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"Two ideal bullets, identical in shape, size and mass, strike the same target with the same speed just before the collision. Force meters at the target register two times the force value for bullet A compared to bullet B, yet none of the force meters are faulty.Explain the reason behind this"

I really don't understand this. aren't the bullets supposed to have the same force even after collision since they have the same shape, size, mass and speed? am I missing something here? the only answer i can come up with is that one bullet has twice acceleration than the other.

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I think I saw a video about this problem... but I couldn't find a link. I believe it has to do with where the bullet hits the "free" object. Whether it hits on a line through the center of mass, or whether it hits near an edge (and thus has both momentum and angular momentum.) Maybe someone else can find the link.

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  • $\begingroup$ For a video answer search for "veritasium bullet block answer" $\endgroup$ – George Herold Mar 16 '15 at 16:58
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Bullet A would need twice the (de)acceleration than bullet B in order to have twice the force of impact. I though of 2 scenarios that would explain the results.

  1. If the target is suspended, then bullet A starts the target swinging. If the target is moving away when bullet B strikes, it takes more time to transfer the momentum and the recorded force is less.

  2. The target is composed of 2 materials with different densities. Bullet A strikes the higher density and stops quickly. Bullet B strikes the lower density and takes twice as long to stop.

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