How to understand linear momentum?

[Javier Mont]

I understand for linear momentum the measure of the dificulty to take a body to the rest. But when I do the next exercises, I find that it requires less force for one body compared to the other with same linear momentum.
Can someone explain me what to understand in an intuitive way what to expect when I see a certain magnitud of linear momentum from a body?

 

[BvU]

Hello Javier,

Actually, PF culture asks you post questions like this in the homework forums. Never mind, I'll ask a mentor.
<Moderator's note: thread moved>

Can you check how long it takes for the two bodies to come to rest ? That's where a difference between the two cases comes in !

A line of thought that may help you:

Acceleration is the time derivative of velocity : $\vec a = {d\vec v\over dt}$. For a body of a given mass m, multiply this equation with m: $$m\vec a = m {d\vec v\over dt}$$On the left you have (according to Newton) $\vec F$ and on the right there is $d\vec p\over dt$ (since m is a constant) and you get $$\vec F ={ d\vec p\over dt}$$.

Now the cases in your picture: both bodies come to rest over a distance of 0.5 m and you calculate the force needed to achieve this (assuming it is constant). So you compare the products $\ F\ d\ \$. This product has the dimension of energy and in fact you compare the kinetic energies of the two bodies. Kinetic energy is ${1\over 2}mv^2$ so it's no wonder you results differ by a factor of 50 !

If you calculate the forces needed to stop the two bodies in the same time instead of over the same distance , the result may be more to your liking...

 

[sophiecentaur]

Now the cases in your picture: both bodies come to rest over a distance of 0.5 m and you calculate the force needed to achieve this (assuming it is constant). So you compare the products F d F d \ F\ d\ \ . This product has the dimension of energy and in fact you compare the kinetic energies of the two bodies. Kinetic energy is 12mv212mv2{1\over 2}mv^2 so it's no wonder you results differ by a factor of 50 !

If you calculate the forces needed to stop the two bodies in the same time instead of over the same distance , the result may be more to your liking...

Yes. It is important to realize the difference between Kinetic Energy and Momentum. Early Science failed to distinguish between the two and had a lot of problems of the sort that the OP quotes.

 

[Javier Mont]

Hello Javier,

Actually, PF culture asks you post questions like this in the homework forums. Never mind, I'll ask a mentor.
<Moderator's note: thread moved>

Can you check how long it takes for the two bodies to come to rest ? That's where a difference between the two cases comes in !

A line of thought that may help you:

Acceleration is the time derivative of velocity : $\vec a = {d\vec v\over dt}$. For a body of a given mass m, multiply this equation with m: $$m\vec a = m {d\vec v\over dt}$$On the left you have (according to Newton) $\vec F$ and on the right there is $d\vec p\over dt$ (since m is a constant) and you get $$\vec F ={ d\vec p\over dt}$$.

Now the cases in your picture: both bodies come to rest over a distance of 0.5 m and you calculate the force needed to achieve this (assuming it is constant). So you compare the products $\ F\ d\ \$. This product has the dimension of energy and in fact you compare the kinetic energies of the two bodies. Kinetic energy is ${1\over 2}mv^2$ so it's no wonder you results differ by a factor of 50 !

If you calculate the forces needed to stop the two bodies in the same time instead of over the same distance , the result may be more to your liking...

Thank you very much! You´re right! That difficult to take a body to rest (linear momentum) is the Force*time, no Force*distance. Thanks!

 

[Javier Mont]

Yes. It is important to realize the difference between Kinetic Energy and Momentum. Early Science failed to distinguish between the two and had a lot of problems of the sort that the OP quotes.

Interesting