Coefficient of restitution is a measure of two colliding bodies. It is specific to the given pair of colliding bodies, which can be measured with suitable arrangement.

We, however, need to understand the context of collision in order to define “coefficient of restitution”. There is a difficulty here that two bodies can collide “head – on” or obliquely. As there can be infinite numbers of angles involved between the pair of colliding bodies, it is apparent that we need to specify the nature of collision so that this (coefficient of restitution) property is a unique value property for the pair of colliding bodies in question.

Figure 1: Head-on and oblique collision

Collision

For this reason, coefficient of restitution is defined for “head-on” collision. The next question, however, is that : what is “head – on” collision? In simple words, the paths of bodies and direction of normal force during collision are along the same straight line. Now, the stage is set to define the term as :

The coefficient of restitution (e) of two bodies for “head-on” collision is a constant and is equal to the ratio of velocity of separation and velocity of approach.

e = velocity of separation velocity of approach e = velocity of separation velocity of approach

The very fact that “head – on” collision and motion of bodies take place along a straight line has important bearing on the evaluation of this constant. If we look closely at the above ratio, then it is very easy to understand following aspects of this quantity :

In order to understand the terms involved in the ratio, we consider an example. One block designated as “1” approaches another block designated as “2” on a smooth surface. For collision to occur, it is clear that v 1 i v 1 i > v 2 i v 2 i . And the velocity of approach, is :

Figure 2: "Head - on" Collision

Collision

v approach = v 1 i − v 2 i v approach = v 1 i − v 2 i

After collision, the bodies move with different velocities and move away from each other. Therefore, the velocity of separation is given by :

v separation = v 2 f − v 1 f v separation = v 2 f − v 1 f

The coefficient of restitution is :

e = v 2 f − v 1 f v 1 i − v 2 i e = v 2 f − v 1 f v 1 i − v 2 i

In order to avoid confusion on account of subscripts, it is helpful to follow certain conventions as given here :

If we stick with this scheme, then we can write the expression of coefficient of restitution as :

⇒ e = v 21 f v 12 i ⇒ e = v 21 f v 12 i

The analysis of collision between two bodies involves many unknowns – two masses and four velocities. On the other, we have only one equation of conservation of linear momentum in the case of inelastic collision. Since kinetic energy is not conserved in inelastic collision, we can not use equation of kinetic energy as in the case of elastic collision.

The measurement of coefficient of restitution is helpful to obtain another equation that can be used in conjunction with conservation of linear momentum.

Continuing with the earlier example, involving two blocks, the conservation of linear momentum is written as :

Figure 3: Head-on collision

Collision

m 1 v 1 i + m 2 v 2 i = m 1 v 1 f + m 2 v 2 f m 1 v 1 i + m 2 v 2 i = m 1 v 1 f + m 2 v 2 f

The coefficient of restitution equation is :

e = v 2 f − v 1 f v 1 i − v 2 i e = v 2 f − v 1 f v 1 i − v 2 i

⇒ v 2 f − v 1 f = e v 1 i − v 2 i ⇒ v 2 f − v 1 f = e v 1 i − v 2 i

Our objective here is to write final velocities in terms of initial velocities for two blocks of given masses. We can eliminate “ v 2f v 2f ” from equation of conservation of linear momentum, using its expression from equation of restitution,

⇒ m 1 v i 1 + m 2 v 2 i = m 1 v 1 f + m 2 { e v 1 i − v 2 i + v 1 f } ⇒ m 1 v i 1 + m 2 v 2 i = m 1 v 1 f + m 2 { e v 1 i − v 2 i + v 1 f }

⇒ v 1 f m 1 + m 2 = m 1 v 1 i + m 2 v 2 i - e m 2 v 1 i + e m 2 v 2 i ⇒ v 1 f m 1 + m 2 = m 1 v 1 i + m 2 v 2 i - e m 2 v 1 i + e m 2 v 2 i

⇒ v 1 f m 1 + m 2 = v 1 i m 1 − e m 2 + v 2 i m 2 + e m 2 ⇒ v 1 f m 1 + m 2 = v 1 i m 1 − e m 2 + v 2 i m 2 + e m 2

⇒ v 1 f = m 1 − e m 2 m 1 + m 2 v 1 i + m 2 + e m 2 m 1 + m 2 v 2 i ⇒ v 1 f = m 1 − e m 2 m 1 + m 2 v 1 i + m 2 + e m 2 m 1 + m 2 v 2 i

Similarly, we can eliminate “ v 1f v 1f ” from equation of conservation of linear momentum, using equation of restitution,

⇒ m 1 v 1 i + m 2 v 2 i = m 1 { v 2 f − e v 1 i − v 2 i } + m 2 v 2 f ⇒ m 1 v 1 i + m 2 v 2 i = m 1 { v 2 f − e v 1 i − v 2 i } + m 2 v 2 f

⇒ v 2 f m 1 + m 2 = m 1 v 1 i + m 2 v 2 i + m 1 e v 1 i − e m 1 v 2 i ⇒ v 2 f m 1 + m 2 = m 1 v 1 i + m 2 v 2 i + m 1 e v 1 i − e m 1 v 2 i

⇒ v 2 f m 1 + m 2 = v 2 i m 2 − e m 1 + v i 1 m 1 + e m 1 ⇒ v 2 f m 1 + m 2 = v 2 i m 2 − e m 1 + v i 1 m 1 + e m 1

⇒ v 2 f = m 2 − e m 1 m 1 + m 2 v 2 i + m 1 + e m 1 m 1 + m 2 v 1 i ⇒ v 2 f = m 2 − e m 1 m 1 + m 2 v 2 i + m 1 + e m 1 m 1 + m 2 v 1 i

The real time collision, however, is not “head – on” collision most of the times. In this section, we shall apply the concept of coefficient of restitution to oblique collision.

