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 = u_{1} – u_{2}

And the relative velocity of the two bodies after collision = (v_{2} – v_{1})

e (coefficient of restitution)= v_{2} – v_{1} / u_{1} –u_{2}

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.

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**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 m_{2} = 4 kg at rest. The. ball of mass m_{1} comes to rest after striking. Calculate the velocity of ball of mass m2 after stnkmg and co-efficient of restitution.

**Solution **. = m_{1} u_{1} + m_{2}u_{2} = m_{1} u_{1} + m_{2}u_{2}

2(4) +4 (0) = 2 (0) +4 (v_{2})

v_{2} = 8-0 / 4 = 2 m/s.Ans

e = v_{2} – v_{1} / u_{1} –u_{2} = 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 (u_{2} = 0). If the co-efficient of restitution between the two balls e = 0.5 . calculate the velocities v_{1} and v_{2} of the two balls after impact

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**Solution.** = m_{1} u_{1} + m_{2}u_{2} = m_{1} u_{1} + m_{2}u_{2}

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

or v1 + 2v_{2} =2

given e =0.5 = v_{2} – v_{1} / u_{1} –u_{2} = u_{1} –u_{2} /2-0

v_{2} – v_{1}

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

3v_{2} = 3

or v_{2} = 1m/s.Ans

equation (1)

v_{1} = 2- 2v_{2} = 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.

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**Solution. **m_{1} u_{1} + m_{2}u_{2} = m_{1} u_{1} + m_{2}u_{2}

600 (12) +400 (9) = 600v + 400 v [ v_{1} = v_{2} =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.** m_{1} u_{1} + m_{2}u_{2} = m_{1} u_{1} + m_{2}u_{2}

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.

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**Solution. **m_{1} u_{1} + m_{2}u_{2} = m_{1} u_{1} + m_{2}u_{2}

30(4) +15 (-12) = 3v_{1} +15v_{2}

v_{2} +2v _{1} = – 4

e = v_{2} – v_{1} / u_{1} –u_{2} = v_{2} –v1 / 4+12 = 5/6

v_{2} –v_{1} = 13.34

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

3v_{1} = – 17.34

v_{1} = – 17.34 /3 = – 5.78 m/s.Ans

v_{2} = v_{1} + 13.34 = – 5.78 +13.34 = 7.56 m/s.Ans

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**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 =30^{0}

m_{1} u_{1} cos θ_{1} + m_{2}u_{2} cos θ_{2}= m_{1} u_{1} cos ϕ_{1} + m_{2}u_{2} cos ϕ_{2}

or 20 x 3 cos 30^{0} +40 x 1 x cos 30^{0} = 20v_{1} cos ϕ_{1} + 40v_{2} cos ϕ_{2}

or v1 cos ϕ_{2} + 20v_{1} cos ϕ_{1 }= 4.33

e = v_{2} cos ϕ_{2} – v_{1} cos ϕ_{1} / u_{2} cos θ_{2} – u_{1} cos θ_{1} = v_{2} cos ϕ_{2} – v_{1} cos ϕ_{1} / 3 cos 30^{0} – 1 cos 30^{0}

1/2 = v_{2} cos ϕ_{2} – v_{1} cos ϕ_{1} / 1.732

or v_{2} cos ϕ_{2} – v_{1} cos ϕ_{1} = 0.866

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

3v_{2 }cos ϕ_{2} = 5.196

v_{2} cos ϕ_{2} = 1.732

v_{1} cos ϕ_{1} = v_{2 }cos ϕ_{2} – 0.866 = 1.732 – 0.866

v_{1} cos ϕ_{1} =v_{2 }cos ϕ_{2} – 0.866 = 1.732 – 0.866

v1 cos ϕ_{1} = 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 30^{0} = v1 sin ϕ1

(4) and (5) => v_{1} = 0.866 / cos 60^{0} = 1.732 m/s.Ans

For ball of mass 40 gm :

u2 sin θ_{2} =v_{2} sin ϕ_{2}

1 sin 30^{0} = v_{2} sin ϕ_{2}

0.5 = v_{2} sin ϕ_{2}

From (3) and (6), we get

tan ϕ_{2} = 0.5 / 1.732

ϕ_{2} = 16.1 Ans.

(3) => v_{2} 1.732 / cos ϕ_{2} = 1.732/ cos 16.1 = 1.8027 m/s.Ans