We discuss problems, which highlight certain aspects of the study leading to inelastic collision. The questions are categorized in terms of the characterizing features of the subject matter :

Problem 4 : Three identical balls designated “1”, “2” and “3” are placed 16 meters apart on a smooth horizontal surface. The ball “1” is given an initial velocity 10 m/s towards “2”. The collision between “1” and “2” is inelastic with coefficient of restitution 0.6, whereas collision between “2” and “3” is perfectly inelastic. Find the time elapsed between two consecutive collisions between “1” and “2”.

Figure 5: Three identical balls are placed on a smooth horizontal surface.

Successive collisions

Solution : To answer this question, we need to understand the collision sequence. The ball “1” moving with velocity 10 m/s strikes second ball in - elastically. As a consequence, depending on coefficient of restitution two balls move with different velocities. This collision constitutes the first collision.

Figure 6: First collision is inelastic.

First collision

Let the mass of each particle is “m”. Let velocity of first ball before collision be “v”, velocity of first ball after collision be “ v 1 v 1 ” and velocity of second ball after collision be “ v 2 v 2 ”. Applying conservation of linear momentum,

m v = m v 1 + m v 2 m v = m v 1 + m v 2

⇒ v = v 1 + v 2 ⇒ v = v 1 + v 2

⇒ v 1 + v 2 = 10 ⇒ v 1 + v 2 = 10

Applying equation of restitution,

⇒ v 2 − v 1 = e v = 0.6 X 10 = 6 ⇒ v 2 − v 1 = e v = 0.6 X 10 = 6

Adding two resulting equations,

⇒ 2 v 2 = 16 ⇒ 2 v 2 = 16

⇒ v 2 = 8 m / s ⇒ v 2 = 8 m / s

Substituting this value in the equation of conservation of linear momentum,

⇒ v 1 = 10 − v 2 = 10 − 8 = 2 m / s ⇒ v 1 = 10 − v 2 = 10 − 8 = 2 m / s

Clearly, second ball covers the horizontal distance of 16 m, subsequent to collision at 8 m/s and strikes the third ball.

The second ball strikes third stationary identical ball elastically. As a result, the balls exchange their velocities. It means second ball becomes stationary and third ball moves off towards right with velocity at 8 m/s. First ball, however, keeps moving towards right at 2 m/s and covers the distance of 16 m to strike the second ball again, which has come to a standstill. The time taken by the first ball to hit the second ball again is :

Figure 7: Second collision is perfectly elastic.

Second collision

t = 16 2 = 8 s t = 16 2 = 8 s

This is the time interval between required two consecutive collisions.

Problem 5 : A spherical ball of mass “m” moving at a speed “v” makes a head on collision with an identical ball at rest. The kinetic energy of the balls after collision is 3/4 th of the value before collision. Find coefficient of restitution.

Solution : The spherical balls have equal mass and initial velocity of “target” ball is zero.

Let the mass of each particle is “m”. Let velocity of first ball before collision be “v”, velocity of first ball after collision be “ v 1 v 1 ” and velocity of second ball after collision be “ v 2 v 2 ”.

Figure 8: Velocities before and after collision

Inelastic collision

For identical mass and stationary target, we have :

v 1 = 1 − e 2 v v 1 = 1 − e 2 v

and

v 2 = 1 + e 2 v v 2 = 1 + e 2 v

According to question,

K f = 3 4 K i K f = 3 4 K i

⇒ 1 2 m v 1 2 + v 2 2 = 3 4 X 1 2 m v 2 ⇒ 1 2 m v 1 2 + v 2 2 = 3 4 X 1 2 m v 2

⇒ v 1 2 + v 2 2 = 3 4 m v 2 ⇒ v 1 2 + v 2 2 = 3 4 m v 2

Substituting for final velocities, we have :

⇒ 1 - e 2 4 v 2 + 1 + e 2 4 v 2 = 3 4 m v 2 ⇒ 1 - e 2 4 v 2 + 1 + e 2 4 v 2 = 3 4 m v 2

⇒ 1 − e 2 + 1 + e 2 = 3 ⇒ 1 − e 2 + 1 + e 2 = 3

⇒ 1 + e 2 − 2 e + 1 + e 2 + 2 e = 3 ⇒ 1 + e 2 − 2 e + 1 + e 2 + 2 e = 3

⇒ e 2 = 1 2 ⇒ e 2 = 1 2

⇒ e = 1 2 ⇒ e = 1 2

Problem 6 : The ram of a pile driver has a mass m and is released from rest at a height h above the top of the pile which has the mass 3m. To what height does the ram rebound if the coefficient of restitution between the ram and the pile is e and e>(1/3)? If the pile meets a constant resistance F from the material in which it is lodged, calculate how far into the material it penetrates before coming to rest.

