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I am trying to desgin a little penny balista toy (built with a 3d printer).

The short description is that there is a track with a mass ("hammer") that slides along it. There is a stack of pennies in a hopper at the front. The hammer goes from back to the front, slams into the penny at the end and launches it.

I had a debate with a friend about whether it would go farther if the penny started at the back and was accelerated along the full length of the track along with the hammer (a more difficult design).

My question is, all else being the same, which method would impart more energy to the penny? (my suspicion, based on those "newtons cradle" toys, is that it would be the same).

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2 Answers 2

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If the mass of the penny is negligible comparing to the hammer, The speed of the hammer at the end of the track should be the same in both cases. With the collision in the first scenario, the speed of the penny should be twice of the hammer if the collision is complete elastic. But, for the second move-along scenario, the speed of the penny will be just the same as the hammer at the end of the track.

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what if the mass of the hammer is comparable to the penny (the hammer in this case is just a bit of not very dense plastic). What happens to this equation if the hammer is at or below the mass of the penny? also how reasonable is it for the collision to be "completely elastic" (i term i am not familiar with) in this case? –  Zak Jul 12 '12 at 23:23
    
If the mass is the same, in the collision case the hammer will stop after the collision and the penny will be launched with the same speed just like in a Newton's cradle. In the move-along case, the penny will keep moving with the same speed as the hammer at the end of the track. Depending on construction, the later speed is likely lower since the track need to accelerate twice as much mass. So, the collision case works better anyway. "Elastic" means the energy is conserved. For inelastic collision, some energy will be lost in the collision and that can be a game changer. –  Chun-Chung Chen Jul 13 '12 at 6:28

If the hammer is perfectly matched to the penny in mass, then it would indeed act like a Newton's cradle and transfer all it's energy to the ejected penny. Ignoring friction, both designs could, in principle, result in total transfer of spring's energy to the penny.

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