MCAT Collisions of a wooden block and bullet

  • #1
Thread moved from the technical PF forums so no Homework Help Template is shown
Hi guys,
I'm currently going over Berkerley Review Physics example problem and got confused on a question:
Ex. 4.1a : A sharpshooter fires a gun at a wooden block of mass M. If the bullet, of mass (m), becomes lodged inside the block, then which of the following is NOT true?

The answer was: (C) If the bullet has the same mass as the block, the block will NOT move after the collision, which I deemed as correct as a false statement

But why is (D) a correct statement: The block can never move faster than the bullet's impact speed, after the collision.

I was trying to solve this conceptual problem mathematically using Δv2 = -(Δv1 m1 ) / (m2), and though that if the block was lighter than the bullet, wouldn't the velocity of the block be greater than the bullet?
 

Answers and Replies

  • #2
gneill
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The bullet was lodged inside the block. So what type of collision was it?
 
  • #3
Inelastic collision. So the block will NEVER be lighter than the bullet. Got it Thanks!
 
  • #4
gneill
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Inelastic collision. So the block will NEVER be lighter than the bullet. Got it Thanks!
No, the block can be lighter that the bullet; There's no constraint on the masses. But the nature of the collision dictates how the final velocity of the block will be related to the initial velocity of the bullet. You should be able to write the appropriate equation to find the velocity of the combined bullet+block.
 
  • #5
My bad, I worded that poorly. I meant the final block, m2, must always be greater than m1 because the the final block will include the mass of both the bullet and the block.
 
  • #6
gneill
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My bad, I worded that poorly. I meant the final block, m2, must always be greater than m1 because the the final block will include the mass of both the bullet and the block.
Okay, that is true. Although I wouldn't distinguish a "final block" that's different from the initial block. Presumably the block remains the block and the bullet remains the bullet. It's just that the two are combined into a single joined mass during the collision.

But the important insight is to recognize the type of collision taking place, and to tie the outcome (the final velocity) to the conservation of momentum.
 

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