You can't assume that the first particle stops and all its momentum goes to the second particle.v3r said:The Attempt at a Solution
Wouldn't it just be
pf2 = pi1
No, it's fine. You just need another equation--conservation of momentum. Then you'll have two equations and two unknowns.v3r said:0.5mvf1^2 - 0.5mvi1^2 + 0.5(3m)vf2^2 = 0
0.5(3m)vf2^2 = 0.5mvi1^2 - 0.5mvf1^2
There's two unknowns, vf2 and vf1
Is this setup wrong?
A simple elastic collision is a type of collision between two objects in which the total kinetic energy of the system remains constant. This means that the total energy before the collision is equal to the total energy after the collision.
In a simple elastic collision, the total momentum before the collision is equal to the total momentum after the collision. This is because there is no external force acting on the system, so the total momentum remains constant.
The key assumptions made in a simple elastic collision include: 1) the objects involved are perfectly elastic, meaning they do not deform or lose any energy during the collision, 2) the collision occurs in a frictionless environment, and 3) the objects involved are point masses with no rotational or angular motion.
The coefficient of restitution, denoted by e, is a measure of the elasticity of a collision. In a simple elastic collision, the coefficient of restitution is equal to 1, indicating a perfectly elastic collision. This means that the objects involved bounce off each other without any loss of energy.
Yes, a simple elastic collision can occur between multiple objects as long as the key assumptions are met. In this case, the total kinetic energy and momentum of the system will still be conserved, but the calculations may be more complex as they involve multiple objects.