Question
Show that in one dimension elastic collision, velocity of separation is equal to the velocity of approach.
Solution
Consider two bodies of masses m1 and m2 moving with velocities u1 and u2 along the same straight line.
Let the two bodies collide and after collision v1 and v2 be the velocities of two masses respectively.
Before collision:
Momentum of mass m1 = m1u1
Momentum of mass m2 = m2u2
Total momentum before collision,
p1 = m1 u1 + m2u2
K.E of mass m1 =
K.E of mass m2 =
Therefore,
Total K.E before collision is,
K.E =
+ 
After collision:
Momentum of mass m1 = m1v1
Momentum of mass m2 = m2v2
Total momentum after collision is,
pf = m1v1 + m2v2
K.E of mass m1 =
K.E of mass m2 =
Total K.E after collision, Kf =
+ 
Now, according to the law of conservation of momentum,
That is, relative velocity of approach is equal to relative velocity of separation.
Hence the result.
Let the two bodies collide and after collision v1 and v2 be the velocities of two masses respectively.
Before collision:
Momentum of mass m1 = m1u1
Momentum of mass m2 = m2u2
Total momentum before collision,
p1 = m1 u1 + m2u2
K.E of mass m1 =
K.E of mass m2 =
Therefore,
Total K.E before collision is,
K.E =
+ After collision:
Momentum of mass m1 = m1v1
Momentum of mass m2 = m2v2
Total momentum after collision is,
pf = m1v1 + m2v2
K.E of mass m1 =
K.E of mass m2 =
Total K.E after collision, Kf =
+ Now, according to the law of conservation of momentum,
That is, relative velocity of approach is equal to relative velocity of separation.
Hence the result.
