Physics problem - Momentum and energy

[Josep767]

Homework Statement

A bullet passes through a ballistic block and making an inelastic collision

Relevant Equations

Energy equations, parabolic equations(?), momentum equations.

Does anyone know how to solve this problem? I translate the full statement:

A 200g bullet is shot against a ballistic block passing it through and making an inelastic collision. The initial speed of the bullet is 200meters/second, the mass of the block is 3kg and the string that holds the block measures 2 meters. The bullet is at 1m of the floor and travels 5 meters in the horizontal axis.

a) What is the speed of the bullet and the block right after the collision?
b) What is the lost energy in the collision?
c) What is the maximum angle that makes the block ?

I don't require the full answer, just an algebraic proposal is enough

(sorry for my english, I'm not native speaker)

Thank you, Josep!

 

[PeroK]

You'll need to do thw work yourself. My proposal is to think about momentum.

 

[Josep767]

Well, I've been thinking about the problem for more than 2h, this is my last resort

 

[PeroK]

Well, I've been thinking about the problem for more than 2h, this is my last resort

And you have no ideas?

 

[WWGD]

Just curious, is this some regional language from Spain, maybe Catalan?

 

[Josep767]

Just curious, is this some regional language from Spain, maybe Catalan?

Yes! It is catalan :)!

 

[Josep767]

And you have no ideas?

Well, I suposed that when the bullet leaves the block it makes a parabolic movement from Xo(initial position)=0m and Xf(final position)=5m and Yo(initial height)=1m and Yf(final height)=0m, but the statement doesn't confirm that so I'm not 100% sure

 

[kuruman]

Well I suposed that when the bullet leaves the block it makes a parabolic movement, but the statement doesn't confirm that so I'm not 100% sure

I think that's a good idea even if the statement does not confirm it. Without the speed of the bullet after traversing the block you will not be able to solve this problem.

 

[Josep767]

I think that's a good idea even if the statement does not confirm it. Without the speed of the bullet after traversing the block you will not be able to solve this problem.

Okey, now I am solving the excercise assuming that, once i finish it I'll post it here and we can comment the results and if you agree or not, ty

 

[kuruman]

Okey, now I am solving the excercise assuming that, once i finish it I'll post it here and we can comment the results and if you agree or not, ty

Yes, be sure you show your workings, not just the numerical answers.

 

[Josep767]

Yes, be sure you show your workings, not just the numerical answers.

In this kind of excercise I have to show all the steps and explain them, the only problem is that I'm doing it in my language so I'll have to translate it.

 

[kuruman]

Just explain the meanings of the symbols you use and show all the relevant equations. We can see from the use of the equations whether you did it right or not. We will not be grading your effort.

 

[PeroK]

Well, I suposed that when the bullet leaves the block it makes a parabolic movement from Xo(initial position)=0m and Xf(final position)=5m and Yo(initial height)=1m and Yf(final height)=0m, but the statement doesn't confirm that so I'm not 100% sure

It doesn't need to confirm that.

That gives you a start. With that assumption, what can you say about the bullet after it leaves the block? Hint: projectile motion.

 

[Josep767]

Okey so what I've done is:
I assumed a parrabolic shot from initial X=0 and final X=5, initial Y=1 and final Y=0
I calculated the time, 0.45secs and calculated de horitzontal speed: 11.06m/s
Then I applied the momentum formula
mb(mass of bullet) and mm(mass of the block)
mb·vb+mm·vm(which is 0) =mb·v'b+mm*v'm
I isolated v'm and it gave me 12.56 m/s
To calculate the loss of energy I calculated the initial energy of the bullet(before the collision) which is 4000J
and the final energy of each body with their respective speeds (12.23J for the bullet and 236.6J for the block)
I made the difference between the energies and that gave me the losses.
For the angle I calculated: 0.5mm*v'm^2=m*g*h but the height gave me around 8m, knowing that the length of the string is 2m, wouldn't it do a loop?

