Is impulse negligible in this problem?

[Rikudo]

Homework Statement

A bar is placed with a slight offset Lo on another identical bar that rests on a horizontal frictionless floor. Coefficient of friction between the bars is μ. Initially, both the bars are moving towards a wall with a velocity u. The wall is parallel to the front face of the bars. (See figure)

If collision of the bars with the wall is perfectly elastic , find final velocities of both the bars long time after collisions with the wall .

Relevant Equations

Impulse - momentum

When the upper block collides with the wall, the impulse to it will be:
$$ \int (-N+ f)dt = -2mu$$
Where N is the contact force from the wall.

Since N is really huge, the impulse from friction in the first equation above can be ignored, hence the equation become:
$$ \int -Ndt = -2mu$$

Meanwhile, the impulse to the lower one is going to be:
$$ \int -f dt = m(v-u)$$
Where $v$ is the velocity of the lower mass after the collision.

I don't think we can ignore the friction in this one. But if this is true, then I will not be able to find $v$.

What do you think?

 

[haruspex]

Homework Statement:: A bar is placed with a slight offset Lo on another identical bar that rests on a horizontal frictionless floor. Coefficient of friction between the bars is μ. Initially, both the bars are moving towards a wall with a velocity u. The wall is parallel to the front face of the bars. (See figure)

If collision of the bars with the wall is perfectly elastic , find final velocities of both the bars long time after collisions with the wall .
Relevant Equations:: Impulse - momentum

View attachment 313376
When the upper block collides with the wall, the impulse to it will be:
$$ \int (-N+ f)dt = -2mu$$
Where N is the contact force from the wall.

Since N is really huge, the impulse from friction in the first equation above can be ignored, hence the equation become:
$$ \int -Ndt = -2mu$$

Meanwhile, the impulse to the lower one is going to be:
$$ \int -f dt = m(v-u)$$
Where $v$ is the velocity of the lower mass after the collision.

I don't think we can ignore the friction in this one. But if this is true, then I will not be able to find $v$.

What do you think?

As I showed in the earlier thread, you have to consider what is known, the force or the impulse. If the force is known and bounded then, in the limit as the duration tends to zero, the impulse it delivers tends to zero. But if the impulse is known then, in that limit, the force is unbounded.

 

[Rikudo]

But, In my thread about a rocket ejecting mass, which you can find it here:
https://www.physicsforums.com/threads/a-rocket-ejecting-mass.1044968/

we do not ignore the small change in speed ($dv$), even though the force imparted to the rocket (by ejecting the mass) is small.

This is why I thought I should not ignore the change in the lower mass's velocity either; no matter how small it is.

 

[Orodruin]

Also, looking at the forces during collision is a red herring. The only relevant force analysis is after the collision (or collisions …).

 

[Halc]

The only relevant question seems to be: Is the velocity u small enough and the coefficient of friction between bars enough to prevent the lower bar from also colliding with the wall. If so, the problem is trivial. If the collision still occurs, then one has to actually do algebra, but the only thing that matters is the velocity at which that lower bar hits the wall.

 

[Chestermiller]

Is $\mu$ the coefficient of static friction or the coefficient of kinetic friction. If the former, what is the coefficient of kinetic friction? If the latter, what is the coefficient of static friction? What would happen if the coefficient of static friction were extremely high?

 

[haruspex]

But, In my thread about a rocket ejecting mass, which you can find it here:
https://www.physicsforums.com/threads/a-rocket-ejecting-mass.1044968/

we do not ignore the small change in speed ($dv$), even though the force imparted to the rocket (by ejecting the mass) is small.

This is why I thought I should not ignore the change in the lower mass's velocity either; no matter how small it is.

You have to look at what happens in the limit. In the rocket thread, in the limit, dm and dv are both zero, but what matters is the ratio of dm to dv, and that does not go to zero.
In this thread, you have $\int f.dt=m(v-u)$ and f is bounded, so in the limit $dt\rightarrow 0$, $v=u$.

 

[hutchphd]

I think one should assume there is no stiction ($\mu_s=\mu_k$).

If so, the problem is trivial.

No. Any relative motion of the blocks ($\mu finite$) will dissipate energy internally

One can use energy conservation, knowing that the collisions are elastic and finding the energy dissipated to internal degrees of freedom (i.e. the total work done by frictional forces). By how much do the blocks relatively displace? Indeed the bottom block may or may not bounce off the wall but any relative sliding makes $heat=2\mu N (sliding~displacement)$ regardless

 

[haruspex]

Also, looking at the forces during collision is a red herring. The only relevant force analysis is after the collision (or collisions …).

In the present case, yes, but there can be "impulsive friction". I wouldn't want to give @Rikudo the impression that one can always ignore forces that are not at the obvious point of impact.

 

[hutchphd]

Very good. Can those be characterized in a useful way?

 

[Orodruin]

No. Any relative motion of the blocks (μfinite) will dissipate energy internally

Still trivial under the given conditions.

 

[hutchphd]

Neither case is very difficult but neither is trivial for me. We can compare notes after OP solves.

 

[haruspex]

I think one should assume there is no stiction (μs=μk).

No, static friction does negligible work in practice, and none at all in the usual idealisation (namely, that kinetic friction takes over after an infinitesimal movement). It only creates a threshold which the applied force must reach.

Neither case is very difficult but neither is trivial for me. We can compare notes after OP solves.

The trick is to think about the two velocities immediately after the top bar has bounced. If the lower bar does not reach the wall then it is indeed trivial. However, being told that $l_0$ is very small implies we should assume it does reach the wall, so there is only the slightly more interesting case to consider.

 

[hutchphd]

I believe we each understand the physics. The thing that took a moment for me was to realize that the lower block will have slowed before its reflection from the wall. This actually makes the probem easier. Enough said.
Oh except to say "parts of the impulse are likely not well understood we will assume them small by fiat"

 

[Rikudo]

$$\int -mg \mu d(l) = \frac 1 2 m(v²-u²)$$
Where v is the speed of the lower block just before the 2nd collision.
$$v = \sqrt {u² - 2g\mu l_0}$$

So, there are two answers:
1. $v = 0$
if $u² ≤2g\mu l_0$

2. $v = \sqrt {u² - 2g\mu l_0}$
if $u² ≥2g\mu l_0$

Do I also need to think of the possibilities if $l_0$ is big/small?

 

[haruspex]

$$\int -mg \mu d(l) = \frac 1 2 m(v²-u²)$$
Where v is the speed of the lower block just before the 2nd collision.
$$v = \sqrt {u² - 2g\mu l_0}$$

So, there are two answers:
1. $v = 0$
if $u² ≤2g\mu l_0$

2. $v = \sqrt {u² - 2g\mu l_0}$
if $u² ≥2g\mu l_0$

Do I also need to think of the possibilities if $l_0$ is big/small?

I'm not disagreeing with your answer, but in your posted analysis of case 2 you have not discussed the continuing interaction between the blocks.