Why Do Objects in Elastic Collisions Sometimes Share or Transfer Velocities?

  • Thread starter Thread starter Biker
  • Start date Start date
  • Tags Tags
    Elastic
AI Thread Summary
In elastic collisions, the behavior of two objects can be understood through the principles of conservation of momentum and kinetic energy. When two objects of equal mass collide, if one is stationary, the stationary object can take on the velocity of the moving object, resulting in both moving at the same speed post-collision. This occurs because the total momentum and kinetic energy before and after the collision must remain constant. The equations governing these principles allow for the determination of final velocities based on initial conditions. Understanding these concepts clarifies why objects may share or transfer velocities during elastic collisions.
Biker
Messages
416
Reaction score
52

Homework Statement


So I have just studied this topic and it seems a bit confusing to me.
Lets just say we have elastic collision, Why sometimes both of the objects move at the same speed? and sometimes if one object is stationary and the other object hits it ( They have the same mass) why does it give all its velocity to the 2nd objects.
I have found something that says:
If m1 = m2 and one of them are stationary (m2) then m2 takes m1's momentum.
Like I just want to know why? Is there is something that I can know about this?

2. Homework Equations .
P = mv
 
Last edited:
Physics news on Phys.org
Biker said:
Why sometimes both of the objects move at the same speed? and sometimes if one object is stationary and the other object hits it ( They have the same mass) why does it give all its velocity to the 2nd objects.
Could you state the initial, and final conditions clearly?
 
All of these results can be obtained from the respective equations. The equations for the conservation of energy and conservation of kinetic energy give you these results. The best thing would be to obtain the final velocities of the two objects. This will solve most of your doubts in this regard. This is easy for one-dimensional collisions. The results hold in general.
 
Biker said:

Homework Statement


So I have just studied this topic and it seems a bit confusing to me.
Lets just say we have elastic collision, Why sometimes both of the objects move at the same speed?
I'm not sure what you mean by that. You have to be given the initial speeds of the two objects and, of course, they might happen to have the same. If you are referring to after a collision, the speeds are given by "conservation of mass" and "conservation of (kinetic) energy".

and sometimes if one object is stationary and the other object hits it ( They have the same mass) why does it give all its velocity to the 2nd objects.
I have found something that says:
If m1 = m2 and one of them are stationary (m2) then m2 takes m1's momentum.
Like I just want to know why? Is there is something that I can know about this?

2. Homework Equations .
P = mv
That's the equation for momentum. If object 1 has mass m and speed v, then it has momentum mv but also has kinetic energy (1/2)mv^2. If object 2 has the same mass but initial speed 0, it has momentum and kinetic energy 0. The total momentum is mv and the total kinetic energy is (1/2)mv^2. Now suppose that, after the collision, they have speeds v1 and v2. The total momentum is now mv1+ mv2= m(v1+ v2) and the total kinetic energy is (1/2)mv1^2+ (1/2)mv2^2= (1/2)m(v1^2+ v2^2). By "conservation of momentum" we have m(v1+ v2)= mv and by "conservation of energy" we have (1/2)m(v1^2+ v2^2)= (1/2)mv^2.

Those are two equations to solve for the two speeds, v1 and v2. An obvious first step in the momentum equation is to divide both sides by m to get v1+ v2= v and divide both sides of the energy equation by (1/2)m to get v1^2+ v2^2= v^2. From the first equation, v2= v- v1 and replacing v2 by that in the second equation, v1^2+ (v^2- 2vv1+ v1^2)= 2v1^2- (2v)v1+ v^2= v^2 so we have the quadratic equation 2v1^2- (2v)v1= 2v1(v1- v)= 0. That two solutions- either v1= 0 so that v2= v or v1= v so that v2= 0. The first is the situation in which the first object hits the second, stopping and the second object moves off with the same speed. The second is the situation in which the first object doesn't hit the second object at all!
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top