Number of collisions by a bullet

  • #1
NODARman
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New poster has been reminded to always show their work when starting schoolwork threads.
Homework Statement:
I'm trying to calculate collision number.
Relevant Equations:
.
How to find the collision number if the moving bullet hits a few wooden blocks and every collision takes 10 percent of its speed. In which block will the bullet stay?
 

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  • #2
PeroK
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Homework Statement:: I'm trying to calculate collision number.
Relevant Equations:: .

How to find the collision number if the moving bullet hits a few wooden blocks and every collision takes 10 percent of its speed. In which block will the bullet stay?
Why would a wooden block reduce a projectile's spped by a fixed percentage? Would something hitting the block at ##1 \ m/s## emerge at ##0.9 \ m/s##?
 
  • #3
NODARman
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IDK. That physics task in the book says that the speed of a bullet decreases by 10% after hitting every block. Every block has the same length. The task doesn't contain L, V, or K. There is only a 10% of speed loss. I've just written the formula:
Vn=Vn-1-(Vn-1/10)
I think this formula works in general, but I don't need it here.
 
  • #4
PeroK
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IDK. That physics task in the book says that the speed of a bullet decreases by 10% after hitting every block. Every block has the same length. The task doesn't contain L, V, or K. There is only a 10% of speed loss. I've just written the formula:
Vn=Vn-1-(Vn-1/10)
I think this formula works in general, but I don't need it here.
That might take a lot of blocks! Are you sure the book doesn't say that it loses 10% of its speed going through the first block? Not every block?
 
  • #5
NODARman
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Yeah, yeah, the first block. I forgot to say that because I don't really understand what it means (and also how to solve it).
Thanks.
 
  • #6
PeroK
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Yeah, yeah, the first block. I forgot to say that because I don't really understand what it means (and also how to solve it).
Thanks.
What do you think might be the constant factor in each collision?
 
  • #7
NODARman
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"The bullet is hitting a few wooden blocks which are placed at a different distance from each other. In which block will the bullet stick if after exiting the first block, it loses 10% of its initial speed."
(I've translated as I could)

So, if after the fist hit the bullet lost 10% of its initial speed, it means the V1=V0-(V0/10) (this will be the second speed of the bullet which will hit the second block). Is that mean that every hit reduces the speed by 10% of V0 and not Vn=Vn-1-(Vn-1/10).

If we put the numbers in the first case, then the speed would be, let's say: 100, 90, 80, 70, etc.
The second case: 100, 90, 81, 78, etc.
 
  • #8
PeroK
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"The bullet is hitting a few wooden blocks which are placed at a different distance from each other. In which block will the bullet stick if after exiting the first block, it loses 10% of its initial speed."
(I've translated as I could)

So, if after the fist hit the bullet lost 10% of its initial speed, it means the V1=V0-(V0/10) (this will be the second speed of the bullet which will hit the second block). Is that mean that every hit reduces the speed by 10% of V0 and not Vn=Vn-(Vn/10).

If we put the numbers in the first case, then the speed would be, let's say: 100, 90, 80, 70, etc.
The second case: 100, 90, 81, 78, etc.
This is not the correct method. You must consider why it loses speed. What causes the bullet to lose speed? Hint Newton's second law.
 

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