The normal force during collision is along a straight line even in an oblique collision. However, motions of the colliding bodies are not along the direction in which collision normal force acts. In this case, we analyze motion with coordinate system having axes parallel and perpendicular to the direction of normal force. This set up facilitates application of the concept of the coefficient of restitution in the direction of normal force.

Figure 4: Oblique collision

Collision

We consider an oblique collision in which normal force acts in “y” direction as shown in the figure above. We can analyze collision using following aspects of oblique collision,

1: We use equation of restitution in the direction of normal force . Recall that coefficient of restitution is defined for "head - on" collision. This implies that it is defined for motion in the direction of normal force acting during collision.

2: The component of normal force during collision is zero in the perpendicular direction. This means that the components of velocity of each body are unchanged before and after collision in the direction perpendicular to normal force.

3: Since net external force on the system of two particles is zero (collision normal forces are internal forces for the system of particles), we can use equation of conservation of linear momentum in component form in directions parallel and perpendicular to normal force.

m 1 v 1 i x + m 2 v 2 i x = m 1 v 1 f x + m 2 v 2 f x m 1 v 1 i x + m 2 v 2 i x = m 1 v 1 f x + m 2 v 2 f x

m 1 v 1 i y + m 2 v 2 i y = m 1 v 1 f y + m 2 v 2 f y m 1 v 1 i y + m 2 v 2 i y = m 1 v 1 f y + m 2 v 2 f y

Example

Problem : A ball of mass “m” hits a floor with a speed “v” making an angle “θ”. If “e” be the coefficient of restitution, then find the angle of reflection of the ball.

Figure 5: A ball of mass “m” hits a floor with a speed “v” making an angle “θ”.

Collision

Solution : Force of collision acts normal to the floor at the point, where ball strikes it. There is no component of collision force parallel to the surface. As such, component of velocity parallel to the surface before and after remains same.

Figure 6: A ball of mass “m” hits a floor with a speed “v” making an angle “θ”.

Collision

Let “θ’” be the angle of reflection. Then, components of velocity before and after are equal,

v sin θ = v' sin θ' v sin θ = v' sin θ'

We, now, consider motion in the normal direction. Here, the coefficient of restitution is given by :

e = v 21 f v 12 i e = v 21 f v 12 i

e = v 2 f − v 1 f v 1 i − v 2 i = 0 − v 1 f y v 1 i y - 0 e = v 2 f − v 1 f v 1 i − v 2 i = 0 − v 1 f y v 1 i y - 0

⇒ v 1 f y = − e v 1 i y = − e v cos θ ⇒ v 1 f y = − e v 1 i y = − e v cos θ

and

tan θ ′ = v 1 f x v 1 f y = v sin θ e v cos θ = tan θ e tan θ ′ = v 1 f x v 1 f y = v sin θ e v cos θ = tan θ e

In this section, we consider a case of tennis ball falling from a height and undergoing repetitive collisions with a horizontal hard surface. The ball looses some kinetic energy (hence speed) after each successive strike due to inelastic collision between ball and horizontal surface. As a consequence, the height attained by the ball diminishes as the number of collision increases.

Our objective here is to find an expression for the speed of the ball and the height attained subsequent to nth collisions.

Let coefficient of friction between ball and the surface be “e”. Also, let the initial height of the ball be “ h 0 h 0 ” and speed before the first strike be “ v 0 v 0 ”. For the analysis of motion, we consider upward vertical direction as positive. Now, applying equation of motion in vertical downward direction, we have :

v 0 2 = 0 + 2 − g − h 0 v 0 2 = 0 + 2 − g − h 0

⇒ v 0 = 2 g h 0 ⇒ v 0 = 2 g h 0

and

⇒ h 0 = v 0 2 2 g ⇒ h 0 = v 0 2 2 g

Figure 7: The ball looses speed and height with successive collision.

Repetitive collisions

Ist strike

Let “ v 1 v 1 ” be the velocity of the ball after first strike. The velocity after 1st collision is obtained from consideration of equation of restitution,

v 2 f − v 1 f = v 1 i − v 2 i v 2 f − v 1 f = v 1 i − v 2 i

Putting values, we have :

⇒ 0 − v 1 = e − v 0 ⇒ 0 − v 1 = e − v 0

⇒ v 1 = e v 0 ⇒ v 1 = e v 0

Height attained with this velocity is given as :

h 1 = v 1 2 2 g = e 2 v 0 2 2 g = e 2 h 0 h 1 = v 1 2 2 g = e 2 v 0 2 2 g = e 2 h 0

2nd strike

Let “ v 2 v 2 ” be the velocity of the ball after second strike. The velocity after 2nd collision is obtained from consideration of equation of restitution,

v 2 f − v 1 f = v 1 i − v 2 i v 2 f − v 1 f = v 1 i − v 2 i

Putting values, we have :

0 − v 2 = e X v 1 0 − v 2 = e X v 1

⇒ v 2 = e v 1 = e X e v 0 = e 2 v 0 ⇒ v 2 = e v 1 = e X e v 0 = e 2 v 0

Height attained with this velocity is given as :

⇒ h 2 = v 2 2 2 g = e 4 v 0 2 2 g = e 4 h 0 ⇒ h 2 = v 2 2 2 g = e 4 v 0 2 2 g = e 4 h 0

nth strike

Continuing in the same fashion, we can get expressions for nth strike. Let “ v n v n ” be the velocity of the ball after nth strike. Then,

⇒ v n = e n v 0 ⇒ v n = e n v 0

Let “ h n h n ” be the height attained. Then,

h n = e 2 n h 0 h n = e 2 n h 0