# Impact of Elastic Bodies

If the balls of different materials are allowed to fall on a marble floor, they will rebound to different heights due to their elasticity. Elasticity is the property of bodies by virtue of which they rebound after impact. The body, which rebounds to a greater height, is known as more elastic, than a body which rebounds to a lesser height. Inelastic body is one which does not rebound at all.

Whenever two elastic bodies collide with each other, they tend to compress each other. Immediately after this, the two bodies attempt to regain its original shape, due to their elasticity. This process, of regaining the original shape, is called restitution. The important terms, used in collision, are defined as:

1.   Time of compression. The time taken by two bodies in compression, after the instant of collision, is known as time of compression.

2.   Time of restitution. The time taken by two bodies to regain the original shape, after compression, is known as time of restitution.

3.   Time of collision. The sum of time of compression and time of restitution is known as time of collision or period of collision or period of impact.

4.   Law of conservation of momentum. It states that if the resultant of the external forces acting on a system is zero, the momentum of the system remains constant. This means that the total momentum of the system before collision is equal to the total momentum of the system after collision. The system may consist of one body or two bodies or more.

5.13.1. Types of Impacts. Impact means the collision of two bodies which occurs in a
very small interval of time and during which the two bodies exert very large force on each other. The important types of impacts are:

(i) Direct impact

(ii) Indirect (oblique) impact.

5.13.2. Direct Impact of Two Bodies. The two bodies A and B are moving in a horizontal line before collision with velocities ul and u2 in the same direction i.e., along x-axis as shown in Fig. 5.71 (a). If u1 > u2, the body A will strike the body B and collision will take place. Let C is the point of collision of the two bodies as shown in Fig 5.71 (b). The point C is also known as the point of contact. The line joining the centres of these two bodies and passing through the point of contact is known as line of impact. Hence here the line O1-C-O2 is called line of impact.

The collision between two bodies is known as direct impact if the two bodies before
impact, are moving along the line of impact.

5.13.4. Co-efficient of Restitution. It is defined as the ratio of velocity of separation
(of the two moving bodies which collides with each other) to their velocity of approach. It is also defined as the ratio of the relative velocities of colliding bodies after impact to their relative velocity before impact. It is denoted by symbol ‘e’. The relative velocities are measured along the line of impact, which is the common normal to the colliding surfaces.

According to Newton’s Law of collision of elastic bodies, “the velocity of separation, of the two moving bodies which collide with each other, bears a constant ratio to their velocity of approach”. And the constant of proportionality is known as co-efficient of restitution.

Fig. 5.73

Fig. 5.73 shows two bodies A and B.

Let u1 = Velocity of A before collision along x-axis

v1 = Velocity of A after collision along x-axis

u2 = Velocity of B before collision along x-axis

v2 = Velocity of B after collision along x-axis.

The body A will collide with body B if velocity of A is more than that of B. Hence velocity of approach (or relative velocity of colliding bodies before impact)

= Initial velocity of A – Initial velocity of B = (u1 – u2)

After collision, the separation of the two bodies will take place if final velocity of B is
more than that of A.

Hence velocity-or separation (or relative velocity of colliding bodies after impact)

= Final velocity of B – Final velocity of A = (v2 – v1)

Now according to Newton’s Law of collision of elastic bodies,

For most of bodies, the value of e lies between 0 and 1. For perfectly elastic bodies e = 1 and for perfectly plastic bodies e = 0.

To determine the velocities after the impact (i.e., velocity v1 and v2) the equation (5.56) is not sufficient to determine the two unknowns. One more equation is needed. This equation is law of conservation of momentum i.e., total initial momentum is equal to total fined momentum.

Note 1. The velocity of approach or separation of the two bodies, which are moving in the same direction before or after the impact, is the difference of their velocities.

2. For the two bodies moving in the opposite direction, the velocity of approach or separation is the sum of their velocities.

3. Newton’s law of collision of elastic bodies, also holds good for indirect impact, i.e.,

4. If the velocity in a direction is + ve then the velocity in the opposite direction will be -ve.

Problem 5.106. Ball A of mass 1 kg moving with a velocity of 2 m/s, strikes directly on (I ball B of mass 2 kg at rest. The ball A, after striking, comes to rest. Find the velocity of ball B after striking and co-efficient of restitution.

