Lead pellet shot vertically into a clay block

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The discussion revolves around calculating the final velocity and distance moved by a clay block after being struck by a lead pellet. The final velocity was initially calculated to be approximately 10.5 m/s, but there are concerns about energy loss during the impact, which complicates the problem. Participants clarify that the bullet remains embedded in the clay block, and they discuss the assumption that the block will experience gravitational acceleration after the impact. There is also a suggestion to use a force diagram to analyze the forces involved. Ultimately, the conversation emphasizes the importance of understanding energy transfer and the assumptions made in the problem.
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Homework Statement


image.jpg


Homework Equations


PCM ∑pi=∑pf
v2=u2+2as

The Attempt at a Solution


First, using the PCM equation, I found the final velocity of the block, which is about 10.5 ms-1.
Then by using the SUVAT formula, i would be able to find the distance moved by the block, but I don't have time nor acceleration. Any hints?
 
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Janiceleong26 said:

Homework Statement


View attachment 90608

Homework Equations


PCM ∑pi=∑pf
v2=u2+2as

The Attempt at a Solution


First, using the PCM equation, I found the final velocity of the block, which is about 10.5 ms-1.
How do you figure this? Won't the block with the embedded bullet rise until gravity causes it to fall back to earth?
Then by using the SUVAT formula, i would be able to find the distance moved by the block, but I don't have time nor acceleration. Any hints?

Maybe you need a different SUVAT formula.
 
Sorry If i sound ignorant, because I'm a chemist not a physicist...but I know some. Is it assumed that the clay block takes all of the impact energy? I mean, do you have to account for energy loss due to the pellet entering, compressing, and displacing the clay matter? This energy transfer would realistically create a smaller rise than modeled if this loss isn't accounted for. But, it would also make the problem stupid difficult to solve...so I imagine there is an assumption that energy isn't lost due to this? Or am I making this over complicated?

If losses aren't to be considered, try finding a way to relate the forces where they balance each other. Maybe start with a force diagram of all forces present. Find the relationship of transferring energy, and see how the upward force at the moment of impact compares to the gravitational force pulling down on the block.

P.S. Haven't did problems like this is a LONG time, so if my rambling doesn't make sense, feel free to ignore it lol...I'm in the right book just not sure if I'm on the same page
 
Janiceleong26 said:
which is about 10.5 ms-1
Not quite right. The bullet stays in the clay.
Janiceleong26 said:
I don't have time nor acceleration
I think you can safely assume this is in Earth's surface level gravitational field.
SteamKing said:
Won't the block with the embedded bullet rise until gravity causes it to fall back to earth?
I assume Janice meant final in the sense of 'at the end of the impact phase'.
chad kimsey said:
do you have to account for energy loss due to the pellet entering, compressing, and displacing the clay matter?
During impact, certainly, which is why Janice did not use conservation of energy to get the 10.5 m/s.
 
haruspex said:
Not quite right. The bullet stays in the clay.

I think you can safely assume this is in Earth's surface level gravitational field.

I assume Janice meant final in the sense of 'at the end of the impact phase'.

During impact, certainly, which is why Janice did not use conservation of energy to get the 10.5 m/s.

Oh...so the mass of the block would now be 100g ? Ok, now I get v=10m/s exactly.
And by using v2=u2+2as, I'm able to get the answer which is 5.1m, but why do we assume that the acceleration of the block is g?
 
haruspex said:
Not quite right. The bullet stays in the clay.

I think you can safely assume this is in Earth's surface level gravitational field.

I assume Janice meant final in the sense of 'at the end of the impact phase'.

During impact, certainly, which is why Janice did not use conservation of energy to get the 10.5 m/s.
I mean, I can't imagine this situation.. How does the clay block stays stationary in air without any "support" ?
 
Janiceleong26 said:
I mean, I can't imagine this situation.. How does the clay block stays stationary in air without any "support" ?
Before the impact? It only says it is stationary at the moment of impact. Maybe it was thrown up in the air and the bullet struck it at the highest point. It really doesn't matter how it came to be.
Janiceleong26 said:
why do we assume that the acceleration of the block is g?
The block plus bullet are now in free fall.
 
haruspex said:
Before the impact? It only says it is stationary at the moment of impact. Maybe it was thrown up in the air and the bullet struck it at the highest point. It really doesn't matter how it came to be.

The block plus bullet are now in free fall.
I see I see, ok thank you !
 

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