 

[PeroK]

View attachment 251772

Okey so what I've done is:
I assumed a parrabolic shot from initial X=0 and final X=5, initial Y=1 and final Y=0
I calculated the time, 0.45secs and calculated de horitzontal speed: 11.06m/s
Then I applied the momentum formula
mb(mass of bullet) and mm(mass of the block)
mb·vb+mm·vm(which is 0) =mb·v'b+mm*v'm
I isolated v'm and it gave me 12.56 m/s
To calculate the loss of energy I calculated the initial energy of the bullet(before the collision) which is 4000J
and the final energy of each body with their respective speeds (12.23J for the bullet and 236.6J for the block)
I made the difference between the energies and that gave me the losses.
For the angle I calculated: 0.5mm*v'm^2=m*g*h but the height gave me around 8m, knowing that the length of the string is 2m, wouldn't it do a loop?

I agree with your numbers - the ones that I've checked.

Yes, the ratio of the masses is only 1/15 and I thought the block would go very fast.

Here's how to check a problem like this in your head(!)

Note that to travel 5m the bullet is not going very fast, so we could estimate this problem by imagining it lost all its momentum in the collision and got stuck in the block. This should give a similar result.

The ratio of the masses is only 1/15, so the block will go approx 1/16 th of the initial speed of the bullet. That's about 12.5 m/s. (200/16)

The block would to the same height if it went straight up: 1.25s (12.5/10) at an average of 6.25 m/s (half of 12.5) is about 8m.

My guess is that the mass of the bullet should be $20g$. That changes things.

 

[Josep767]

Now I need to calculate the angle that the block would do, but I'm pretty much stuck.

 

[PeroK]

Now I need to calculate the angle that the block would do, but I'm pretty much stuck.

Try $20g$ for the bullet.

 

[PeroK]

I do not think the text says that the bullet leaves the block. The collision is said inelastic which usually means that the colliding bodies move together after collision. A ballistic pendulum is used to determine the momentum of a bullet in the way that the bullet stays inside the block, and the maximum height of the block after collision is detected.

That may have been lost in transalation. In any case, the bullet is far too heavy.

 

[jbriggs444]

I was going to suggest the same. A quick trip to Google and

Depending on the gun, the mass of a bullet usually ranges between 0.02 kilograms and 0.04 kilograms. The mass of a bullet depends on the caliber and type of gun used. A bullet can be fired from a revolver, pistol, rifle, machine gun or other weapon.

 

[Josep767]

Ye I know, the bullet weight is 0.2kg i checked it twice because it seemed to heavy for me. If you translate the excercise with google translator you can see that the bullet goes through the block.

 

[PeroK]

Ye I know, the bullet weight is 0.2kg i checked it twice because it seemed to heavy for me. If you translate the excercise with google translator you can see that the bullet goes through the block.

Try with $20g$ anyway. With $200g$ you won't get an answer.

 

[DEvens]

The bullet has to go through the block. The string holding the block is 2 meters, and the bullet travels 5 meters in the horizontal direction. The block has only a maximum possible horizontal travel of +/- 2 meters.

Also, if it travel 5 meters in 0.45 seconds, how fast is it moving?

And 200 grams is way heavy for a bullet. It's kind of heavy for a crossbow bolt. But if it is the value in the problem, it is the value in the problem.

 

[PeroK]

The horizontal distance of 5m may be just there to confuse you. It cannot be relevant. You are not asked to allow for drag, and in the collision only the horizontal momentum matters.

I assumed that was after it exited the block: $5m$ horizontal and $1m$ vertical determines the exit speed of the bullet. But, it does seem odd to add that to the problem.

 

[haruspex]

I assumed that was after it exited the block: $5m$ horizontal and $1m$ vertical determines the exit speed of the bullet. But, it does seem odd to add that to the problem.

Ah, yes, you must be right. It all makes sense then. In particular, it means that despite a mass of 200g, not enough momentum is transferred to make the bob loop-the-loop.

 

[PeroK]

Ah, yes, you must be right. It all makes sense then. In particular, it means that despite a mass of 200g, not enough momentum is transferred to make the bob loop-the-loop.

See post #15.

 

[kuruman]

Try with $20g$ anyway. With $200g$ you won't get an answer.

I agree. With 20 g the answer is much more reasonable and the block rises to a height that is considerably less than the length of the string.