Question Regarding Impulse and momentum

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    Impulse Momentum
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Homework Help Overview

The discussion revolves around the effectiveness of rubber versus lead bullets in knocking down a bear, focusing on the concepts of impulse and momentum. Participants explore the implications of using bullets with the same momentum but different material properties.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants question the differences in behavior between rubber and lead upon impact, particularly regarding momentum and energy transfer. There are discussions about the implications of collision time and the resulting forces involved.

Discussion Status

The conversation is ongoing, with various interpretations being explored. Some participants have offered insights into the nature of momentum and its vector properties, while others are seeking clarification on specific concepts like change in momentum (Δp) and the effects of bouncing.

Contextual Notes

There is a noted confusion regarding the definitions and implications of momentum and impulse, as well as the assumptions about the mass of the bullets. Participants are also navigating language barriers, which may affect their understanding of the physics concepts discussed.

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Homework Statement


Which will be more effective for knocking a bear down,A Rubber bullet or a Lead bullet of same momentum?and WHY?


Homework Equations



Equation for impulse of a body= F x t = MVf-MVi (change in momentum)

The Attempt at a Solution



I have no clue so please tell me and the proper and explanatory reason as well
 
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What's the most obvious difference between what happens when a lump of rubber hits something hard and when a lump of lead (of similar mass) does?
 
haruspex said:
What's the most obvious difference between what happens when a lump of rubber hits something hard and when a lump of lead (of similar mass) does?

Lump of lead hurts :D
 
kashan123999 said:
Lump of lead hurts :D

:biggrin:

but suppose it's a wall …

walls don't hurt! :wink:​
 
The rubber will hurt more (assuming rubber can rebound)
Don't anthropormphise walls, they don't like it!
 
technician said:
The rubber will hurt more (assuming rubber can rebound)
Don't anthropormphise walls, they don't like it!

btw what is meant by "KNocking down" the bear? sorry i am a non native speaker

Let ME Put it again in a more coherent way,I need the explanation in view of IMPULSE of two objects,
THE first explanation our teacher told us,"The lead bullet will be preferable because the time of collision will be shorter than rubber bullet hence force will be more" As we know F=Δp/t hence time of collision is INVERSELY proportional to force,Lead bullet has a shorter time of collision thus it will be more preferable...Now another teacher told us,"As we know momentum of two objects are same,As lead bullet has more mass than rubber bullet,thus Its Velocity will be low,Conversly,Rubber bullet has less mass,thus its velocity will be large because P=mv → p/m=v,mass inversely proportional to velocity,Thus because of more velocity of the rubber bullet it will have MORE kinetic eNERGY so it will be able to knock down the bear...Now i know only one of it is right,but which one and WHY? please help me
 
hi kashan123999! :smile:
kashan123999 said:
btw what is meant by "KNocking down" the bear? sorry i am a non native speaker

it means causing the bear to fall to the ground

(ie, not because it's injured but because the force actually pushes it over)​

the second (energy) explanation is just plain wrong …

knocking the bear over is all about momentum, not energy :frown:​

the first (force) explanation is also just plain wrong …

knocking the bear over is all about momentum, not force :frown:

also, the first explanation isn't really an explanation, since it doesn't mention what ∆p is in each of the two cases​

what is ∆p ? :smile:
 
tiny-tim said:
hi kashan123999! :smile:


it means causing the bear to fall to the ground

(ie, not because it's injured but because the force actually pushes it over)​

the second (energy) explanation is just plain wrong …

knocking the bear over is all about momentum, not energy :frown:​

the first (force) explanation is also just plain wrong …

knocking the bear over is all about momentum, not force :frown:

also, the first explanation isn't really an explanation, since it doesn't mention what ∆p is in each of the two cases​

what is ∆p ? :smile:


sorry,i am a total retard at physics,btw momentum is same of both the objects,can you tell me the correct explanation and which is more preferable please
 
kashan123999 said:
… momentum is same of both the objects …

i didn't ask you about p,

i asked you about ∆p (the change in p)

anyway, you haven't answered my original question …

what happens to each object (lead and rubber) if they hit a wall?​
 
  • #10
I would expect that it depends on the mass of the bullets. Are we supposed to assume that both have the same mass, too?
 
  • #11
tiny-tim said:
i didn't ask you about p,

i asked you about ∆p (the change in p)

anyway, you haven't answered my original question …

what happens to each object (lead and rubber) if they hit a wall?​

Final momentum of both objects would be zero after striking,intially they would have same momentum as well x... I don't know whether i am right,not confident :D

tHE LEAd bullet will penetrate,while the other won't... Am I right sire?
 
  • #12
mfb said:
I would expect that it depends on the mass of the bullets. Are we supposed to assume that both have the same mass, too?

