> if mass is assumed to be **constant,** the *velocity* of the centre of mass
> of the system has to be different after the collision for the kinetic
> energy to be different.
> 
> However, if the momentum of the system is conserved, *the velocity* of
> the centre of mass of the system **should remain the same**.

1) mass is **not** constant and velocity is different: in a completely inelastic collision the two objects (A: m =1, B  m = 2) stick together and mass becomes A+B = 3


![enter image description here][1]

Suppose that $v_a = 6m/s$ and $v_b = 0  \rightarrow E_k = 0.5 * 6^2 = 18, p_a = 1 * 6 = 6$ 
 
After collision velocity would be anyway **lower** as KE should be distributed among more mass, but some KE is lost in the crash. How much?

Momentum is conserved: $ p_{ab} = 6$ , from this datum you can calculate its velocity:
$$v_{ab} = v_{cm}= \frac{6}{3} = 2$$ and $E_k = 0.5 * 2^2 *3 = 6 \rightarrow E_a = 2 + E_b = 4$.

Some energy has been transferred to B, but two thirds of the energy have been lost.

Velocity of center of mass is the same, although KE has changed. 

Please note that momentum is conserved because *we are assuming that on the surface of contact there is **no friction**.*

> ...*how can there be a change in kinetic energy of the system if there is no change in momentum?*

A change of KE without a change of momentum is not only possible but very frequent, because as you noted **p=mv** momentum varies **linearly** and *KE quadratically*. You canget the same product by a wide range of factors: **6** = 6*1, = 3*2, = 2*3, = 1*6, = 0.5*12, etc., these are all momenta

All these factors give same values for m*v, but as the figure for v must be squared, you get all different values between momentum and energy, therefore the same factors give momentum = **6**, but KE =3, =6, =9, =18, =72, etc

I hope this clarified all your doubts

 [1]: https://i.sstatic.net/oYclE.png.