# Block bombarded by particles {colision}

## Homework Statement:

A trouble with quantity of motion, impulse and force.

## Relevant Equations:

All below

I thought i could apply a conservation of momentum in this case,

Apparently, this is not right, so i dont know what to do now.

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Delta2
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I suggest you have a look at the formulas here https://en.wikipedia.org/wiki/Elastic_collision#One-dimensional_Newtonian and do the proper approximations for the case that ##m_2=\delta m<<m=m_1##.

According to one approximation I can see for this case (namely that ##m+\delta m\approx m## )we ll have $$v_{n+1}=(1-\frac{\delta m}{m})v_{n}+\frac{2\delta m}{m}v_0$$. What do you get for ##v_n## ( the velocity after the n-th collision) if you solve this recursive equation?

LCSphysicist
I suggest you have a look at the formulas here https://en.wikipedia.org/wiki/Elastic_collision#One-dimensional_Newtonian and do the proper approximations for the case that ##m_2=\delta m<<m=m_1##.

According to one approximation I can see for this case (namely that ##m+\delta m\approx m## )we ll have $$v_{n+1}=(1-\frac{\delta m}{m})v_{n}+\frac{2\delta m}{m}v_0$$. What do you get for ##v_n## ( the velocity after the n-th collision) if you solve this recursive equation?
I use wolframalpha to solve this, because i am little tired now, but we have:

I think this is not the answer, because if n tends to infinity {just to verify and to comparison with the problem}, Vn will diverge and so will not tends to Vo

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Delta2
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I see, maybe the approximations I have suggested are not so good. What do you get in wolfram alpha if you want to solve the recursive equation $$v_{n+1}=bv_{n}+av_0$$ for ##1>b>0## and ##1>a>0## cause I don't seem to get something valid there.

LCSphysicist
Delta2
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The exact recursive equation I get (without approximations) is
$$v_{n+1}=\frac{m-\delta m}{m+\delta m}v_n+\frac{2\delta m}{m+\delta m}v_0$$

LCSphysicist
haruspex
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Homework Statement:: A trouble with quantity of motion, impulse and force.
Relevant Equations:: All below

View attachment 264180

I thought i could apply a conservation of momentum in this case,

View attachment 264183

View attachment 264182
Apparently, this is not right, so i dont know what to do now.
How do you get the ##2V+v_0##? Does your equation make sense when ##V=v_0##?
You must need to use the elasticity somewhere.

LCSphysicist
haruspex
Homework Helper
Gold Member
The exact recursive equation I get (without approximations) is
$$v_{n+1}=\frac{m-\delta m}{m+\delta m}v_n+\frac{2\delta m}{m+\delta m}v_0$$
You are doing to much on behalf of the OP. Please just provide hints and point out errors, at least to begin with.

LCSphysicist and Delta2
Delta2
Homework Helper
Gold Member
Sometimes in physics a successful result involves the art of making successful approximations. I think this is the case here, it's all about making the proper approximations here stemming from the fact that ##\delta m<<m##.

Last edited:
LCSphysicist
How do you get the ##2V+v_0##? Does your equation make sense when ##V=v_0##?
You must need to use the elasticity somewhere.
I wonder i could change to the reference of the block, but this is leading me to some contradictions. I dont know why, see example:

A ball is going with 50m/s, and the block 30m/s.
Now the ball has 20m/s in block frame, and go back with -20m/s
But in the block frame, the ball has
+20 + 30 = +50
-20 + 30 = +10, what is wrong, because the ball need to back off.
Considering the mass of the block very large with respect to the ball.
Maybe i just can use this when the two blocks are approximating?

The exact recursive equation I get (without approximations) is
$$v_{n+1}=\frac{m-\delta m}{m+\delta m}v_n+\frac{2\delta m}{m+\delta m}v_0$$
In this cases that have something a >> b, i always try to put b/a and so i cut this terms, because is nearly zero, but in this case seems that dont match, i will try by another approximations.

etotheipi
Gold Member
2019 Award
The exact recursive equation I get (without approximations) is
$$v_{n+1}=\frac{m-\delta m}{m+\delta m}v_n+\frac{2\delta m}{m+\delta m}v_0$$
It's possible to solve this by rearranging it to $$(m + \delta m)v_{n+1} - (m - \delta m)v_n = 2v_0 \delta m$$and then solving the non-homogenous recurrence relation. To do this you postulate ##v_n = \lambda^n## for the homogenous part, and ##v_n = k## for the non-homogenous part. I won't show details since we can't give full solutions, however you end up with something like ##v_n = v_0(1-e^{-n\ln{(\frac{m+ \delta m}{m - \delta m})}})## which you can then get into the required form along with ##\ln(1+x) \approx x##.

There's likely a nicer way, but it gets the job done .

LCSphysicist and Delta2
Yes, now i got it. THanks all of you.

I am just stuck in this,
I wonder i could change to the reference of the block, but this is leading me to some contradictions. I dont know why, see example:

A ball is going with 50m/s, and the block 30m/s.
Now the ball has 20m/s in block frame, and go back with -20m/s
But in the block frame, the ball has
+20 + 30 = +50
-20 + 30 = +10, what is wrong, because the ball need to back off.
Considering the mass of the block very large with respect to the ball.
Maybe i just can use this when the two blocks are approximating?
There is any error in see by this way?

Delta2
Homework Helper
Gold Member
I am just stuck in this,

There is any error in see by this way?
The problem just states that the particles bounce back in the negative x-direction, i am not sure if we can interpret this as that they bounce back with equal and opposite velocity (relative to the block). If we interpret it this way then because the collision is elastic and hence kinetic energy is conserved this would imply that the particle transfers zero kinetic energy to the block, so the velocity of the block would remain constant.

LCSphysicist
haruspex
Homework Helper
Gold Member
Now the ball has 20m/s in block frame, and go back with -20m/s
Then either momentum is not conserved or the block has turned into a brick wall.

neilparker62
Homework Helper
Elastic collison so ##\Delta P = 2\mu\Delta v## where ##\mu = \frac{m\times M}{m+M}\approx m## if m<<M and ##\Delta v=v_0-v_n##. Then ##P_{n+1}=P_n+\Delta P##.

LCSphysicist