Is Linear Momentum Conserved in This Case? (sliding box hits a floor tile)

[Heisenberg7]

Homework Statement

A box with edge 1.5 m is sliding across a floor with speed of 4 m/s. The leading lower edge of the box strikes an upraised floor tile with a completely inelastic impact. What is the angular velocity of the box immediately after the box hits the tile? The mass of the box is 9 kg.

Relevant Equations

$\vec{P} = const$

Hello,

I actually solved this problem using conservation of angular momentum, but I was wondering if linear momentum is conserved. Here's my thought process:
(block + tile system) The block is going to hit the tile with some force $\vec{F}$. Due to the Newton's third law, that force is also going to act on the block itself. These are internal forces, so they play no role in our investigation. Now, here comes the problem. When the block hits the tile, the tile has to move slightly which could possibly cause an external force due to the floor. This implies that linear momentum of our block + tile system is not conserved.

There is another problem though. The tile itself is a part of the floor. So, if we were to consider the block + floor system, the linear momentum would be conserved, wouldn't it?

 

[kuruman]

Yes, the linear momentum of the block and the tile and the house to which the tile is attached and the Earth to which the house is attached is conserved. You have to have an isolated system, i.e. no external forces.

 

[jbriggs444]

When the block hits the tile, the tile has to move slightly which could possibly cause an external force due to the floor. This implies that linear momentum of our block + tile system is not conserved.

It depends on the coefficient of friction between block and tile.

Let us assume for the moment an ideal case with a frictionless tile and a frictionless floor. So 100% of the horizontal force on the block comes from the tile and 100% of the vertical force on the block comes from the floor.

If the horizontal force from tile on block were the only force present during the collision then the result would be a block rotating about its center of mass in such a way that its lower right corner is horizontally motionless. The center of mass would continue horizontally at a reduced speed while the lower right corner rotated down and back beneath it.

But that is not what happens. The floor gets in the way of that rotation.

The vertical force of floor on block will be exactly sufficient to prevent any vertical motion of the lower right corner of the block.

The combined effect of the two forces is that the lower right corner of the block becomes motionless.

Since neither the impulsive horizontal force from the tile nor the impulsive vertical force from the floor have any moment arm about the lower right corner of the block, it is convenient to adopt an axis of rotation at that point. Angular momentum about that axis is conserved across the collision.

We can do a conservation of momentum argument (as you have already done) and solve for the resulting rotation rate about the lower right corner.

But let us return to my original assertion about a coefficient of friction.

With the results of our computation in hand, we know how the block should be moving in the wake of the collision. It is a rotation about a fixed point at the lower right corner. We know the rotation rate.

So we are in a position to calculate the instantaneous velocity of the block's center of mass.

It will have non-zero horizontal velocity. It will have non-zero vertical velocity. Because the center of mass is rotating about the fixed point at the lower right corner.

Since we know horizontal momentum before and after, we know how much horizontal impulse has been applied.

Since we know vertical momentum before and after, we know how much vertical impulse has been applied.

The ratio, vertical:horizontal gives the required coefficient of friction to avoid slipping against the tile and thereby not needing the floor to act as a back stop.