Verrine said:Homework Statement
A bullet goes through a block and covers a distance D with a velocity of V0=200m/s. Calculate the velocity that the same bullet has AFTER passing through another block of the same material but which has half the width of the initial block.
Homework Equations
The Attempt at a Solution
Ok sure. I have an attempt, the problem is that my answer doesn't coincide with the answer in the textbook.PeroK said:You have to make an attempt at solving this problem yourself. It is your homework, after all! That said, are you sure that's all the information you have?
Verrine said:Ok sure. I have an attempt, the problem is that my answer doesn't coincide with the answer in the textbook.
For the first block I used the fact that the variation of kinetic energy is equal with the work force, so: 0-m*V0^2/2 =Fr *d => Fr=m*V0^2/2d
For the second block: mV^2/2 - mV0^2/2= Fr * d/2, so V=244 m/s
But the answer is wrong ;(. And yes, this is all they give me.
PeroK said:Is ##D## the thickness of the block?
Are you to assume that the bullet goes through the second block after the first? Or, that it goes through the second block in a separate experiment?
Where did you get the final velocity of ##0## from?
I'm not sure I follow your calculations, but you seem to have the bullet speeding up as a result of going through the second block!?
gleem said:perhaps you should post the question as the books state it.
Verrine said:The book is not in English. I had to translate it. But trust me, this is what it says.
Yes! This is the answer! Can you please explain how you did it? :DPeroK said:What's the answer? We might then work out what the question is!
My guess is ##141m/s##.
Verrine said:Yes! This is the answer! Can you please explain how you did it? :D
Verrine said:It goes through the second block in a separate experiment. To be sincere, I don't know myself what D actually stands for. What they say is that the bullet passes through the first block and it covers a certain distance, which is D. I also saw the D as the thickness of the block, but I am not entirely sure if that is also how they see it :(. Sorry
For the calculus: we had mV^2/2 - mV0^2/2=Fr*d/2 => mV^2/2=mV0^2/2 + mV0^2/2 * d/2 => V^2=6V0^2 /4 => v=244 m/s
PeroK said:You need to think about why you got a final velocity greater than the initial velocity here.
Verrine said:I simply tried to solve the equations with what formulas I knew... I actually have no idea what is going on here. The problem is confusing and I don't really see the connection between the velocity of the bullet and the thickness of the block. Sorry. If you could explain it step by step, I would be really grateful.
Yes, I know about Kinetic energy and work.PeroK said:There are two ways to solve this. The better way is to look at the energy. Do you know about Kinetic Energy and the Work done by a Force?
If not, then you can also use the equations of motion.
Verrine said:Yes, I know about Kinetic energy and work.
I think I got it. If I do mv^2/2 -mv0^2/2=Fr*d/2, then mv^2/2 =mv0^2/4. So v=v0 /rad 2 =v0*rad2 /2 =141m/s :):) Thank you . :DPeroK said:What's the initial KE of the bullet? And what force does work to stop it in this case?
Think about the signs there. Should V be more or less than V0? Will it gain or lose KE?Verrine said:For the second block: mV^2/2 - mV0^2/2= Fr * d/2