# coefficient of restitution formula for inelastic collision perfectly elastic collisions in physics

know all coefficient of restitution formula for inelastic collision perfectly elastic collisions in physics ?

**Collisions**

**Elastic Collisions** :

If there is no change of kinetic energy during a collision it is called an elastic collision. The collision between subatomic particles is generally elastic. The collision between two steel or glass balls is nearly elastic. Inelastic Collisions: If there is a loss of kinetic energy during a collision, it is called an inelastic collision. Since there is always some loss of kinetic energy in any collision, collisions are generally inelastic. If the loss is negligibly small, the collision is very nearly elastic. Perfectly elastic collisions are not possible. If two bodies stick together, after colliding, the collision is perfectly inelastic, e.g. a bullet striking a block of wood and being embedded in it. The loss of kinetic energy usually results in heat or sound energy. In may be remembered that the total momentum remains conserved in both elastic and inelastic collisions. Further, since the interacting forces become effectively zero after the collision, the potential energy remains the same both before and after the collision.

One-dimensional or Head-on Collision Consider two bodies of masses m1 and m2 moving with velocities u1 and u2 in the same straight line (with u1 > u2) colliding with each other. Let v1 and v2 be their respective velocities after the collision. If velocities u1, u2, v1 and v2 are all along the same straight line, the collision is known as one-dimensional or head-on collision (Fig. 4.10) From the law of conservation of momentum

If momentum along positive x-axis is taken to be positive, the momentum along the negative x-axis is taken to be negative.

Two-dimensional or Oblique Collision If the velocities of the colliding bodies are not along the same straight line, the collision is known as two dimensional or oblique collision

In this case, we apply the law of conservation of momentum separately for x and y components of momenta. The components of momentum along the positive x-axis and positive y-axis are taken to be positive and components of momentum along negative x-axis and negative y-axis are taken to be negative.

Momentum conservation of x-components gives

Momentum conservation of y-components gives

**Coefficient**** of Restitution**

Newton proved experimentally that, when two bodies collide, the ratio of the relative velocity after collision to the relative velocity before collision is constant for the two bodies. This constant is known as coefficient of restitution and is denoted by letter e.

(i) For a perfectly elastice collision, e = 1.

(ii) For a perfectly in inelastic collision, e = 0, because the two bodies stick together and hence v2 = v1.

(iii) Perfectly elastic or perfectly inelastic collisions do not occur in nature. Hence, for any collision, e lies

between 0 and 1.

(iv) For a head-on collision

(v) For an oblique collision

**Velocities after Head-on Elastic Collision**

Refer to Fig. below again. From the law of conservation of momentum, we have

The coefficient of restitution is defined as

**Perfectly Elastic Collision**

For perfectly elastic collision, e = 1. Putting e = 1 in Eqs. (3) and (4) we get

**Special cases**

(i) If both bodies have the same mass, then

m1 = m2 = m

In this case,

v1 = u2

and v2 = u1

This means that in a one-dimensional elastic collision between two bodies of equal mass, the bodies merely exchange their velocities after the collision.

(ii) If one of the bodies, say m2, is initially at rest, then u2 = 0

In this case,

If, in addition, m1 = m2 = m, these equations give v1 = 0 v2 = u1 Thus, if a body suffers a one-dimensional elastic collision with another body of the same mass at rest, the first body is stopped dead, but the second begins in move with the velocity of the first. However, if the body at rest, namely B, is much more massive than the colliding body A, i.e. m2 > m1, such that m1 is negligibly small, then

Thus, if a very light body suffers an elastic collision with a very heavy body at rest, the velocity of the lighter body is reversed on collision, while the heavier body remains practically at rest. A common example of this type of collision is the dropping of a hard steel ball on a hard concrete floor. The ball rebounds and regains its original height from where it was dropped while the much more massive ground remains at rest. Finally, if the body at rest is much lighter than the colliding body, i.e. if m2 < m1, we have

i.e. the velocity of the massive body remains practically unchanged on collision with the lighter body at rest and the lighter body acquires nearly twice the initial velocity of the massive body.

(iii) Kinetic energy delivered by incident body to a stationary body in perfectly elastic head-on collision. Kinetic energy of m1 before collision is

(iv) Change in kinetic energy of a system in a perfectly inelastic head-on collision. In a perfectly inelastic collision, the two stick together after the collision. Hence v1 = v2 and e = 0. Putting e = 0 in Eqs. (1) and (2), we get

Total K.E. of the system before collision is

In general, if u2 ≠ 0, we have

#### Oblique Impact on a Fixed Horizontal Plane

Consider a body of mass m moving with a velocity u making an angle α with the normal ON to a fixed horizontal floor as shown in Fig. 4.12. After collision with the horizontal plane, the body is deflected with a velocity v making an angle β with the normal. Since the horizontal plane is fixed, it remains at rest. Hence the impact takes place along the normal. The normal component of u is ucosα along – y direction and the normal component of v is v cosβ along the +y direction. Now

Since the impulsive force acts along the normal, the momentum along the normal is not conservved. Since the component of the impulsive force along the horizontal is zero, the momentum along the horizontal is conserved. Hence

For a perfectly elastic collision, e =1 and Eqs. (3) and (4) give

v = u and β = α

i.e. for a perfectly elastic collision, the body rebounds from a fixed surface with the same speed and at the same angle on the other side of the normal. Direct Impact on a Fixed Plane If the body falling normally on a fixed plane rebounds after impact, then, in this case α = β = 0. Using this in Eqs. (3) we get

v = eu

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