Sol. Given:

Mass of ball A,m1 = 1 kg

Initial velocity of ball A, u1 = 2 m/s

Mass of ball B, m2 = 2 kg

Initial velocity of ball B, u2 = 0

Final velocity of ball A, v1 = 0

This is a case of direct impact.

Let v2 = Velocity of ball B after impact, and

e = Co-efficient of restitution.

Total initial momentum =m1u1 + m2u2 = 1 × 2 + 2 × 0= 2 kg m/s.

Total final momentum = m1u1 + m2v2 = 1 × 0 + 2 × v2 = 2v2 kg m/s.

According to the law of conservation of momentum,

Total initial momentum = Total final momentum

2 = 2 × v2

Problem 5.107. A body of mass 50 kg, moving with a velocity of 6 m/s collides directly with a stationary body of mass 30 kg. If the two bodies become coupled so that they move on together after the impact, what is their common velocity.

Sol. Given:

Mass of first body, m1 = 50 kg

Initial velocity of first body,         u1 = 6 m/s

Mass of second body, m2 = 30 kg

Initial velocity of second body, u2 = 0.

Total mass of two bodies = (m1 + m2) = (50 + 30) = 80 kg.

Let V = Common velocity of the two bodies after impact.

Total momentum before impact   = m1u1 + m2u2 = 50 × 6 + 30 × 0 = 300 kg m/s

Total momentum after impact = (m1 + m2) × V

= (50 + 30 ) × V = 80 V kg m/s

But total momentum before impact = Total momentum after impact

or 300 = 80 × V

Problem 5.108. A bullet of mass 50 gm is fired into a freely suspended target to mass 5 kg. On impact, the target moves with a velocity of 7 m/s along with the bullet in the direction of firing. Find the velocity of bullet.

Sol. Given:

Mass of target, m2 = 5 kg

Initial velocity of bullet = u1

Initial velocity of target, u2 = 0

Total mass of bullet and target = 5 + 0.05 = 5.05 kg

Final velocity of bullet and target = 7 m/s

Total initial momentum (i.e., momentum before impact)

= m1 × u1 + m2 × u2 = 0.05 × u1 + 5 × 0

= 0.05 u1 kg m/s

Total final momentum i.e., momentum after impact)

= Total mass × Common velocity = (5.05) × 7 kg m/s

Equating the initial momentum to final momentum, we get

Problem 5.109. A ball of mass 20 kg moving with a velocity of 5 m/s strikes directly another ball of mass 10 kg moving in the opposite direction with a velocity of 10 m/s. If the coefficient of restitution is equal to 516, then determine the velocity of each ball after impact.

Sol. Given:

Mass of first ball, m1 = 20 kg

Initial velocity of first ball, u2 = 5 m/s

Mass of second ball, m2 = 10 kg

Initial velocity of second ball, u2 = – 10 m/s

(Negative sign is due, to opposite direction)

Let v1 = Velocity of first ball after impact

v2 = Velocity of second ball after impact

Total momentum before impact

=m1u1 + m2u2 = 20 × 5 + 10 × (- 10) = 100-100 =0

Total momentum after impact

= m1v1 + m2v2 = 20 v1 + 10 v2

Equating the total momentum after impact and before impact, we get

(Minus sign indicates that the velocity of first ball after impact will be in opposite direction).

Problem 5.110. Prove that the two perfect elastic bodies of equal masses exchange velocities in the case of direct central impact.

Sol. Given:

Two perfect elastic bodies of equal mass

Let m1 = Mass of first body

m2 = Mass of second body

u1 = Velocity of first body before impact

u2 = Velocity of second body before impact

v1 = Velocity of first body after impact

v2 = Velocity of second body after impact

Given: m1 =m2

e = 1                                                                 (for perfect elastic body)

u1 – u2 = v2 – v1

According to the law of conservation of momentum,

Total initial momentum = Total final momentum

or m1u1 + m2u2 = m1v1 + m2v2

But m1 = m2 (Given)

u1 + u2 = v1 + v2                                        …(ii)

Adding equations (i) and (ii), we get

2u1 = 2v2                                                    u1 = v2                                           … (iii)

Substituting the value of u1 in equation (ii), we get

v2+u2=vl+v2                 or               u2=v1                                               …(iv)

From equation (iii), it is clear that the velocity of first body before impact is equal to the velocity of the second body after impact.

Equation (iv) shows that the velocity of second body before impact is equal to the velocity of first body after impact.

Hence the two perfect elastic bodies of equal masses exchange velocities after impact.

