Number of collisions by a bullet

AI Thread Summary
The discussion centers on calculating the number of collisions a bullet makes with wooden blocks, where each collision reduces its speed by 10%. Participants debate whether the speed loss applies to the bullet's initial speed or its current speed after each collision. Clarification is sought on the mechanics of the bullet's speed reduction and the implications of Newton's second law. The conversation highlights confusion over the problem's parameters and the correct formula to use for calculating the bullet's speed after multiple collisions. Understanding the physics behind the speed loss is crucial for solving the problem accurately.
NODARman
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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?
 
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NODARman said:
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##?
 
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.
 
NODARman said:
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?
 
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.
 
NODARman said:
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?
 
"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.
 
NODARman said:
"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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