Engineering Mechanics

Chapter 27 : Collision of Elastic Bodies „„„„„ 557

12 12

99

2or 2

77

MU U

=+Mv Mv =+v v ...(i)

We also know from the law of collision of elastic bodies that

(v 2 – v 1 ) = e (u 1 – u 2 ) = 0.75

9

714

UU

U

⎛⎞

⎜⎟−=

⎝⎠

or 21

9

14

U

vv=+

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

(^1111)
99 9
23or0
714 7
UU U
=+vvvv⎛⎞⎜⎟+ = + =
⎝⎠
Thus the first ball will come to rest after impact. Ans.
Example 27.3. The masses of two balls are in the ratio of 2 : 1 and their velocities are in
the ratio of 1 : 2, but in the opposite direction before impact. If the coefficient of restitution be 5/6,
prove that after the impact, each ball will move back with 5/6th of its original velocity.
Solution. Given : Mass of first ball (m 1 ) = 2 M ; Mass of second ball (M 2 ) = M ; Initial
velocity of first ball (u 1 ) = U ; Initial velocity of second ball (u 2 ) = – 2U (Minus sign due to opposite
direction) and coefficient of restitution (e) =
5
6
Let v 1 = Final velocity of the first ball, and
v 2 = Final velocity of the second ball.
We know from the law of conservation of momentum that
m 1 u 1 + m 2 u 2 = m 1 v 1 + m 2 v 2
2 M × U + M (– 2U) = 2Mv 1 + Mv 2
or 0 = 2Mv 1 + Mv 2
∴ v 2 = – 2v 1 ...(i)
We also know from the law of collision of elastic bodies that
(v 2 – v 1 ) = e (u 1 – u 2 )^55 –2()
62
U
=−=⎡⎤⎣⎦UU ...(ii)
Substituting the value of v 2 from equation (i)
11 1
55
2() or
26
U
⎡⎤⎣⎦−− =vv v=−×U
Minus sign indicates that the direction of v 1 is opposite to that of U. Thus the first ball will
move back with
5
th
6
of its original velocity. Ans.
Now substituting the value of v 1 in equation (i),
2
55
2– 2
66
vUU=− ⎛⎞⎜⎟× =+ ×
⎝⎠
Plus sign indicates that the direction of v 2 is the same as that of v 1 or opposite to that of u 2.
Thus the second ball will also move back with^5 th
6
of its original velocity. Ans.