# Elastic and non elastic.

1. Mar 5, 2016

### Biker

1. The problem statement, all variables and given/known data
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. Relevant equations.
P = mv

Last edited: Mar 5, 2016
2. Mar 5, 2016

### Qwertywerty

Could you state the initial, and final conditions clearly?

3. Mar 5, 2016

### CrazyNinja

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.

4. Mar 5, 2016

### HallsofIvy

Staff Emeritus
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".

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!