What if they have same mass? and what if they have different masses? what if rubber bullet has LESS mass than LEAD bullet?
 
Last edited:
  • #13
kashan123999 said:
Final momentum of both objects would be zero after striking …

nooo …

won't the rubber bounce? :wink:​
 
  • #14
tiny-tim said:
nooo …

won't the rubber bounce? :wink:​

Yes maybe,Just a touch little sir :D so ? where are you leading me btw...to critical thinking i guess :D
 
  • #15
kashan123999 said:
Yes maybe,Just a touch little sir :D

maybe?! :rolleyes:

rubber is famous for bouncing!​

btw, don't call me "sir" … i'm only a little goldfish! o:)

so ? where are you leading me btw...to critical thinking i guess :D

'fraid so!

ok, think … what is ∆p if it bounces? :smile:
 
  • #16
tiny-tim said:
maybe?! :rolleyes:

rubber is famous for bouncing!​

btw, don't call me "sir" … i'm only a little goldfish! o:)



'fraid so!

ok, think … what is ∆p if it bounces? :smile:

The Rubber will,yes it will bounce and the Δp = Final - original momentum=Final will be less than original, thus Delta P will be negative :approve::shy:


and well you are a sir to me,you seem to be a pretty mature Person,I am a teenage boy :D
 
  • #17
kashan123999 said:
Δp = Final - original momentum=Final will be less than original, thus Delta P will be negative :approve::shy:

no, momentum is a vector, so you must add (or subtract) it like a vector

a vector has a magnitude and a direction

suppose the speed of the rubber (mass 1 kg) initially is 50 m/s north, and finally is 45 m/s south …

what is the total change in momentum (p2 - p1) ?
 
  • #18
tiny-tim said:
no, momentum is a vector, so you must add (or subtract) it like a vector

a vector has a magnitude and a direction

suppose the speed of the rubber (mass 1 kg) initially is 50 m/s north, and finally is 45 m/s south …

what is the total change in momentum (p2 - p1) ?

5m/s north?
 
  • #19
Think of it like this:
When the rubber bounces there are 2changes of momentum:
First...when the rubber bullet is stopped
Second ...when the rubber bullet rebounds
These 2 changes in momentum produce a force on the bear/ wall
 
  • #20
kashan123999 said:
5m/s north?

no, 45 m/s south is minus 45 m/s north :wink:
 
  • #21
tiny-tim said:
no, 45 m/s south is minus 45 m/s north :wink:

:@:@:@ I don't know :(
 
  • #22
- the rubber bullet stops, the bear gets a kick in the original direction of motion of the bullet
- the bullet bounces back, the bear gets a kick in [which direction]?

And how can you add those two kicks?
 
  • #23
(just got up :zzz:)

what is 50 m/s north minus minus 45 m/s north ? :smile:
 
  • #24
tiny-tim said:
(just got up :zzz:)

what is 50 m/s north minus minus 45 m/s north ? :smile:


95 m/s north :@
 
  • #25
kashan123999 said:
95 m/s north :@

yes :smile: … and that's what we mean by vector addition …

we find the change in momentum by subtracting the 45 south vector from the 50 north vector …

since they're in opposite directions, that means we subtract minus 45 from 50, giving 95 (times the mass) as the total change in momentum
 
  • #26
tiny-tim said:
yes :smile: … and that's what we mean by vector addition …

we find the change in momentum by subtracting the 45 south vector from the 50 north vector …

since they're in opposite directions, that means we subtract minus 45 from 50, giving 95 (times the mass) as the total change in momentum

ahan yes.......... so :D where is my answer :PD:D:D:D:D:D::D::D
 
  • #27
when two bodies of rubber and of lead, with the same initial momentum mv, hit a wall,

what, in each case, is the change in momentum?
 
  • #28
tiny-tim said:
when two bodies of rubber and of lead, with the same initial momentum mv, hit a wall,

what, in each case, is the change in momentum?

don't know :D
 
  • #29
kashan123999 said:
don't know :D
Lead mass m with initial momentum mv, stopped dead by wall. What is its final momentum. So what is its change in momentum?
Rubber mass m with initial momentum mv, bounces back from wall at speed u. What is its final momentum? (Careful with the sign. Remember it's now traveling in the opposite direction.) So what is its change in momentum?
 
  • #30
haruspex said:
Lead mass m with initial momentum mv, stopped dead by wall. What is its final momentum. So what is its change in momentum?
Rubber mass m with initial momentum mv, bounces back from wall at speed u. What is its final momentum? (Careful with the sign. Remember it's now traveling in the opposite direction.) So what is its change in momentum?

Lead... Change in momentum= final - intial = 0-mv=-mv i guess? opposite to the direction of original propagation...

Rubber bullet... mu-mv = m(u-v) ??
 

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