Inelastic collision

I like science because elastic collisions are me and my friend and in elastic is you and me

Formula

The formula for the velocities after a one-dimensional collision is:

$v_{a}={\frac {C_{R}m_{b}(u_{b}-u_{a})+m_{a}u_{a}+m_{b}u_{b}}{m_{a}+m_{b}}}$

$v_{b}={\frac {C_{R}m_{a}(u_{a}-u_{b})+m_{a}u_{a}+m_{b}u_{b}}{m_{a}+m_{b}}}$

where

v_a is the final velocity of the first object after impact

v_b is the final velocity of the second object after impact

u_a is the initial velocity of the first object before impact

u_b is the initial velocity of the second object before impact

m_a is the mass of the first object

m_b is the mass of the second object

C_R is the coefficient of restitution; if it is 1 we have an elastic collision; if it is 0 we have a perfectly inelastic collision, see below.

In a center of momentum frame the formulas reduce to:

$v_{a}=-C_{R}u_{a}$

$v_{b}=-C_{R}u_{b}$

For two- and three-dimensional collisions the velocities in these formulas are the components perpendicular to the tangent line/plane at the point of contact.

The normal impulse is:

$J_{n}={\frac {m_{a}m_{b}}{m_{a}+m_{b}}}(1+C_{R})(u_{b}-u_{a})$

Giving the velocity updates:

$\Delta {\vec {v_{a}}}={\frac {J_{n}}{m_{a}}}{\vec {n_{a}}}$

$\Delta {\vec {v_{b}}}={\frac {J_{n}}{m_{b}}}{\vec {n_{b}}}$

Perfectly inelastic collision

A completely inelastic collision between equal masses

A perfectly inelastic collision occurs when the maximum amount of kinetic energy of a system is lost. In a perfectly inelastic collision, i.e., a zero coefficient of restitution, the colliding particles stick together. In such a collision, kinetic energy is lost by bonding the two bodies together. This bonding energy usually results in a maximum kinetic energy loss of the system. It is necessary to consider conservation of momentum: (Note: In the sliding block example above, momentum of the two body system is only conserved if the surface has zero friction. With friction, momentum of the two bodies is transferred to the surface that the two bodies are sliding upon. Similarly, if there is air resistance, the momentum of the bodies can be transferred to the air.) The equation below holds true for the two-body (Body A, Body B) system collision in the example above. In this example, momentum of the system is conserved because there is no friction between the sliding bodies and the surface.

$m_{a}u_{a}+m_{b}u_{b}=\left(m_{a}+m_{b}\right)v\,$

where v is the final velocity, which is hence given by

$v={\frac {m_{a}u_{a}+m_{b}u_{b}}{m_{a}+m_{b}}}$

Another perfectly inelastic collision

The reduction of total kinetic energy is equal to the total kinetic energy before the collision in a center of momentum frame with respect to the system of two particles, because in such a frame the kinetic energy after the collision is zero. In this frame most of the kinetic energy before the collision is that of the particle with the smaller mass. In another frame, in addition to the reduction of kinetic energy there may be a transfer of kinetic energy from one particle to the other; the fact that this depends on the frame shows how relative this is.

With time reversed we have the situation of two objects pushed away from each other, e.g. shooting a projectile, or a rocket applying thrust (compare the derivation of the Tsiolkovsky rocket equation).

Partially inelastic collisions

Partially inelastic collisions are the most common form of collisions in the real world. In this type of collision, the objects involved in the collisions do not stick, but some kinetic energy is still lost. Friction, sound and heat are some ways the kinetic energy can be lost through partial inelastic collisions.

External links

• Petit, Regis. "The Art of Billiards Play". Retrieved 30 July 2012. Gives the general vector equations of a collision between two bodies of any speed.