What is the impulse exerted by a bullet on a block?

AI Thread Summary
The discussion revolves around calculating the impulse exerted by a bullet on a block after the bullet passes through it. The bullet's mass is 17.6 g, and it travels at an initial speed of 565 m/s, exiting at 524 m/s. Participants clarify that impulse is equivalent to the change in momentum, which can be calculated using the bullet's mass and its initial and final velocities. There is confusion regarding whether the mass of the block should be included in the impulse calculation, but it's established that only the bullet's mass is relevant since the block does not stick to the bullet. The conversation emphasizes that impulse can be negative depending on the chosen sign convention, as impulse is a vector quantity.
NasuSama
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Homework Statement



A bullet of mass m = 17.6 g traveling horizontally at speed vo = 565 m/s strikes a block of mass M = 678 g sitting on a frictionless, horizontal table. This time, however, it comes out the other side of the block at speed v = 524 m/s.

Calculate J, the magnitude of the impulse exerted by the bullet on the block.

Homework Equations



Hm...

Impulse = Momentum = Ft = mv_f² - mv_i²

The Attempt at a Solution



I think that the impulse is same as the momentum. I think that is...

Impulse = (M + m)v_f² - mv_i²

But it seems like I'm on the wrong path.
 
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The momentum is mV. When a force F acts for time Δt on a point mass m the momentum will change and the change of momentum equals to the impulse of the force: FΔt=mVf-mVi

ehild
 
ehild said:
The momentum is mV. When a force F acts for time Δt on a point mass m the momentum will change and the change of momentum equals to the impulse of the force: FΔt=mVf-mVi

ehild

Does this means that only the mass of the bullet takes in account of this situation?
 
It means that the momentum is proportional to the velocity instead of the square of the velocity as you wrote.

ehild
 
NasuSama said:

Homework Statement



A bullet of mass m = 17.6 g traveling horizontally at speed vo = 565 m/s strikes a block of mass M = 678 g sitting on a frictionless, horizontal table. This time, however, it comes out the other side of the block at speed v = 524 m/s.

Calculate J, the magnitude of the impulse exerted by the bullet on the block.

Homework Equations



Hm...

Impulse = Momentum = Ft = mv_f - mv_

The Attempt at a Solution



I think that the impulse is same as the momentum. I think that is...

Impulse = (M + m)v_f - mv_i

But it seems like I'm on the wrong path.

This is the revised work.
 
ehild said:
It means that the momentum is proportional to the velocity instead of the square of the velocity as you wrote.

ehild

You didn't answer my question. Does the mass of the bullet takes in account of the momentum or the mass of both objects?
 
Lacking approach and explanation here. Not helpful. -__-
 
NasuSama said:
Lacking approach and explanation here. Not helpful. -__-

Sorry.


ehild
 
Waiting for other response. Otherwise, it seems that I need to answer myself.
 
  • #10
Did you try using I = pfinal - pinitial?
 
  • #11
bdh2991 said:
Did you try using I = pfinal - pinitial?

Yes, I did, but it seems that I made the wrong approach. I believe that is...

(m + M)v_f² - (m + M)v_i²
= - (m + M)v_i²
 
  • #12
It doesn't state that part of the mass of the block sticks to the bullet so you should just be able to use the bullet mass and the final and inital velocities for change in momentum
 
  • #13
bdh2991 said:
It doesn't state that part of the mass of the block sticks to the bullet so you should just be able to use the bullet mass and the final and inital velocities for change in momentum

Then, it's mv_f - mv_i, where m is the mass of the small bullet.
 
  • #14
Then, this is the negative impulse. We have m(v_f - v_i)
 
  • #15
bdh2991 said:
It doesn't state that part of the mass of the block sticks to the bullet so you should just be able to use the bullet mass and the final and inital velocities for change in momentum

This is what I have:

I = mv_f - mv_i

If I only use the mass of the bullet as the part of this equation, then we have...

I = 0.0176 * (524 - 565) = -0.722

Is that true?
 
  • #16
NasuSama said:
This is what I have:

I = mv_f - mv_i

If I only use the mass of the bullet as the part of this equation, then we have...

I = 0.0176 * (524 - 565) = -0.722

Is that true?

All you were supposed to calculate was I = Δp , where Δp is linear momentum and I is impulse...

Your work seems correct to me. Mass of the block was only to confuse you. This is because from the given parameters , you cannot solve for deceleration in bullet and time period..
 
  • #17
sankalpmittal said:
All you were supposed to calculate was I = Δp , where Δp is linear momentum and I is impulse...

Your work seems correct to me. Mass of the block was only to confuse you. This is because from the given parameters , you cannot solve for deceleration in bullet and time period..

But it seems that the answer is not right. I entered it, but not right.
 
  • #18
NasuSama said:
But it seems that the answer is not right. I entered it, but not right.

Well consider right x-axis to be positive and left to be negative. (Choice of sign convention lay in user's choice , but then he has to stick to that convention throughout the problem.) So by Newton's third law , force which block exerts on bullet will be equal and opposite to force exerted by bullet on block. Since Δt is constant , so impulse exerted by block on bullet will be negative of that by bullet on block. Thus I=Δp , simply.

Why you say , this isn't correct ?
 
  • #19
sankalpmittal said:
Well consider right x-axis to be positive and left to be negative. (Choice of sign convention lay in user's choice , but then he has to stick to that convention throughout the problem.) So by Newton's third law , force which block exerts on bullet will be equal and opposite to force exerted by bullet on block. Since Δt is constant , so impulse exerted by block on bullet will be negative of that by bullet on block. Thus I=Δp , simply.

Why you say , this isn't correct ?

Based on what I know, impulse can't be negative. It's force times time. Even though impulse is same as the momentum, impulse can't have negative magnitude. I spoke with my instructor about this, and he said that impulse can't be negative.
 
  • #20
Adding into what I said before, impulse is Force times time. Though it's the same as momentum, the impulse can't be negative. That is what is different from the momentum.
 
  • #21
NasuSama said:
Adding into what I said before, impulse is Force times time. Though it's the same as momentum, the impulse can't be negative. That is what is different from the momentum.

Of course , impulse can be negative if you consider its direction based on your sign convention. Impulse is a vector quantity , and vector quantities can be negative , although its magnitude is always positive.

http://answers.yahoo.com/question/index?qid=20080113150901AAmY7fE
http://answers.yahoo.com/question/index?qid=20100303191234AAUuV5V
http://www.thestudentroom.co.uk/showthread.php?t=912889

Also consult professor google...
 

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