Application of momentum conservation in inelastic collisions

[burian]

Homework Statement

A small block of mass M moves on a frictionless surface on an inclined plane, as shown in the figure. The angle of the incline suddenly changes from 60 to 30 at point B. The block is initially rest at A

Relevant Equations

$$ \frac{mv^2}{2} = mgh$$

momentum conservation

So, what I did was suppose the mass of ramp is $ M_r$ and let velocity at B of block be v, then, after inellastic collsion both bodies v' velocity

at B ,

$$M\vec{v}= M_r \vec{v'}+ M \vec{v'}$$

or,

$$ \frac{M}{M +M_r} \vec{v}= \vec{v'}$$
Now,

Suppose I take the limit as mass of ramp goes to infinity, then I get $\vec{ v'}=0$

that means, after the collision, the block is temporarily at rest and then slides to ground.So, we can just neglect upper half and do energy conservation in lower half ramp to get velocity at final position.

However, my answer came out wrong for some reason, I wish to ask if my concepts are correct?

answer I got:

$$| v|= \sqrt{60}$$

actual answer:
$$| v|= \sqrt{105}$$

 

[PeroK]

Velocity and momentum are vector quantities.

 

[burian]

@PeroK I have put the vector arrows on top but how does go to my mistake?

 

[PeroK]

@PeroK I have put the vector arrows on top but how does go to my mistake?

What do you think happens when the mass $m$ hits the lower ramp?

 

[PeroK]

PS What would happen if the second ramp were horizontal and the first ramp at $30°$ to it?

 

[burian]

What do you think happens when the mass $m$ hits the lower ramp?

as shown with equations, it should have it's velocity drop to 0 because of the large mass of block

 

[burian]

PS What would happen if the second ramp were horizontal and the first ramp at $30°$ to it?

I can't guess

 

[PeroK]

as shown with equations, it should have it's velocity drop to 0 because of the large mass of block

Velocity is a vector quantity. You need to think about its components.

 

[PeroK]

I can't guess

That's what you need to work out using the velocity vector.

 

[burian]

Velocity is a vector quantity. You need to think about its components.

Why? can't I just write momentum conservation in a vector form

Are you suggesting there is impulse in one of the directions? I considered that there would be an impulse when the block hits the second surface but neglected it coz the time period shouldbe small where they collide

 

[anuttarasammyak]

actual answer:v|=105

The answer you refer seems the speed just before the ramp C. After the ramp C the block goes horizontally with reduced speed by factor of $\frac{\sqrt{3}}{2}$ as well as at ramp B.

 

[haruspex]

Are you suggesting there is impulse in one of the directions? I considered that there would be an impulse when the block hits the second surface but neglected it coz the time period shouldbe small where they collide

"The angle of the incline suddenly changes"

The thing about sudden impacts is that although the duration is brief (and in general unknown) the forces can be very large and unknown. So the integral of the force over the duration, i.e. the net change in momentum, can be significant.
To make @PeroK's point more starkly, what if the first ramp were vertical and the second horizontal?

 

[rude man]

Homework Statement:: A small block of mass M moves on a frictionless surface on an inclined plane, as shown in the figure. The angle of the incline suddenly changes from 60 to 30 at point B. The block is initially rest at A
Relevant Equations:: $$ \frac{mv^2}{2} = mgh$$

momentum conservation

View attachment 267810

So, what I did was suppose the mass of ramp is $ M_r$ and let velocity at B of block be v, then, after inellastic collsion both bodies v' velocity

at B ,

$$M\vec{v}= M_r \vec{v'}+ M \vec{v'}$$

or,

$$ \frac{M}{M +M_r} \vec{v}= \vec{v'}$$
Now,

Suppose I take the limit as mass of ramp goes to infinity, then I get $\vec{ v'}=0$

that means, after the collision, the block is temporarily at rest and then slides to ground.So, we can just neglect upper half and do energy conservation in lower half ramp to get velocity at final position.

However, my answer came out wrong for some reason, I wish to ask if my concepts are correct?

answer I got:

$$| v|= \sqrt{60}$$

actual answer:
$$| v|= \sqrt{105}$$

what is the question?
plus: momentum is conserved within a system (your block) if and only if there are no externally applied forces.
Burt there is gravity and the upward push on the block.
So is momentum conservation applicable to solve whatever your question is?

 

[haruspex]

So is momentum conservation applicable to solve whatever your question is?

At the ramp boundary, yes.

 

[rude man]

Yes.
So we can assume the question is velocity before & after moment of impact.

 

[PeroK]

Why? can't I just write momentum conservation in a vector form

A vector equation gives you two or three separate equations, one for each vector component. Momentum may be conserved in one direction but not in another, as in this case.

Consider throwing a ball horizontally. When it hits the floor the horizontal momentum is approximately conserved, but the vertical momentum is approximately reversed. The two components change in very different ways as a result of the bounce.

 

[burian]

"The angle of the incline suddenly changes"

The thing about sudden impacts is that although the duration is brief (and in general unknown) the forces can be very large and unknown. So the integral of the force over the duration, i.e. the net change in momentum, can be significant.
To make @PeroK's point more starkly, what if the first ramp were vertical and the second horizontal?

what is the question?
plus: momentum is conserved within a system (your block) if and only if there are no externally applied forces.
Burt there is gravity and the upward push on the block.
So is momentum conservation applicable to solve whatever your question is?

