Does the force applied to an object depend on its mass and velocity?

[Karan Punjabi]

I'm in a whole confusion that I want to understand momentum. If i consider object having mass 600 kg moving with a velocity 1 m/s and if another object with mass 60 kg moving with velocity 10 m/s the we say both objects have same momentum . so is it so like 60 kg mass has 10 times less inertia than 600 kg mass so if that object has 10 times more velocity then both mass and velocity balance each other and it is eqaully hard to stop both of them? Am I right?

 

[S.G. Janssens]

Am I right?

Yes, I think you are, provided that by "equally hard" you mean that the impulse (i.e. the product of force and duration of application of said force, assuming it to be constant) required to bring the objects to a stop has the same magnitude for both objects.

 

[Karan Punjabi]

Yes, I think you are, provided that by "equally hard" you mean that the impulse (i.e. the product of force and duration of application of said force, assuming it to be constant) required to bring the objects to a stop has the same magnitude for both objects.

Yes by EQUALLY HARD i mean that only...so that's why we say mass × velocity

 

[S.G. Janssens]

Then from what you wrote I don't see any problem with your understanding.

 

[Karan Punjabi]

Then from what you wrote I don't see any problem with your understanding.

Yeah... I am understanding but i don't know why i am not satisfied.

 

[HallsofIvy]

Yes, although you also need to consider the potential energy of the objects. A 60 kg object moving at100 m/s has 300000 joules kinetic energy and a 600 kg object moving at 10 m/s has 30000 joules kinetic energy. The difference comes up in what you mean by "equally hard" as Krylov suggested. Same force and (time) duration of application but the force must be applied over a greater distance for the 60 kg object.

 

[S.G. Janssens]

For me, satisfaction often comes from solving problems, for example by doing some well-designed exercises. Maybe you tried that already?

 

[Karan Punjabi]

For me, satisfaction often comes from solving problems, for example by doing some well-designed exercises. Maybe you tried that already?

Yes i did some too

 

[Karan Punjabi]

Yes, although you also need to consider the potential energy of the objects. A 60 kg object moving at100 m/s has 300000 joules kinetic energy and a 600 kg object moving at 10 m/s has 30000 joules kinetic energy. The difference comes up in what you mean by "equally hard" as Krylov suggested. Same force and (time) duration of application but the force must be applied over a greater distance for the 60 kg object.

I didn't caught you in the last point that force should be applied over a greater distance?

 

[S.G. Janssens]

I didn't caught you in the last point that force should be applied over a greater distance?

I think what @HallsofIvy justly pointed out, is that there are different ways to quantify how hard it is to stop the objects. In terms of impulse, stopping the objects is equally hard, but of course it is energetically more expensive to stop $m_1$ because initially the kinetic energy $T_1$ of object 1 ($60$ kg, $100$m/s) is ten times the kinetic energy $T_2$ of object 2 ($600$kg, $10$m/s).

Since work done equals (constant) force times distance, and the change in kinetic energy is equal to the work done, you find that the distances traveled by the two masses since the moment the force was applied, are
$$
d_1 = \frac{T_1}{F} > \frac{T_2}{F} = d_2
$$
so indeed $F$ must be applied over a greater distance for object 1.

 

[Karan Punjabi]

I think what @HallsofIvy justly pointed out, is that there are different ways to quantify how hard it is to stop the objects. In terms of impulse, stopping the objects is equally hard, but of course it is energetically more expensive to stop $m_1$ because initially the kinetic energy $T_1$ of object 1 ($60$ kg, $100$m/s) is ten times the kinetic energy $T_2$ of object 2 ($600$kg, $10$m/s).

Since work done equals (constant) force times distance, and the change in kinetic energy is equal to the work done, you find that the distances traveled by the two masses since the moment the force was applied, are
$$
d_1 = \frac{T_1}{F} > \frac{T_2}{F} = d_2
$$
so indeed $F$ must be applied over a greater distance for object 1.

Ohk...yeah that's a matter of kinetic energy i know...but still the force applied will be same