(^560) A Textbook of Engineering Mechanics
∴ Loss of kinetic energy, during impact
EL = E 1 – E 2 ( 11 22 222 2) ( 11 2 2)
1
2
=+−+⎡⎤⎣⎦mu m u mv m v
Multiplying the numerator and denominator of the right hand side by (m 1 + m 2 ),
()
()()
22
1211 22
12
1
L 2
E mmmu mu
mm
=++⎡
- ⎣
−+()mmmv mv1211 22()^22 + ⎤⎦
22 2 2 2 2
11 122 121 2 2
12
1
()
2( )
mu mmu mmu m u
mm
=+++⎡ - ⎣
22 2 2 2 2
–(mv 11 +++mmv 122 mmv 211 m v2 2)⎤⎦
{}
22 2 2 2 2
11 2 2 12 1 2
12
1
(()
2( )
mu m u mm u u
mm
=+++⎡ - ⎣
{ }
22 2 2 2 2
–(mv 11 m v2 2 mm v12 1( v 2 )
++ +⎤
⎦
{
2
11 2 2 1 2 1 2
12
1
()–(2)
2( )
mu m u mm uu
mm
=+⎡ - ⎣
}
2
++mm u12 1(– ) (2u 2 mm uu 1212 )
{
2
–(mv 11 +m v2 2) –(2mm vv1 21 2)
}
2
++mm v12 1(– ) (2v 2 mm vv 1212 )⎤⎦
{}
22
11 2 2 1 2 1 2
12
1
()(–)
2( )
mu m u mm u u
mm
=++⎡ - ⎣
- {(m 1 v 1 + m 2 v 2 )^2 + m 1 m 2 (v 1 – v 2 )^2 }]
We know that in a direct impact,
Initial momentum = Final momentum
i.e. m 1 u 1 + m 2 u 2 = m 1 v 1 + m 2 v 2
(m 1 u 1 + m 2 u 2 )^2 = (m 1 v 1 + m 2 v 2 )^2 ...(Squaring both sides)
Therefore loss of kinetic energy due to impact,
(^) ()() ()
22
12 1 2 12 1 2
12
1
- L 2
Emmuummvv
mm
=−−⎡⎤
+ ⎢⎥⎣⎦
Now substituting (v 1 – v 2 ) = e (u 1 – u 2 ) in the above equation,(^) ()() ()
(^222)
12 1 2 12 1 2
12
1
L 2
Emmuummeuu
mm
=−−−⎡⎤
- ⎢⎥⎣⎦
(^) ()()()
12 2 2
12
12
1
2
mm
uu e
mm
=−−
Note. The loss of kinetic energy may also be found out by calculating the kinetic energy of the
system before impact, and then by subtracting from it the kinetic energy of the system after impact.
Example 27.5. A ball impinges directly on a similar ball at rest. The first ball is reduced
to rest by the impact. Find the coefficient of restitution, if half of the initial kinetic energy is lost
by impact.