Collision and time to stop a block

In summary: Can you see where there may be some concerns?You seem to be equating the time part of the specific impulse with the time it takes the block to stop, but the impulse force is not applied during the entire motion of the block, and you seem to have equated the impulse force to the friction. You also don't account, in Q1, for the fact the wall is delivering 2p momentum to the right at each collision. In short: what is wrong with using the impulse/momentum relation here is "everything".
  • #1
Happiness
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[Moderator note: Misplaced homework moved from General Physics, so no template shown]

Screen Shot 2016-07-23 at 2.32.53 am.png

Screen Shot 2016-07-23 at 2.33.16 am.png


For the part on finding the total time the block spend moving, the answer is ##\frac{v_0}{\mu g}## (attached below).

I get a different answer if I use the impulse-momentum theorem:

The total change in momentum ##\Delta p=-mv_0##. The force ##F## slowing the block down is always ##-\mu Mg##. Thus the time ##t=\frac{\Delta p}{F}=\frac{mv_0}{\mu Mg}##.

Q1: Why are the answers different?

Also, I believe the time for the case where the ball sticks to the block on the first hit should be ##\frac{\Delta p}{F}=\frac{mv_0}{\mu (M+m)g}##, because the force on the ball-block combined mass is ##-\mu (M+m)g##. Q2: Am I right?

Screen Shot 2016-07-23 at 2.33.43 am.png

Screen Shot 2016-07-23 at 2.33.59 am.png


EDIT: I've found the answer to my first question. It is not correct to use the total change in momentum ##\Delta p=-mv_0## to find time ##t=\frac{\Delta p}{F}##.
 
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  • #2
Well done.
You were using the wrong force in the impulse/momentum theory.

Per your second question: please show your reasoning.
 
  • #3
Simon Bridge said:
Well done.
You were using the wrong force in the impulse/momentum theory.

Per your second question: please show your reasoning.
 
  • #4
Simon Bridge said:
Per your second question: please show your reasoning.

The weight of the combined mass is ##(M+m)g##. Thus the friction is ##-\mu (M+m)g##.
 
  • #5
Simon Bridge said:
You were using the wrong force in the impulse/momentum theory.
I would have said the error was in overlooking the additional rightwards momentum gained from each bounce off the wall.
Happiness said:
The weight of the combined mass is ##(M+m)g##. Thus the friction is ##-\mu (M+m)g##.
This is a bit awkward. It depends how you interpret the ball sticking to the block. If it sticks in such a way that the force between them is purely horizontal then you need to consider the friction on each separately, and we are told the ball has no friction.
 
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  • #6
@haruspex ; I won't always respond with a list of everything that is wrong when asked "what's wrong with this?". I'll try to point the querant in a direction that will reveal what is going on.

I'm not really clear on Happiness' reasoning throughout...

ie. the interpretation that the ball sticks to the block in such a way as to increase friction for the combination is neither here nor there for the use of the impulse-momentum formula. It's sort-of a side note. Here the impulse delivered to the m+M combination is the change in momentum ... we imagine momentum is conserved though so the momentum of the ball and of the block remains the same in total. So I wanted to see reasoning, that happiness was using, around that.

The ball momentum does change though ... since mu=(m+M)v and v<u so the impulse would be maybe give ##m(u-v) = (m+M)a\Delta t## or something?
It looks like happiness is saying that (m+M)a = \mu(m+M) or something ... ie the force on the block etc is the friction. But friction is not the only force acting on the block ... for a short time the impulse force acts on the block too since: ##I=\int_0^{\Delta t}F_{I}\;dt = \Delta p_{ball}## or whatever the actual reasoning is.
But whatever you use, that has to be the initial accelerating impulse, not the friction slowing the block ... so there is quite a lot to draw happiness' attention to ... and that worthy is right there reading this while I talk like he isn't... so: hi @Happiness , how you doing?

Can you see where there may be some concerns?
You seem to be equating the time part of the specific impulse with the time it takes the block to stop, but the impulse force is not applied during the entire motion of the block, and you seem to have equated the impulse force to the friction. You also don't account, in Q1, for the fact the wall is delivering 2p momentum to the right at each collision. In short: what is wrong with using the impulse/momentum relation here is "everything".

It looks to me like you are doing physics by trying to guess the right equations and just plugging numbers in.
You are discovering that this is not going to work. ie. the ##\Delta t## in two different equation may be referring to two different changes in time for different contexts. It is important to understand the equations you are using as well.
 
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  • #7
Simon Bridge said:
I won't always respond with a list of everything that is wrong when asked "what's wrong with this?
Me neither, but Happiness had specifically asked what was wrong with the reasoning. If I reverse-engineer the reasoning from the equation, I believe it's this: time block spent moving=(momentum lost by system)/(force of friction on system while block moves). The error in that is that it ignores the impulses from the wall at each bounce.
Simon Bridge said:
ie. the interpretation that the ball sticks to the block in such a way as to increase friction for the combination is neither here nor there for the use of the impulse-momentum formula.
Sure, but my comment here was in respect of Happiness' second question, namely, why is the frictional force in the sticking case still μMg and not μ(M+m)g.
 
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  • #8
@haruspex fair enough and points well made. :)
 

Related to Collision and time to stop a block

1. What is a collision and how does it occur?

A collision is an event in which two objects come into contact with each other. This can occur when the objects are moving or stationary. In the context of a block, a collision can occur when the block comes into contact with another object, such as a wall or another block.

2. How is the time to stop a block calculated during a collision?

The time to stop a block during a collision is calculated by using the block's velocity, mass, and the force of the collision. This can be done using the formula t = mv/F, where t is the time, m is the mass, v is the velocity, and F is the force of the collision.

3. What factors can affect the time to stop a block during a collision?

The time to stop a block during a collision can be affected by various factors, such as the mass and velocity of the block, the force of the collision, and the surface material of the objects involved. In addition, external factors such as friction and air resistance can also play a role in the time to stop a block during a collision.

4. Can the time to stop a block be predicted accurately?

The time to stop a block during a collision can be predicted with a certain level of accuracy using mathematical equations and simulations. However, it may not always be possible to accurately predict the exact time to stop a block due to the complex nature of collisions and the many variables involved.

5. How can the time to stop a block be reduced during a collision?

The time to stop a block during a collision can be reduced by decreasing the mass or velocity of the block, or by reducing the force of the collision. Additionally, using materials with low friction can also help reduce the time to stop a block during a collision.

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