Problem 5.111. Three perfectly elastic balls A, Band C of masses 2 kg, 6 kg and 12 kg
are moving in the same direction with velocities 12 m/s, 4 m/s and 2 m/s respectively. If the ball
A strikes with the ball B, which in turns, strikes with the ball C, prove that the balls A and B will he brought to rest by the impact.

Sol. Given:

For perfectly elastic balls, e = 1

Mass of first ball, m1 = 2 kg

Mass of second ball, m2 = 6 kg

Mass of third ball, m3 = 12 kg

Initial velocity of first ball, u1 = 12 m/s

Initial velocity of second ball, u2 = 4 m/s

Initial velocity of third ball, u3 = 2 m/s

(a) First of all consider the impact of first and second ball

Let v1 = Final velocity of first ball after impact

v2 = Final velocity of second ball after impact

Total momentum of 1st and 2nd ball before impact

= Total momentum of 1st and 2nd ball after impact

m1u1 + m2u2 = m1v1 + m2v2

or 2 × 12 + 6 × 4 = 2 × v1 + 6 × v2

Substituting this value of v2 in equation (ii),

8 – v1 = 8       or       v1 = 0.

The velocity of ball A after impact is zero. Hence the ball A will be brought to rest by the impact of A and B. Ans.

The velocity of ball B after impact with A will be (v2 = 8 m/s),

(b) Now consider the impact of second and third ball. The second ball is now moving with a velocity of 8 m/s and strikes the third ball which is moving with ‘a velocity of 2 m/s.

Now Initial velocity of second ball, u2* = 8 m/s

Initial velocity of third ball, u3 = 2 m/s

Let v2* = New final velocity of second ball after striking the third ball.

v3= Final velocity of third ball

Law of conservation of momentum gives:

Total momentum of 2nd and 3rd ball before impact

= Total momentum of 2nd and 3rd ball after impact

or m2 × u2* + m3u3 = m2 × v2* + m3 × v3

or 6 × 8 + 12 × 2 = 6 × v2* + 12v3

or 48 + 24 = 6v2* + 12v3             or         72 = 6v2 * + 12v3

or 6v2* + 12v3 = 72            or            v2* + 2v3 = 12                    … (iii)

Using equation (5,56), we get

(v3 – v2*) = e (u2* – u3) = 1(8 – 2) = 6

v3 – v2*=6                                                 … (iv)

Adding equations (iii) and (iv), we get

3v3= 12 + 6 = 18

Substituting the value of v3 in equation (iv),

6-v2*=6         or            v2*=6-6=0.

Hence the velocity of second ball, after impact with third ball, is zero. Hence the second ball will also be brought to rest after impact with third ball C. Ans.

Problems on Indirect Impact

Problem. 5.112. A ball of mass 1 kg, moving with a velocity of 6 m/s, strikes on a ball
of mass 2 kg moving with a velocity of 2 m/s. At the instant of impact, the velocities of the two balls are parallel and inclined at 300 to the line joining their centres. If co-efficient of restitution is 1, find:

(i)                           the velocity and direction in which the 1 kg ball will move after impact.

(ii)                        the velocity and direction in which the 2 kg ball will move after impact.

Sol. Given:

Mass of first ball, m1 = 1 kg

Mass of second ball, m2 = 2 kg

Initial velocity of first ball, u1 = 6 m/s

Initial velocity of second ball, u2 = 2 m/s

Angle made by first ball with line of impact, θ1 = 300

Angle made by second ball with line of impact, θ2 = 30°

Co-efficient of restitution,

v1 =Velocity of first ball after impact

u2 = Velocity of second ball after impact

ɸ1 = Angle made by first ball after impact with line of impact

ɸ2 = Angle made by second ball after impact with line of impact.

Velocity of each ball normal to the line of impact remains unchanged. This means that the components of velocities of each ball normal to line of impact before and after impact is same.

For ball A,

Initial normal component = Final normal component

6 sin 30° = v1 sin ɸ1                 or         v1 sin ɸ1 = 6 sin 30° = 3                   … (i)

For ball B,

Final normal component = Initial normal component

v2 sin ɸ2 = 2 sin 30° = 1                                                          … (ii)

According to law of conservation of momentum,

Total initial momentum along the line of impact

= Total final momentum along the line of impact.

m1 × u1 cos θ1 + m2 × u2 cos θ2 = m1 × v1 cos ɸ1 + m2 × v2 cos ɸ2.

