# Collisions

An instance of one moving object or person striking violently against another.Elastic collision: An elastic collision is an encounter between two bodies in which the total kinetic energy of the two bodies after the encounter is equal to their total kinetic energy before the encounter. Elastic collisions occur only if there is no net conversion of kinetic energy into other forms.

If two bodies of masses m1 and m2, moving with speeds u1 and u2 with u1 > u2 the same straight line, collide with the each other, their respective speeds v1 and v2 after collision are given by

By the law of conservation of momentum

m1u1 + m2u2 = m1v1 + m2v2

m1 (u1 – v1) = m2 (v2 – u2) … (1)

By the law of conservation of kinetic energy

$$\frac{1}{2}{{m}_{1}}~{{\left( {{u}_{1}} \right)}^{2}}+\frac{1}{2}{{m}_{2}}{{\left( {{u}_{2}} \right)}^{2}}=\frac{1}{2}{{m}_{1}}{{\left( {{v}_{1}} \right)}^{2}}+\frac{1}{2}{{m}_{2}}{{\left( {{v}_{2}} \right)}^{2}}$$

m1 (u12 – v12) = m2 (v22 – u22)

m1 (u1 + v­1) (u1 – v1) = m2 (v2 + u2) (v2 – u2) … (2)

Divide the equation (2) by equation (1)

We get    u1 + v1 = v2 + u2 … (3)

V2 = u1 + v1 – u2

Substitute this in the equation (1) we get

m1 (u1 – v1) = m2 (u1 + v1 – u2 – u2)

m1u1 – m2u1 + 2m2u2 = m2v1 + m1v1

$${{v}_{1}}=\frac{-{{u}_{1}}\left( {{m}_{1}}-~{{m}_{2}} \right)}{{{m}_{1}}~+~{{m}_{2}}}+\frac{2{{m}_{2}}{{u}_{2}}}{{{m}_{1}}+~{{m}_{2}}}$$

$$~{{v}_{1}}=\frac{\left( {{m}_{2}}-~{{m}_{1}} \right){{u}_{1}}}{{{m}_{1}}+~{{m}_{2}}}+\frac{2{{m}_{2}}{{u}_{2}}}{{{m}_{1}}+{{m}_{2}}}$$

Now from equation (3) we can write v1 = u2 + v2 –u1

Substitute this in the equation 1 and solve as done above then you will get the value of v2 as

$${{v}_{2}}=\left( \frac{2{{m}_{1}}}{{{m}_{1}}+{{m}_{2}}} \right){{u}_{1}}+\left( \frac{{{m}_{2}}-{{m}_{1}}}{{{m}_{1}}+{{m}_{2}}} \right){{u}_{2}}$$

Inelastic collision: An inelastic collision, in contrast to an elastic collision, is a collision in which kinetic energy is not conserved due to the action of internal friction. In collisions of macroscopic bodies, all kinetic energy is turned into vibrational energy of the atoms, causing a heating effect, and the bodies are deformed.

If two bodies of masses m1 and m2, moving with speeds u1 and u2 along the same straight line, collide and stick together, the speed v of the composite body is given by

$$v=\frac{{{m}_{1}}{{u}_{1}}+{{m}_{2}}{{u}_{2}}}{{{m}_{1}}+{{m}_{2}}}$$

The kinetic energy of the system of bodies after collision is less than that before collision. The loss in kinetic energy appears in the form of heat and sound energy.

The coefficient of restitution (e):

The coefficient of restitution (COR) is a measure of the “restitution” of a collision between two objects: how much of the kinetic energy remains for the objects to rebound from one another vs. how much is lost as heat, or work done deforming the objects.

$$e=\frac{\left| {{v}_{1}}-{{v}_{2}} \right|}{{{u}_{1}}-{{u}_{2}}}$$

Suppose a body is dropped from a height h0 and it strikes the ground with a speed v0. Let, after the inelastic collision, the speed with which it rebounds be v1 and h1 the height to which it rises, then

$$e=\frac{{{v}_{1}}}{{{v}_{0}}}=\frac{\sqrt{2g{{h}_{1}}}}{\sqrt{2g{{h}_{0}}}}={{\left( \frac{{{h}_{1}}}{{{h}_{2}}} \right)}^{\frac{1}{2}}}$$

If after n collisions with the ground, the speed is vn and the height to which the body rises is hn, then

$${{e}^{n}}=\frac{{{v}_{n}}}{{{v}_{0}}}={{\left( \frac{{{h}_{n}}}{{{h}_{0}}} \right)}^{\frac{1}{2}}}$$

Height after nth bounces is hn = eh0. Speed of rebound after nth bounce is vn = en √(2ghₒ). Total distance travelled before the body comes to rest

$${{h}_{0}}\left[ \frac{1+{{e}^{2}}}{1-{{e}^{2}}} \right]$$