Engineering Mechanics

Chapter 27 : Collision of Elastic Bodies „„„„„ 561

Solution. Given : Initial velocity of second body (u 2 ) = 0 (because it is at rest) and final
velocity of the first body (v 1 ) = 0 (because it comes to rest by the impact)

Let m 1 = Mass of the first body,

m 2 = m 1 = Mass of the second body,

...(Q both the balls are similar)

u 1 = Initial velocity of the first body,

v 2 = Final velocity of the second body, and

e = Coefficient of restitution.

We know that kinetic energy of the system before impact,

22

111 22

11

22

Emu mu=+ 112

1

2

= mu ...()Qu 2 = 0

and kinetic energy of the system after impact,

22

211 22

11

22

Emvmv=+ 122

1

2

= mv ...()Q v 1 = 0

∴ Loss of kinetic energy during impact

EL = E 1 – E 2 =

22

11 12

11

22

⎛⎞⎛⎞⎜⎟⎜⎟mu − mv

⎝⎠⎝⎠

Since half of the initial K.E. is equal to loss of K.E. by impact, therefore

∴

222

11 11 12

1111

2222

⎛⎞⎛⎞⎛⎞⎜⎟⎜⎟⎜⎟mu =−mu mv

⎝⎠⎝⎠⎝⎠

2

11 22

2 11 12

mu

=−mu mv

∴ mu 1122 = 2 mv 12

or uv 1222 = 2 ...(i)

We know from the law of conservation of elastic bodies that

(v 2 – v 1 ) = e (u 1 – u 2 )

v 2 – 0 = e (u 1 – 0) ...(Q v 1 = 0 and u 2 = 0)

∴ v 2 = eu 1 ...(ii)

Substituting the value of v 2 in equation (i),

()

2222

ueu eu11 1== 22

or ee^2 ==^12 or 0.707 Ans.
Example 27.6. A sphere of mass 1 kg, moving at 3 m/s, overtakes another sphere of mass
5 kg moving in the same line at 60 cm/s. Find the loss of kinetic energy during impact, and show
that the direction of motion of the first sphere is reversed. Take coefficient of restitution as 0.75.

Solution. Given : Mass of first sphere (m 1 ) = 1 kg; Initial velocity of first sphere (u 1 ) = 3 m/s;
Mass of second sphere (m 2 ) = 5 kg ; Initial velocity of second sphere (u 2 ) = 60 cm/s = 0.6 m/s and
coefficient of restitution (e) = 0.75.