# Two identical cubical blocks, moving in the same direction with a common speed v, strike a third such block initially at rest on a horizontal frictionless surface. What is the motion of the blocks after the collision? Does it matter whether or not the two initially moving blocks were in contact? Does it matter whether these two blocks were glued together? Assume that the collisions are (a) completely inelastic or (b) elastic.

Jitender Pal
7 years ago

Momentum of the body p is equal to the mass of the body m times velocity of the body v.
So, p = mv
In accordance to the principle of conservation of energy, the final momentum of the system is equal to the initial momentum of the system.
(a)
In the inelastic collision, consider u is the initial velocity of two block system, vf is the final velocity of the system. As the third block is rest, therefore initial velocity of the third block will be zero.
So, before collision, the initial momentum pi of the system will be,

In accordance to conservation momentum law, the initial momentum pi of the system will be equal to final momentum pf of the system.
Yes, it will mater, because of the mass distribution of the two blocks is different.
Yes, it will matter whether these two blocks were glued together because after the collision they will travel along with each other.

Yes, it will mater, because of the mass distribution of the two blocks is different.
No, it will matter whether these two blocks were glued together because after the collision they will travel differently with different direction.