Why conservation of momentum?

The teacher is assuming that the bullet passes through instantaneously. In other words, the bullet moves so quickly that there is no time for friction to act. Hence, the momentum that the bullet loses is entirely transferred to the block, and none is transferred to the ground via friction.

An appropriate follow-up question would be, is this a reasonable assumption to make? Let's take friction into account and try to estimate how fast the block would actually be moving when the bullet exits.

We know the bullet is traveling at $\frac{500\text{ m/s}+100\text{ m/s}}{2}\approx300\text{ m/s}$ on average through the block. If the block is wood and cubic in shape, then the block is only ~0.3 m wide.* Hence, the bullet would pass through the block in a time of:

$$t=\frac{d}{v}=\frac{0.3\text{ m}}{300\text{ m/s}}=0.001\text{ s}$$

Your teacher calculated that the block reaches a speed of 0.8 m/s as a result of the collision with the bullet. That's a high estimate, because it ignores friction with the ground, which slows the block. But let's go ahead and assume that, while the bullet is in the block, the block's average speed is $\frac{0+0.8\text{ m/s}}{2}\approx0.4\text{ m/s}$. At this speed, and for a time of 0.001 s, the block would only travel a distance of

$$d=vt=(0.4\text{ m/s})(0.001\text{ s})=0.0004\text{ m}$$

while in contact with the bullet. If the coefficient of friction is ~0.16, and if we use $g=10\text{ m/s}^2$, then the block's final speed when the bullet exits would be:

$$\Delta KE=Fd\cos(180°)=(-0.16*100\text{ N}*0.0004\text{ m})=-0.0064 \text{ J}$$

$$\Delta KE=\frac{1}{2}mv_f^2-\frac{1}{2}mv_i^2$$

$$-0.0064\text{ J}=\frac{1}{2}(10\text{ kg})(v_f^2-(0.8\text{ m/s})^2)$$

$$v_f\approx0.799\text{ m/s}$$

There are plenty of things wrong with this calculation (for example, I'm assuming the block immediately reaches a max speed of 0.8 m/s upon collision with the bullet, which isn't true). But as an order of magnitude estimate, it's reasonable enough to show that friction doesn't have much impact on the block while the bullet is inside.

*The density of wood is ~500 kg/m3, so $l=V^{1/3}=(\frac{m}{\rho})^{1/3}=(\frac{10\text{ kg}}{500\text{ kg/m}^3})^{1/3}\approx0.3$.

You are right: you should treat the momentum transfer as an impulse. Assume that the bullet takes no time at all to get through the block, and that the momentum it loses is transferred instantaneously to the block. Then, you need to solve the rest of the problem.

The way I would approach the rest of the problem is to use the block's instantaneous momentum to and mass to calculate how much kinetic energy was transferred to the block, and then calculate the coefficient of friction. Force x distance = work, which = the energy the block got from the bullet; and force in this case is the coefficient of friction times the * weight* of the block. (Make sure you don't mistake the mass for the weight!)

What if the friction force was not external?

If you consider the system {Bullet + Block + Ground} then you have three phases:

  • Phase 1: The bullet has kinetic energy, block and ground are at rest.

  • Phase 2: The bullet has lost some kinetic energy, which has been transfered to the block as kinetic energy

  • Phase 3: The block has lost energy. It has been transfered as heat to the ground. Knowing the initial energy, and the distance it took to dissipate it, you can calculate the coeffcient.