Figure 9: The ram of a pile driver is released from rest at a height "h".

Inelastic collision

Solution : We need to know the velocities of ram and pile immediately after the inelastic collision. Let the velocity of the ram just before strike be "v", velocity of the ram after the strike be " v 1 v 1 " velocity of the pile after strike be " v 2 v 2 ".

Let the downward vertical y-direction be positive. We can determine velocity of the ram just before it hits the pile. Applying equation of motion for constant acceleration,

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

In order to determine velocities immediately after collision, we shall apply conservation of linear momentum in vertical direction (downward direction as positive)

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

Putting values, we have :

⇒ m v + 0 x 3 m = m v 1 + 3 m v 2 ⇒ m v + 0 x 3 m = m v 1 + 3 m v 2

⇒ v = v 1 + 3 v 2 ⇒ v = v 1 + 3 v 2

Now, applying equation of restitution, we have :

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

⇒ e v = v 2 − v 1 ⇒ e v = v 2 − v 1

Thus, we have two unknowns “ v 1 v 1 ” and “ v 2 v 2 ”. Also we have two equations. Adding two equations,

⇒ 1 + e v = 4 v 2 ⇒ 1 + e v = 4 v 2

The velocity of pile, therefore, is :

⇒ v 2 = 1 + e v 4 ⇒ v 2 = 1 + e v 4

Putting this value in the first equation, we get the velocity of the ram after collision,

⇒ v 1 = v − 3 v 2 = 1 − 3 e v 4 ⇒ v 1 = v − 3 v 2 = 1 − 3 e v 4

Putting expression of “v”, as obtained earlier, in above two expressions, we have :

⇒ v 2 = 1 + e v 4 ⇒ v 2 = 1 + e v 4

⇒ v 2 = 1 + e 2 g h 4 ⇒ v 2 = 1 + e 2 g h 4

and

⇒ v 1 = 1 − 3 e 2 g h 4 ⇒ v 1 = 1 − 3 e 2 g h 4

Note that 1 - 3e < 0 as e > 1/3. It means that the velocity of ram is opposite to the assumed positive reference direction. This further means that it rebounds after the strike in upward direction.

Figure 10: Velocities of colliding bodies.

Inelastic collision

⇒ v 1 = 3 e − 1 2 g h 4 ⇒ v 1 = 3 e − 1 2 g h 4

Let h' be the height attained by the ram after the strike. For the upward motion of ram,

0 = v 1 2 − 2 g h ′ 0 = v 1 2 − 2 g h ′

⇒ h ' = v 1 2 2 g = 3 e - 1 2 X g h 2 g X 16 = 3 e - 1 2 X h 16 ⇒ h ' = v 1 2 2 g = 3 e - 1 2 X g h 2 g X 16 = 3 e - 1 2 X h 16

Now, we consider the motion of pile into the material. There are two forces acting on the pile (i) constant resistance “F” and (ii) force due to gravity “3mg". As given, the net force acts up to decelerate the pile. The deceleration is :

a = F − 3 m g 3 m a = F − 3 m g 3 m

Let "y" be the vertical distance moved by the pile. Applying equation of motion,

Figure 11: The pile is decelerated due to resistance offered by the material.

Motion of pile

0 = v 2 2 − 2 X F − 3 m g 3 m X y 0 = v 2 2 − 2 X F − 3 m g 3 m X y

⇒ y = v 2 2 X 3 m 2 F − 3 m g ⇒ y = v 2 2 X 3 m 2 F − 3 m g

⇒ y = 1 + e 2 X 6 m g h 2 F − 3 m g X 16 = 1 + e 2 X 3 m g h 16 F − 3 m g ⇒ y = 1 + e 2 X 6 m g h 2 F − 3 m g X 16 = 1 + e 2 X 3 m g h 16 F − 3 m g