More on the problem posted yesterday. Such a simple setup, and it’s generated a lot of positive discussion in the physics blogohedron (including twitter), which is a sign that I’m not the only one who thinks it’s a great demo.

As I stated, the kinetic energy of a particle of momentum p is given by p^/2m (since KE = 1/2 mv^2 and p = mv). We can estimate how much of the initial kinetic energy is lost; since momentum is the same before and after the collision, all we need to know is the mass of “moving things” before and after. The ratio of the initial and final KE is then just m/(m+M), where m is the mass of the projectile and M is the mass of the target. If this is a 10g bullet and a 1kg block of wood, then only 1% of the kinetic energy remains after the collision.

Let’s assume the block of wood went up about a meter in the video. To do that the speed is about 4.4 m/s right after the collision, which gives the bullet+block system a kinetic energy of 9.8 Joules (which is what you’d better expect to 2 digits for a ~1 kg system rising a meter). That means the bullet had a kinetic energy of 970 Joules! It was also traveling at ~440 m/s prior to impact.

Elastic collisions are different. KE is conserved, and the incident particle recoils. This gives you a momentum equation and a kinetic equation to solve, with the final velocities being the two unknowns, so you can solve for them in terms of the initial velocity of the projectile. Do a little math and you get the results.

If we solve for the final speed of the two objects but use the same masses and speeds, the block will start up at a speed of 8.7 m/s — almost twice as fast, so there is almost 4 times the kinetic energy (38 Joules). The projectile (more like a super-ball rather than a bullet, and probably a harder target) will end up going -431 m/s. The negative sign means its going down, as you expect — it will recoil, just not quite as fast. That’s 930 Joules, and 930 + 38 is about 970 (the same to the two digits I kept in the calculations)

If you play around with those formulas, you can see the target always goes in the direction of the projectile, but the projectile can recoil if the target is more massive, come to a stop if it has the same mass as the target, or will continue on in the same direction if it is more massive. I hope that matches intuition.

The rotating case is a little more complicated to work out, since you have more variables. Since the ball recoils and the block rotates, that’s another set of equations, and there’s the issue of the ball not bouncing straight down, which literally (and I literally mean literally; thanks a lot, OED) brings another dimension to the problem, though this effect is probably small for this problem and could be ignored.