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3. Motion and Force eLearn.Punjab

Elastic Collision in One Dimension

Consider two smooth, non-rotating balls of masses m1, and m2, moving initially with velocities v’1
and v‘2 respectively, in the same direction. They collide and after collision, they move along the
same straight line without rotation. Let their velocities after the collision be v1 and v2 respectively,
as shown in Fig. 3.9.

We take the positive direction of the velocity and momentum to the
right. By applying the law of conservation of momentum we have

m1v1 + m2v2 = m1v'1 + m2v'2
m1(v1 - v'1) = m2 (v'2 - v2 ) ......... (3.14)

As the collision is elastic, so the K.E is also conserved. From the
conservation of K.E we have

1 m1v12 + 1 m2 v 2 2 = 1 m1v '12 + 1 m2 v '2 2
2 2 2 2

or m1(v12 - v'12 ) = m2 (v'22 - v22 )

or m1(v1 + v'1) (v1 - v'1) = m2 (v'2 + v2 ) (v'2 - v2 ) ......... (3.15)

Dividing equation 3.15 by 3.14

(v1 + v'1) = (v'2 + v2 ) ......... (3.16) Fig.3.9
or (v1 - v2 ) = (v'2 - v'1) = - (v'1 - v'2 )

We note that, before. collision (v1 - v2) is the velocity of first ball relative to the second ball. Similarly
(v’1 - v’2) is the velocity of the first ball relative, to the second ball after collision. It means that
relative velocities before and after the collision have the same magnitude but are reversed after the

collision. In other words, the magnitude of relative velocity of approach is equal to the magnitude

of relative velocity of separation.

In equations 3.14 and 3.16, m1, m2, v1 and v2 are known quantities. We solve these equations to find
the values of v’1 and v’2 which are unknown. The results are

v '1 = m1 − m2 v1 + 2m2 v2 ......... (3.17)
m1 + m2 m1 + m2

v '2 = 2m1 v1 + m2 − m1 v2 ......... (3.18)
m1 + m2 m1 + m2

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