1 × 6 cos 30° + 2 × 2 cos 30° = 1 × v1 cos ɸ1 + 2 × v2 cos ɸ1

6 × 0.866 + 4 × 0.866 = v1 cos ɸ2 + 2v2 cos ɸ2

8.66 = v1 cos ɸ1 + 2v2 cos ɸ2

v1 cos ɸ1 + 2v2 cos ɸ2 = 8.66                                                … (iii)

The co-efficient of restitution for this case is given by the equation

(v2 cos ɸ2 – v1 cos ɸ1) = e (u1 cos θ1 -u2 cos θ2)

Substituting the value of v2 cos ɸ2 in equation (iii),

v1 cos ɸ1 × 2 × 3.464 = 8.66

v1 cos ɸ1 = 8.66 – 2  3.464 = 1.732                                                … (vi)

Dividing equation (i) by equation (vi), we get

Problem 5.113. A ball of mass 2 kg, moving with a velocity of 20 m/s, strikes on a ball of mass 2 kg moving with a velocity of30 m/s. At the instant of impact, the velocities of the balls are inclined at (Ill angle of 30° and 60° to the line joining their centres, as shown in Fig. 5.75. If co-efficient of restitution is 0.9, then find:

(i)                           the magnitude and direction of first ball after impact

(ii)                        the magnitude and direction of second hall after impact.

Sol. Given:

Mass of first ball, m1 = 2 kg

Mass of second ball, m2 = 2 kg

Initial velocity of first ball, u1 = 20 m/s

Initial velocity of second ball, u2 = 30 m/s

Angle made by first ball with line of impact,

θ1= 30°

Angle made by second ball with line of impact,

θ2 = 60°

Co-efficient of restitution,

e = 0.9

Fig. 5.75

Let v1 = Velocity of first ball after impact

v2 = Velocity of second ball after impact

ɸ1 = Angle made by first ball with line of impact after impact

ɸ2 = Angle made by second ball after impact.

The components of velocity of each ball perpendicular to the line of impact before and alter impact is same.

For ball A,

Final normal component = Initial normal component

v1 sin ɸ1 = 20 sin 30° = 10                                                  … (i)

For ball B.

Final normal component = Initial normal component

v2 sin ɸ2 = 30 sin 60° = 25.98                                                 … (ii)

Now according to the law of conservation of momentum

Total momentum along the line of impact before collision

= Total momentum along the line of impact after collision

m1 × u1 cos θ1 + m2 × u2 cos θ2 = m1 × v1 cos ɸ1 + m2 × v2 cos ɸ2.

u1 cos θ1 = 20 × cos 30° = 17.32

u2 cos θ 2 = – 30 × cos 60° = – 15

(Minus sign is due to opposite direction)

Substituting these values in the above equation

2 × 17.32 + 2 × (- 15) = 2 × v1 cos ɸ1 + 2 × v2 cos ɸ2

17.32 – 15 = v1 cos ɸ1 + v2 cos ɸ2

2.32 = v1 cos ɸ1 + v2 cos ɸ2

v1 cos ɸ1 + v2 cosɸ2 = 2.32                                          … (iii)

The co-efficient of restitution for the indirect impact between two bodies is given by

v2 cos ɸ2 – v1 cos ɸ1 = e [u cos θ1 – u2 cos θ2] = 0.9 [17.32 – (-15)] .

(u cos θ1 = 17.32 and u2 cos θ2 =- 15)

= 0.9 × 32.32

v2 cos ɸ2 – v1 cos ɸ1 =29.08                                    … (iv)

Adding equations (iii) and (iv), we get

2v2 cos ɸ2 = 2.32 + 29.03 = 31.40

Substituting the above value in equation (iii), we get

v1 cos ɸ1 + 15.7 = 2.32

v1 cos ɸ1 = 2.32 – 15.7 = – 13.38

Minus sign shows that final velocity of first will be in the opposite direction. Numerically,

v1 cos ɸ1 = 13.38

v1 cos ɸ1 = 13.38                                                       … (vi)

Dividing equation (i) by equation (vi), we get

5.13.5. Loss of Kinetic Energy During Impact. When the two bodies collide with
each other, the loss of kinetic energy takes place due to impact. This loss of energy- may be obtained by finding out the kinetic energy of the two bodies before and after the impact. The loss of kinetic energy during impact is equal to the difference of the two kinetic energies.