Usually we say can apply momentum consv in vertical direction coz the time period gravity acts over is small

 

[burian]

A vector equation gives you two or three separate equations, one for each vector component. Momentum may be conserved in one direction but not in another, as in this case.

Consider throwing a ball horizontally. When it hits the floor the horizontal momentum is approximately conserved, but the vertical momentum is approximately reversed. The two components change in very different ways as a result of the bounce.

Mhm I see, I wam looking for a description of this impuslive force which happens

 

[haruspex]

Usually we say can apply momentum consv in vertical direction coz the time period gravity acts over is small

You did not answer my question.
If the first ramp were vertical and fhe second horizontal, what would you expect the velocity of the block to be after meeting the second ramp?

 

[burian]

You did not answer my question.
If the first ramp were vertical and fhe second horizontal, what would you expect the velocity of the block to be after meeting the second ramp?

I would guess that it becomes zero. But maybe it may bounce as well hmm

 

[PeroK]

I would guess that it becomes zero. But maybe it may bounce as well hmm

The bouncing you are expected to ignore!

 

[burian]

The bouncing you are expected to ignore!

hmm I think I understand now, along normal the momentum not conserved but along surface it is. But one thing though, I still seek a more elaborate description of what exactly is happenign which makes this normal impulse not negligible in comparison to gravity in collisions

 

[PeroK]

hmm I think I understand now, along normal the momentum not conserved but along surface it is. But one thing though, I still seek a more elaborate description of what exactly is happenign which makes this normal impulse not negligible in comparison to gravity in collisions

Let's see some analysis of the problem from you. Can you solve this using the velocity vector?

See my hint about modelling the collision in post #5.

 

[jbriggs444]

Let's see some analysis of the problem from you. Can you solve this using the velocity vector?

See my hint about modelling the collision in post #5.

Suppose that the collision is slow. Let us say that it acts for a duration $\Delta t$. Over the course of that duration the force of gravity has time to act. The amount of momentum delivered by a force F over a time $\Delta t$ is given by $\Delta p=F \Delta t$.

If the collision is faster, gravity has less time to act. The amount of momentum delivered by gravity is lower. For an ideal collision that takes no time at all, gravity has time to deliver no momentum at all.

It can be useful to think of forces as trickles of momentum flowing into an object over time. An impulse can be thought of as a lump delivery -- a chunk of momentum arriving all at once.

By analogy, think of your kitchen sink. The drain is leaking and water is trickling out. [That's gravity]. You dump a half-gallon pitcher of water in all at once. [That's the impulse from the collision]. You can measure the water in the sink before and after. As long as you measure fast, it's it is a plus half a gallon difference. You can ignore the trickle going down the drain.

 

[burian]

PS What would happen if the second ramp were horizontal and the first ramp at $30°$ to it?

I think I get the idea now,

$$ v'\sin \theta = \sqrt{2 gh}$$
is the final equation that I had got
here the $$\theta$$ is angle of ramp, I got this equation by finding component of velocity post collision parallel to ramp. And v' is the final velocity (i.e: velocity after it gets out of ramp)

Now that I have this, could you tell me how to describe the impulse which occurs to component perpendicular to ramp?

 

[PeroK]

I think I get the idea now,

$$ v'\sin \theta = \sqrt{2 gh}$$
is the final equation that I had got
here the $$\theta$$ is angle of ramp, I got this equation by finding component of velocity post collision parallel to ramp. And v' is the final velocity (i.e: velocity after it gets out of ramp)

Now that I have this, could you tell me how to describe the impulse which occurs to component perpendicular to ramp?

That doesn't look right.

The idea is that the object loses all its momentum in the direction normal to the ramp it collides with and none of its momentum in the direction tangential to the ramp it collides with.

Therefore, you have to calculate the components of velocity (normal and tangential to the ramp) at the time of collision. The velocity after the collision is the tangential component only.

That's the idea. Now you have to do the calculations.

 

[burian]

Hmm I did this by drawing a diagram after collision I'm not sure how the diagram would be right at collision. But I drew this

I can't figure out how to draw it right at collision, I'm thinking that it may chance of tipping over , how do I know if tips over or not?

 

[PeroK]

Hmm I did this by drawing a diagram after collision I'm not sure how the diagram would be right at collision. But I drew this

View attachment 268246

I can't figure out how to draw it right at collision, I'm thinking that it may chance of tipping over , how do I know if tips over or not?

In that diagram you really want to show the components of $v$ before the collision. And, especially how $v'$ relates to them.

You're not supposed to worry about the object bouncing or tipping over!

 

[haruspex]

I think I get the idea now,

$$ v'\sin \theta = \sqrt{2 gh}$$
is the final equation that I had got
here the $$\theta$$ is angle of ramp, I got this equation by finding component of velocity post collision parallel to ramp. And v' is the final velocity (i.e: velocity after it gets out of ramp)

It's always worth running a sanity check. Does your expression give a sensible answer when the first surface is at right angles to the second?

You know that what is preserved is the component of momentum parallel to the second surface, so resolve the momentum before impact into components parallel and normal to the second surface.