Question Regarding Impulse and momentum

[kashan123999]

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

 

[haruspex]

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?

 

[kashan123999]

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

 

[tiny-tim]

Lump of lead hurts :D

but suppose it's a wall …

walls don't hurt! ​

 

[technician]

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

 

[kashan123999]

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

 

[tiny-tim]

hi kashan123999!

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 ​

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

knocking the bear over is all about momentum, not force

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 ?

 

[kashan123999]

hi kashan123999!

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 ​

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

knocking the bear over is all about momentum, not force

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 ?

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

 

[tiny-tim]

… 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?​

 

[mfb]

I would expect that it depends on the mass of the bullets. Are we supposed to assume that both have the same mass, too?

 

[kashan123999]

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?

 

[kashan123999]

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?

 

[tiny-tim]

Final momentum of both objects would be zero after striking …

nooo …

won't the rubber bounce? ​

 

[kashan123999]

nooo …

won't the rubber bounce? ​

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

 

[tiny-tim]

Yes maybe,Just a touch little sir :D

maybe?!

rubber is famous for bouncing!​

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

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

'fraid so!

ok, think … what is ∆p if it bounces?

 

[kashan123999]

maybe?!

rubber is famous for bouncing!​

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



'fraid so!

ok, think … what is ∆p if it bounces?

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 :shy:


and well you are a sir to me,you seem to be a pretty mature Person,I am a teenage boy :D

 

[tiny-tim]

Δp = Final - original momentum=Final will be less than original, thus Delta P will be negative :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) ?

 

[kashan123999]

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?

 

[technician]

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

 

[tiny-tim]

5m/s north?

no, 45 m/s south is minus 45 m/s north

 

[kashan123999]

no, 45 m/s south is minus 45 m/s north

:@:@:@ I don't know :(

 

[mfb]

- 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?

 

[tiny-tim]

(just got up :zzz:)

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

 

[kashan123999]

(just got up :zzz:)

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

95 m/s north :@

 

[tiny-tim]

95 m/s north :@

yes … 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

 

[kashan123999]

yes … 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

 

[tiny-tim]

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?

 

[kashan123999]

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

 

[haruspex]

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?

 

[kashan123999]

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) ??

 

[tiny-tim]

hmm …

have you done conservation of momentum?

if so, what is the equation for conservation of momentum for rubber bouncing off a wall on ice (ie, no foundations, so the wall can slide along the ice, but the wall is initially stationary)?

 

[haruspex]

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) ??

I was trying hard not to confuse you on this. You can think in terms of speeds or in terms of velocities. Because all the movement is in the same straight line, but in opposite directions, the difference is how the sign is handled. In my previous post I defined u as a speed, but you seem to be using velocities, so I'll stick with that.
If u and v are velocities measured in the same direction then the momenta are mu and mv, and the change in momentum is mu-mv, as you wrote. But note that u will be negative. So if you compare the magnitude of -mv (change in momentum for the lead) with that of m(u-v) (change in momentum for the rubber), which is larger?

 

[kashan123999]

I don't know :( i am confused as hell,Tommorow is my physics exam,please tell me the answer,i can develop critical skills later on :D I am a pakistani so that really makes me a rote learner and far far behind the world sorry :D so,what is the detailed answer?

 

[tiny-tim]

have you done conservation of momentum?

 

[kashan123999]

have you done conservation of momentum?

yep,initial momentum = final :@ change in momentum = 0

 

[tiny-tim]

yep,initial momentum = final

ok, so try applying total initial momentum = total final momentum to rubber bouncing off a wall on ice (ie, no foundations, so the wall can slide along the ice, but the wall is initially stationary)