# Lead pellet shot vertically into a clay block

1. Oct 22, 2015

### Janiceleong26

1. The problem statement, all variables and given/known data

2. Relevant equations
PCM ∑pi=∑pf
v2=u2+2as

3. 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?

2. Oct 22, 2015

### SteamKing

Staff Emeritus
How do you figure this? Won't the block with the embedded bullet rise until gravity causes it to fall back to earth?
Maybe you need a different SUVAT formula.

3. Oct 22, 2015

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

4. Oct 23, 2015

### haruspex

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.

5. Oct 23, 2015

### Janiceleong26

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?

6. Oct 23, 2015

### Janiceleong26

I mean, I can't imagine this situation.. How does the clay block stays stationary in air without any "support" ?

7. Oct 23, 2015

### haruspex

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.

8. Oct 23, 2015

### Janiceleong26

I see I see, ok thank you !!