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

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Homework Help Overview

The problem involves a bullet of mass 17.6 g traveling at a speed of 565 m/s striking a block of mass 678 g on a frictionless table and exiting at a speed of 524 m/s. The objective is to calculate the impulse exerted by the bullet on the block.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the relationship between impulse and momentum, with some suggesting that impulse is equivalent to the change in momentum. There are questions about whether to consider only the bullet's mass or both masses in the momentum calculations. Some participants express confusion regarding the correct application of formulas and the implications of negative impulse.

Discussion Status

There is ongoing exploration of the problem, with various interpretations of the impulse calculation being discussed. Some participants have provided guidance on using the change in momentum formula, while others are questioning the assumptions made about the mass involved in the calculations. The discussion reflects a mix of attempts to clarify concepts and resolve misunderstandings.

Contextual Notes

Participants note that the problem does not specify whether part of the block's mass sticks to the bullet, leading to uncertainty about how to approach the impulse calculation. There is also mention of sign conventions affecting the interpretation of impulse as 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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