. Hence it is perfectly all right if we do not consider the signs in this question.

Alright I think I got it! So when the velocities are unknown, and yet I cannot predict the direction of the boron and neutron so I need not to consider the sign. Is that what you meant? Thanks!

 

[Kajal Sengupta]

Yes, that is what I meant. Assigning a sign to the velocity would mean that you already know how the particles will move after collision which is not true. If at all they are moving in opposite directions then that should come out as a consequence of calculation, which it does in conservation of momentum calculation for the examples like splitting of nucleus, firing of bullet and rifle etc.

 

[sophiecentaur]

Assigning a sign to the velocity would mean that you already know how the particles will move after collision which is not true.

Do not worry. It doesn't matter what sign you assign the final velocities that you put in your equation. If you choose to call them +v₁ and a +v₂ (moving to the right, say) then, once you have solved the equation, the result will be that one of the vs will turn out to be negative. If you happen to choose -v₁ and -v₂ then the outcome of the equation will give the 'other v' as negative. You don't need to 'know' initially; the maths does it for you as long as you go through the algebra with no slip ups.