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Coefficient of Restitution

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”. The constant of proportionality is known as coefficient of restitution (e).

Relative velocity of the two bodies [Fig. 7.2(a)] before collision = u1 – u2

And the relative velocity of the two bodies after collision = (v2 – v1)

 

e (coefficient of restitution)= v2 – v1 / u1 –u2

 

For perfectly elastic body,e = 1

For perfectly plastic body, e = 0

For general elastic body 0 < e < 1.

Inelastic body is one which does not rebound at all after impact.

In general there will be loss of kinetic energy

Loss in K.E. = Total K.E. before impact- Total K.E. after impact.

 

 

SOLVED EXAMPLES

 

Example 7.1. A ball of mass m1 = 2 kg moving with a velocity of 4 m/s strikes directly on a ball o( mass m2 = 4 kg at rest. The. ball of mass m1 comes to rest after striking. Calculate the velocity of ball of mass m2 after stnkmg and co-efficient of restitution.

 

Solution .                 = m1 u1 + m2u2 = m1 u1 + m2u2

2(4) +4 (0) = 2 (0) +4 (v2)

v2 = 8-0 / 4 = 2 m/s.Ans

e = v2 – v1 / u1 –u2 = 2-0 /4-0 = 2/ – =1/2 = 0.5

e = 0.5.Ans

 

Example 7.2. A ball of mass m1 = 1 kg moving with a velocity u1 = 2 m /s, strikes directly on another ball of mass m2 = 2 kg at rest (u2 = 0). If the co-efficient of restitution  between the two balls e = 0.5 . calculate the velocities v1 and v2 of the two balls after impact

 

Solution. =              m1 u1 + m2u2 = m1 u1 + m2u2

1(2) + 2 (0) = 1 (v1) +2 (v2)

or                                     v1 + 2v2 =2

given                                            e =0.5 = v2 – v1 / u1 –u2 = u1 –u2 /2-0

v2 – v1

 

By adding equations (1) and (2), we get

3v2 = 3

or                     v2 = 1m/s.Ans

equation (1)

v1 = 2- 2v2 = 2-2 x 1 = 0 m/s.Ans

 

 Example 7.3. A vehicle of mass 600 kg and moving with a velocity of 12 m/s strikes another vehicle of mass 400 kg moving with velocity 9 m/s in the same direction. Due to the impact both the vehicles get coupled and move together. Find the common velocity with which the two vehicles move after impact.

 

Solution.    m1 u1 + m2u2 = m1 u1 + m2u2

600 (12) +400 (9) = 600v + 400 v [ v1 = v2 =v common velocity )

v = 600 x 12 +400 x 9 / 600 +400 = 10.8m/s.Ans

 

Example 7.4. A body of mass 100 kg, moving with a velocity of 9 m/s, collides directly with a stationary body of mass 50 kg. If the two bodies become coupled so that they move together after impact, what is their commun velocity.

 

Solution.     m1 u1 + m2u2 = m1 u1 + m2u2

100(9) +50 (0) = 100v +50v

v  = 900 +0 / 150 = 900 /150 = 6 m/s.Ans

 

Example 7.5. A ball of mass 30 kg moving with a velocity of 4 m/s strikes directly another ball of mass 15 kg moving in the opposite direction with a velocity of 12 m/s. If the co-efficient of restitutio/! is equal to 5/6, then determine the velocity of each ball after impact.

 

Solution.    m1 u1 + m2u2 = m1 u1 + m2u2

30(4) +15 (-12) = 3v1 +15v2

v2 +2v 1 = – 4

e = v2 – v1 / u1 –u2 = v2 –v1 / 4+12 = 5/6

v2 –v1 = 13.34

 

By subtracting equation (2) from (1), we get

3v1 = – 17.34

v1 = – 17.34 /3 = – 5.78 m/s.Ans

v2 = v1 + 13.34 = – 5.78 +13.34 = 7.56 m/s.Ans
 

Example 7.6. A ball of mass 20 gm, moving with a velocity of 3 m/s strikes on a ball of mass 40 gm moving with a velocity of 1 m/s. The velocities of the two balls are parallel and

inclined at 30″ to the line joining their centres nt the instant of impact. if the coefficient of

restitution is !,find (i) the direction and magnitude of velocity of ball of mass 20 gm after impact

1/2 (ii) the direction and magnitude of velocity of ball of mass 40 gm.

 

Solution.              θ1 = θ2 =300

m1 u1 cos θ1 + m2u2 cos θ2= m1 u1 cos ϕ1 + m2u2 cos ϕ2

or                        20 x 3 cos 300 +40 x 1 x cos 300 = 20v1 cos ϕ1 + 40v2 cos ϕ2

or                                      v1 cos ϕ2  + 20v1 cos ϕ1 = 4.33

e = v2 cos ϕ2  – v1 cos ϕ1 / u2 cos θ2 – u1 cos θ1 = v2 cos ϕ2  – v1 cos ϕ1 / 3 cos 300 – 1 cos 300

1/2 = v2 cos ϕ2 – v1 cos ϕ1 / 1.732

or     v2 cos ϕ2 – v1 cos ϕ1 = 0.866

 

By adding equation (1) and (2), we get

3v2 cos ϕ2 = 5.196

v2 cos ϕ2 = 1.732

v1 cos ϕ1 = v2 cos ϕ2 – 0.866 = 1.732 – 0.866

v1 cos ϕ1 =v2 cos ϕ2 – 0.866 = 1.732 – 0.866

v1 cos ϕ1 = 0.866

 

Now normal component of velocity of each ball is also conserved.

. . For ball of mass 20 gm (vertical component of velocities)

u1 sin θ1 = v1 sin ϕ1

3 sin 300  = v1 sin  ϕ1

(4) and (5) =>                    v1  = 0.866 / cos 600 = 1.732 m/s.Ans

 

For ball of mass 40 gm :

u2 sin θ2 =v2 sin ϕ2

1 sin 300 = v2 sin ϕ2

0.5 = v2 sin ϕ2

From (3) and (6), we get

tan ϕ2 = 0.5 / 1.732

ϕ2 = 16.1 Ans.

(3) =>    v2 1.732 / cos ϕ2 = 1.732/ cos 16.1 = 1.8027 m/s.Ans