Consider two bodies A and B having a direct impact.

Let m1 = Mass of the first body,

u1 = Velocity of first body before impact

v1 = Velocity of first body after impact

m2, u2 and v2 = corresponding values of mass, initial velocity and final velocity of second body

EL = Loss of energy

e = Co-efficient of restitution.

Kinetic energy of first body before impact

Problem 5.114. A vehicle of mass 600 kg and moving with a velocity of 12 m/s strikes another vehicle of mass 400 kg, moving at 9 m/s in the same direction. Both the vehicles get coupled together due to impact. Find the common velocity with which the two vehicles will move. Also find the loss of kinetic energy due to impact.

Sol. Given:

Mass of first vehicle, m1 = 600 kg

Initial velocity of first vehicle, u1 = 12 m/s

Mass of second vehicle, m2 = 400 kg

Initial velocity of second vehicle, u2 = 9 m/s

When the two vehicles get coupled, then total mass of the two vehicles,

M = m1 + m2 = 600 + 400 = 1000 kg

(i) Let V = Common velocity of the two vehicles after impact

Total momentum before impact

= m1u1 + m2u2 = 600 × 12 + 400 × 9

= 7200 + 3600

= 10800 kg m/s                                               … (i)

Total momentum after impact

= (m1 + m2) × V

= (600 + 400) × V

= 1000 V kg m/s                                                                     … (ii)

According to law of conservation of momentum,

Total momentum before impact = Total momentum after impact

10800 = 1000 × V

Problem 5.115. A bullet of mass 100 gm is fixed into a freely suspended target of mass 10 kg. Due to impact, the bullet gets imbedded in the target and the target with bullet moves with a velocity of 7 m/s. Find the velocity of the bullet and the loss of kinetic energy.

Sol. Given:

Mass of bullet,

Mass of target, m2 = 10 kg

Common velocity of bullet and target = 7 m/s

Initial velocity of target, u2 = 0

Let u1 = Initial velocity of bullet

EL = Loss of kinetic energy due to impact

Total momentum before impact

= m1u1 + m2u2

= 0.1 × u1 + 10 × 0 = 0.1 u1                                                                       … (i)

Total momentum after impact

= (m1 + m2) × Common velocity

= (0.1 + 10) × 7 = 70.7 kg m/s                                                            … (ii)

Equating the total momentum before impact to the total momentum after impact, we get

0.1 u1 =70.7

The above equation gives the loss of K.E. due to direct impact in terms of masses, initial velocity and co-efficient of restitution.

Problem 5.118. The co-efficient of restitution between two spheres of masses 1 kg and 5 kg is 0.75. The sphere of mass 1 kg, moving with a velocity of 3 m/s, strikes the sphere of mass 5 kg moving in the same direction with a velocity of 60 cm/s. Find the velocities of the two spheres after impact and also loss of kinetic energy during impact.

Sol. Given:

Co-efficient of restitution = 0.75

Mass of 1st sphere, m1 = 1 kg

Mass of 2nd sphere, m2 = 5 kg

Initial velocity of 1st sphere, u1 = 3 m/s

Initial velocity of 2nd sphere, u2 = 60 cm/s = 0.6 m/s

Let v1 = Final velocity of 1st sphere

v2 = Final-velocity of 2nd sphere

Law of conservation of momentum gives:

Total momentum before impact

= Total momentum after impact

m1u1 + m2u2 = m1v1 + m2v2

1 × 3 + 5 × 0.6 = 1 × v1 + 5v2

3 + 3 = v1 + 5v2          or           v1 + 5v2 = 6                     … (i)

Now using equation (5.56), we get

5.13.6. Impact of a Body on a Fixed Plane. There are two types of impact of a body on a fixed plane. These are:

(i) Direct impact of a body on a fixed plane, and

(ii) Indirect impact of a body on a fixed plane.

5.13.7. Direct Impact of a Body on a Fixed Plane. The fixed plane is at rest before
impact and after impact. The mass of the fixed plan, is very large. Consider a body having a direct impact on a fixed plane.

Let u = Initial velocity of the body,

v = Final velocity of the body

e = Co-efficient of restitution.

The velocity of approach between body and fixed plane

= Initial velocity of body – velocity of fixed plane = u – 0 = u.

The velocity of separation

= Final velocity of body – Velocity of fixed plane = u – 0 = u.

Co-efficient of restitution is given by,