Momentum paradox: Why can't we write it as p=m+v ?

[shivaprasadvh]

You forgot to multiply it by 1/m

Oh you it to be multiplied by 1/m. Thanks for that...

 

[A.T.]

Kinetic Energy and momentum are related.

Everything is "related" somehow. The question was about specifically about momentum, and why it has the formula mv.

 

[jbriggs444]

So like kinetic energy is dependent on equation mv which we call momentum, say speed and velocity are also dependent on equation of mv. So KE was just one example.

For a fixed non-zero momentum (mv) I can have any speed I like (greater than zero), any velocity I like (other than zero) and any energy I like (greater than zero).

So does it really make sense to say that any of those quantities are dependent on mv?

[Thanks to @Hemant for the correction on allowed speeds]

 

[Dale]

Hence the final velocity with which it hits the ground is more

Actually, since you dropped them from the same height the final velocity is the same.

 

[shivaprasadvh]

Actually, since you dropped them from the same height the final velocity is the same.

Got it final velocity is same but as the mass is different so the depth with which each penetrates is less and more.

 

[shivaprasadvh]

Everything is "related" somehow. The question was about specifically about momentum, and why it has the formula mv.

Through observations, there are several effects which vary with mass and vary with velocity but are the same if the product mv is same. This is observed by physicist. Then they termed this multiplication mv as momentum. It did not happen that momentum = mass * velocity was derived first and then they observed.

 

[shivaprasadvh]

For a fixed non-zero momentum (mv) I can have any speed I like (other than zero), any velocity I like (other than zero) and any energy I like (other than zero).

So does it really make sense to say that any of those quantities are dependent on mv?

For a fixed non zero momentum = Does this mean the mass and velocity of the object remains same throughout or mass and velocity vary such a way that their momentum remains constant?

 

[jbriggs444]

For a fixed non zero momentum = Does this mean the mass and velocity of the object remains same throughout or mass and velocity vary such a way that their momentum remains constant?

You pick a momentum. I am free to pick a mass and a velocity whose product is equal to that momentum.

 

[sysprog]

The repeated addition thing is not the definition of multiplication. It is the easiest way to teach multiplication to little children who only know addition. The repeated addition thing only even works for multiplication by integers.

And that only if they're non-zero integers.

 

[A.T.]

Through observations, there are several effects which vary with mass and vary with velocity but are the same if the product mv is same.

Which effects specifically?

 

[sophiecentaur]

For a fixed non zero momentum = Does this mean the mass and velocity of the object remains same throughout or mass and velocity vary such a way that their momentum remains constant?

Not sure how you were intending to vary the mass of something . You could throw some away, I guess but then the total momentum would be unchanged (the same as their centre of mass. The fact that you can write a valid equation doesn't mean that it's possible to experiment with varying all the quantities in the equation.

 

[sysprog]

For a fixed non-zero momentum (mv) I can have any speed I like (greater than zero), any velocity I like (other than zero) and any energy I like (greater than zero).

So does it really make sense to say that any of those quantities are dependent on mv?

[Thanks to @Hemant for the correction on allowed speeds]

The kinetic energy of an object is proportional to the mass of the object and to the square of the object's velocity:

$p=mv$ so $v=\frac p m$, and $K=\frac 1 2mv^2$, so $K=\frac 1 2m(\frac p m)^2=\frac {p^2} {2m}$.

That means that for a given fixed non-zero momentum, just as a greater mass means a lesser velocity, and a greater velocity means a lesser mass, the kinetic energy 'more sensitively' depends on the mass, and the mass 'less sensitively' depends on the kinetic energy.

 

[jbriggs444]

$p=mv$ so $v=\frac p m$, and $K=\frac 1 2mv^2$, so $K=\frac 1 2m(\frac p m)^2=\frac {p^2} {2m}$.

That means that for a given fixed non-zero momentum, just as a greater mass means a lesser velocity, and a greater velocity means a lesser mass, the kinetic energy 'more sensitively' depends on the mass, and the mass 'less sensitively' depends on the kinetic energy.

I am not getting that. You've shown that, for fixed momentum, kinetic energy is inversely proportional to mass. Surely that means that each is equally sensitive to the other.

 

[sysprog]

I am not getting that. You've shown that, for fixed momentum, kinetic energy is inversely proportional to mass. Surely that means that each is equally sensitive to the other.

That's true of mass and velocity $-$ half the mass means twice the velocity, and half the velocity means twice the mass $-$ but half the mass means four times the kinetic energy (i.e. kinetic energy is 'more sensitive' to decreases in mass than velocity is).

 

[shivaprasadvh]

Not sure how you were intending to vary the mass of something . You could throw some away, I guess but then the total momentum would be unchanged (the same as their centre of mass. The fact that you can write a valid equation doesn't mean that it's possible to experiment with varying all the quantities in the equation.

If a force is applied for a body with mass 1 kg in such a way that it gets an initial velocity of v m / s. Now if the same force is applied to a body mass with 5 kg then what would happen to the initial velocity. Will it be v m/s or more than that or less than that. The initial velocity achieved will be less than v m / s. So it would happen 1 kg will travel more distance than 5 kg. So we can say the KE for first object is more than KE for second object. So here KE depends on variables m and v.

 

[shivaprasadvh]

Which effects specifically?

KE is the effect by product of m and v.

