The discussion revolves around the role of positive and negative signs in elastic collisions, particularly in the context of momentum conservation and the calculation of velocities. Participants explore whether signs should be considered when analyzing perfectly elastic collisions, with specific reference to a problem involving a neutron colliding with a boron nucleus.
There is no clear consensus on whether signs should be considered in elastic collisions. Participants express differing views on the importance of signs in momentum calculations and the implications for solving specific problems.
The discussion highlights the complexity of applying conservation laws in collision scenarios, particularly regarding the assumptions made about direction and the mathematical treatment of velocities. Specific examples and teacher guidance are referenced, but the underlying principles remain debated.
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??Kajal Sengupta said: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.
I think you would need to ask your teacher about that. You may have posed a question or taken an answer out of context.Iris JT said: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??
How about you tell us exactly what that question is?Iris JT said: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??
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'.Iris JT said: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.
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!sophiecentaur said: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.
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 said: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.
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.Kajal Sengupta said:Assigning a sign to the velocity would mean that you already know how the particles will move after collision which is not true.