Ap physics 1 velocity center of mass question

[ldkdkdjdj]

Homework Statement

Can anyone explain how if you add mass to one block colliding with another in an inelastic system, the center of mass velocity of the system changes? Doesn’t the center of mass velocity of a system only change if there is a net external force? In that case the force are equal and opposite, so center of mass velocity of the system doesn’t change, no?

Relevant Equations

Idk

Idk

 

[haruspex]

if you add mass to one block colliding with another in an inelastic system, the center of mass velocity of the system changes?

Does it? Perhaps you need to define the situation more clearly.

 

[ldkdkdjdj]

Does it? Perhaps you need to define the situation more clearly.

From the 2021 physics 1 test:
(c) The disk is now moving at a constant speed v on the surface (frictionless) toward a block of mass Mg, which is at rest on the surface, as shown above. The disk and block collide head-on and stick together, and the center of mass of the disk-block system moves with speed Vcm

i. Suppose the mass of the disk is much greater than the mass of the block. Estimate the velocity of the center of mass of the disk-block system. Explain how you arrived at your prediction without deriving it mathematically.

 

[kuruman]

It seems that you are confusing the constancy of the velocity of the CM during a collision with the dependence of the velocity of the CM on the velocity and mass of the colliding objects. For example
Case I
You have equal masses 6 kg each moving towards each other at 2 m/s. The velocity of the center of mass is $$V_{cm}=\frac{6~(\text{kg})\times 2~(\text{m/s})+6~(\text{kg})\times (-2)~(\text{m/s})}{6~(\text{kg})+6~(\text{kg})}=\frac{12~(\text{kg}\cdot\text{m/s})-12~(\text{kg}\cdot\text{m/s})}{12~(\text{kg})}=0~\text{m/s}.$$Case II
You move 2 kg from one mass to the other keeping the velocities the same. The velocity of the center of mass is $$V_{cm}=\frac{8~(\text{kg})\times 2~(\text{m/s})+4~(\text{kg})\times (-2)~(\text{m/s})}{6~(\text{kg})+6~(\text{kg})}=\frac{16~(\text{kg}\cdot\text{m/s})-8~(\text{kg}\cdot\text{m/s})}{12~(\text{kg})}=\frac{2}{3}~\text{m/s}.$$Of course, in each case the velocity of the center of mass after the collision is the same as before the collision. Here you are asked to estimate the velocity of the CM if most, i.e. almost all but not quite, of the mass is moved from one object to the other.

 

[ldkdkdjdj]

Ohh that makes more sense now, tysmm