If a force is applied for a body with mass 1 kg in such a way that it gets an initial velocity of v m / s. Now if the same force is applied to a body mass with 5 kg then what would happen to the initial velocity. Will it be v m/s or more than that or less than that. The initial velocity achieved will be less than v m / s. So it would happen 1 kg will travel more distance than 5 kg. So we can say the KE for first object is more than KE for second object. So here KE depends on variables m and v.

If I want to give an initial velocity of 10 m / s by applying force F, to an object which is 5 kg. Now if I apply force F to an object with 10 kg will same initial velocity is achieved? (Conditions of friction, direction of force in both the cases are same)?

 

[A.T.]

Through observations, there are several effects which vary with mass and vary with velocity but are the same if the product mv is same.

Which effects specifically?

KE is the effect by product of m and v.

But KE is not always the same, when the the product mv is the same. So you have not answered my question at all.

If a force is applied for a body with mass 1 kg in such a way that it gets an initial velocity of v m / s. Now if the same force is applied to a body mass with 5 kg then what would happen to the initial velocity. Will it be v m/s or more than that or less than that. The initial velocity achieved will be less than v m / s. So it would happen 1 kg will travel more distance than 5 kg. So we can say the KE for first object is more than KE for second object. So here KE depends on variables m and v.

If I want to give an initial velocity of 10 m / s by applying force F, to an object which is 5 kg. Now if I apply force F to an object with 10 kg will same initial velocity is achieved? (Conditions of friction, direction of force in both the cases are same)?

1) "Same force is applied" doesn't say much. You have to specify if it is applied over the same time or over the same distance.

2) "More X means more Y" doesn't say much. You have to specify the exact relationship, not just talk about more or less.

3) Momentum is a vector while KE is a scalar. The total momentum can be zero, while the total KE is some positive value. So the simplistic relationship you are assuming based on a single body doesn't hold in general.

I have no idea why you are so desperate to confuse momentum and KE. It doesn't help in answering the OPs question about the significance of mv at all.

 

[shivaprasadvh]

But KE is not always the same, when the the product mv is the same. So you have not answered my question at all.1) "Same force is applied" doesn't say much. You have to specify if it is applied over the same time or over the same distance.

2) "More X means more Y" doesn't say much. You have to specify the exact relationship, not just talk about more or less.

3) Momentum is a vector while KE is a scalar. The total momentum can be zero, while the total KE is some positive value. So the simplistic relationship you are assuming based on a single body doesn't hold in general.

I have no idea why you are so desperate to confuse momentum and KE. It doesn't help in answering the OPs question about the significance of mv at all.

I am not desperate. I am explaining why it is not m+v and mv.

1. I have mentioned in the bracket condition Conditions of friction, direction of force in both the cases are same.
2.More or less is happening because of the relation with the factor mass and velocity.
3. I am not confusing momentum and KE. KE is outcome of the factor m and v. So product of mv i.e. momentum has an impact on KE.

 

[jbriggs444]

That's true of mass and velocity $-$ half the mass means twice the velocity, and half the velocity means twice the mass $-$ but half the mass means four times the kinetic energy (i.e. kinetic energy is 'more sensitive' to decreases in mass than velocity is).

Well, let us go back to see what your formulas have to say about that.

$p=mv$ so $v=\frac p m$, and $K=\frac 1 2mv^2$, so $K=\frac 1 2m(\frac p m)^2=\frac {p^2} {2m}$.

Look at that last equation:$$K=\frac{p^2}{2m}$$
Double the kinetic energy and you halve the mass.
Double the mass and you halve the kinetic energy.

For fixed momentum... Half the mass means twice the velocity. But half the mass only means two times the kinetic energy. You get a times 4 factor for velocity but a divide by two factor for the mass

Thanks to @etotheipi for catching my earlier error.

 

[A.T.]

So product of mv i.e. momentum has an impact on KE.

The product m²v also "has an impact on KE". Why is momentum defined as mv and not m²v?

 

[etotheipi]

For fixed momentum... Half the mass means four times the velocity. But half the mass only means two times the kinetic energy. You get a times 4 factor for velocity but a divide by two factor for the mass.

I think this should be for fixed momentum, half the mass means double the velocity ($v = p/m$) and half the mass means double the kinetic energy ($T = p^2/2m$)

 

[sysprog]

I should have said that halving the mass and doubling the velocity leaves momentum constant but doubles the kinetic energy.

 

[sysprog]

If I want to give an initial velocity of 10 m / s by applying force F, to an object which is 5 kg. Now if I apply force F to an object with 10 kg will same initial velocity is achieved? (Conditions of friction, direction of force in both the cases are same)?

$F=ma$, so for 5 kg and 10 m/s² the force is 50 N, and the velocity of the 5 kg after 1 second would be 10 m/s. If the mass were 10 kg the same 50 N force would produce an acceleration of 5 m/s², and the velocity of the 10 kg after 1 second would be 5 m/s.

$K=\frac 1 2 mv^2$, so for 5 kg at 10 m/s the kinetic energy would be 250 J , and for 10 kg at 5 m/s the kinetic energy would be 125 J.

$p=mv$, so for either 5 kg at 10 ms/s or 10 kg at 5 m/s the momentum would be the same: 50 kg$\cdot$m/s.

 

[Paul Colby]

For what it's worth the early history of momentum is discussed here. It might be instructive to read what people thought of the concept before it was completely understood.

https://en.wikipedia.org/wiki/Momentum#History_of_the_concept