Solving Momentum Problem With Perfectly Elastic Collisions

  • Thread starter Ed Quanta
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In summary, the problem is to calculate the velocity of the smaller sphere after the bigger sphere hits the ground in a sequence of closely spaced impulsive collisions. The collisions are perfectly elastic and the problem can be solved using momentum and energy conservation laws. The final velocities can be calculated by using the momentum and energy equations, where the velocity of the bigger ball after the first collision is equal but opposite to the initial velocity, and the velocity of the smaller ball can be calculated using the total momentum and energy of the system.
  • #1
Ed Quanta
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Two spheres are perfectly vertically alligned and falling towards the ground. They are separated by a small distance. The bigger sphere is on the bottom, and both spheres are supposed to have the common velocity v just before they strike the ground. I have to calculate the velocity of the smaller sphere, after the bigger sphere hits the ground. All collisions are perfectly elastic, and I am told to solve this as a sequence of closely spaced impulsive collisions. Help or ideas anybody? I am not sure what the sequence of collisions I am calculating are. And I am also not sure how to use either the lab or center of mass frame of references since the lab frame supposes that the velocity of one of the masses is zero which is not true, and the center of mass frame assumes the total momentum is zero, and I am not convinced that this is true.
 
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  • #2
The larger ball strikes the ground first (collision 1) it then collides with the smaller ball (collision 2). The larger ball will rebound with equal but opposite velocity. Use that velocity in the second collison.
 
  • #3
Simply use the momentum and energy conservation laws. Thats enough to solve the problem.
 
  • #4
Ok, but I am having trouble with what is going on in the second collision. I understand that when the larger mass hits the ground it will rebound with equal but opposite velocity. The smaller mass is going to collide with this mass with an equal but opposite velocity. However, since its mass is smaller, the momentum of the larger mass will be greater, and the total momentum will equal v(m-M) where m is the smaller mass, and M is the larger mass. However, this total momentum is non zero so I cannot use the center of mass frame?

-Mv + mv=v(m-M)=v1fM +v2fm

How am I to know the final velocities from this information alone? I want to know v2f, so v2f=(v(m-M)-v1fM)/m. But how am I to know v1f? I understand that momentum and energy are conserved, but how do the velocities change after the collision. It seems to me that they would be the same. By the way, v2f is supposed to equal v(3M-m/M+m).
 
  • #5
Originally posted by Ed Quanta
Ok, but I am having trouble with what is going on in the second collision. I understand that when the larger mass hits the ground it will rebound with equal but opposite velocity. The smaller mass is going to collide with this mass with an equal but opposite velocity. However, since its mass is smaller, the momentum of the larger mass will be greater, and the total momentum will equal v(m-M) where m is the smaller mass, and M is the larger mass. However, this total momentum is non zero so I cannot use the center of mass frame?

-Mv + mv=v(m-M)=v1fM +v2fm

How am I to know the final velocities from this information alone? I want to know v2f, so v2f=(v(m-M)-v1fM)/m. But how am I to know v1f? I understand that momentum and energy are conserved, but how do the velocities change after the collision. It seems to me that they would be the same. By the way, v2f is supposed to equal v(3M-m/M+m).

You say "I understand that momentum and energy are conserved" but you have written only the momentum equation. As tomkeus suggested, use momentum AND ENERGY conservation.
 

What is a perfectly elastic collision?

A perfectly elastic collision is a type of collision in which there is no loss of kinetic energy. This means that the total kinetic energy of the system before and after the collision remains the same.

How is momentum conserved in a perfectly elastic collision?

In a perfectly elastic collision, momentum is conserved because the total momentum of the system before the collision is equal to the total momentum of the system after the collision. This is due to the fact that there is no loss of kinetic energy in a perfectly elastic collision.

What is the equation for solving momentum problems in perfectly elastic collisions?

The equation for solving momentum problems in perfectly elastic collisions is m1v1 + m2v2 = m1v1' + m2v2', where m1 and m2 are the masses of the objects involved in the collision, and v1, v2, v1', and v2' are the initial and final velocities of the objects, respectively.

Can the coefficient of restitution be greater than 1 in a perfectly elastic collision?

No, the coefficient of restitution, which measures the ratio of the relative velocities of two objects before and after a collision, cannot be greater than 1 in a perfectly elastic collision. This is because in a perfectly elastic collision, there is no loss of kinetic energy, so the relative velocity of the objects after the collision cannot be greater than their relative velocity before the collision.

What are some real-life examples of perfectly elastic collisions?

Some real-life examples of perfectly elastic collisions include the collision of two billiard balls on a pool table, the collision of two atoms in a nuclear reaction, and the collision of two molecules in a gas. In these situations, there is no loss of kinetic energy and momentum is conserved.

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