Impulse and Momentum: Ball and cart connected by a cord

[octu]

Homework Statement

[/B]
A 1kg cart, a 0.1kg ball, and an elastic cord connecting the ball and the cart, are placed in a weightless environment; see Figure. The cart has the shape of a symmetric trapezoid with its left and right edges at a 60-deg angle from the vertical. It is also attached with four rollers (two on top and two on bottom) so that it can move freely horizontally. At t = 0, the system is in static equilibrium with the elastic cord fully stretched to exert a force F_s = -1 N (to the left) on the cart. The ball is then released from rest in its t = 0 position and hits the cart at t = 2 seconds. Note that the attachment of the elastic cord is at exactly the same level as the centre of mass of the cart. Answer the following questions. (a) Will the cart move right after the release of the ball (t = 0⁺)? Provide your reason(s). (b) What are the impulsive forces at the instant the ball hits the cart? (c) Assume the ball sticks to the cart surface, what is the speed of the cart right after the ball hits the cart? Neglect the mass of the string.

M = 1kg
m = 0.1kg
θ = 60

Homework Equations

F = ma
p = mv

The Attempt at a Solution

[/B]
a) The cart will want to move after the ball is released because the ball and cart are connected by the cord. The force pulling the ball to the cart is equal and opposite to the force pulling the cart to the ball. However, there is a ledge blocking the cart's motion, so it will stay put.

b) Right of the bat I'm not really sure what's meant by impulsive forces. I thought to find an impulsive force a time frame over which the force acts must be known? Anyways, I figure the ball is pulled in by a force of 1N over two seconds, so the acceleration and velocity after 2s can be given by:

a_m = 1/m
v_m = 2/m

The impulse is given by the change in momentum. The ball goes from v_m = 2/m stationary at the moment of contact, so:

Impulse = 2/m

and impulsive force would be this divided by some time? Not sure.

Well, whatever the impulsive force is, it acts normal to the cart surface and can be broken down into x and y components. the x component is proportional to cos(60) and the y component to sin(60).

Another thing - I feel that the v_m value is incorrect, because wouldn't the force change over time, if the cord is elastic?

c) The ball has momentum mv_m before it hits the cart. A proportion cos(60) of this momentum is carried forward, so:

mv_m = (m + M)*v*cos(60)

mv_m / (( m + M)*cos(60)) = v

where v is the velocity of the ball and cart stuck together.

Any ideas?

 

[haruspex]

The cart will want to move after the ball is released because the ball and cart are connected by the cord. The force pulling the ball to the cart is equal and opposite to the force pulling the cart to the ball. However, there is a ledge blocking the cart's motion, so it will stay put.

The question asks whether the cart will tend to move to the right just after the ball is released. Intuitively, would it? Can you think of a reason why it might?

the ball is pulled in by a force of 1N over two seconds

It is elastic. The force will not be constant.

The ball goes from vm = 2/m stationary at the moment of contact, so

The cart is free to move. Why would the ball become stationary?

whatever the impulsive force is, it acts normal to the cart surface

This is awkward. At this point we have not been told that the ball sticks to the cart, so I suspect your answer is the one expected. But when we factor in that it sticks, the picture gets rather complicated. I do not like this question.

 

[CWatters]

+1

Impulse is the change in momentum. Look it up on Wikipedia.

I don't think you can answer b) without knowing if the collision is elastic or inelastic. I think I would assume its an elastic collision for b) and inelastic for c).

However I don't think you can answer c) without knowing more about the roller since they are rotating? Perhaps you are meant to assume the rollers are massless?

 

[octu]

The question asks whether the cart will tend to move to the right just after the ball is released. Intuitively, would it? Can you think of a reason why it might?

There is a normal force acting on the cart from the wall, equal to the pulling force from the cord. Just after the ball is released, perhaps the pulling force decreases, and the normal force pushes the cart to the right. Is that reasoning valid?

 

[haruspex]

There is a normal force acting on the cart from the wall, equal to the pulling force from the cord. Just after the ball is released, perhaps the pulling force decreases, and the normal force pushes the cart to the right. Is that reasoning valid?

That fits with my intuition, and there is a good reason for it.
In the real world, everything has a bit of elasticity, including the barrier that is preventing the cart moving left. So when the ball is released, the recoil from the barrier will push the cart right.
That said, I am not at all sure what the questioner expects. Maybe they want you to take the barrier as completely rigid.

 

[octu]

The question asks whether the cart will tend to move to the right just after the ball is released. Intuitively, would it? Can you think of a reason why it might?

When I Read the question I took "right after" to mean "immediately after".

This is my attempt at a solution. What do you think? I used the Dirac delta function to find the impulses. Thanks for the help, I appreciate it.

a) The cart will move to the right after the ball is released. When the tension in the cord is relieved, the normal force from the wall will push the cart. b) The ball feels an impulsive force when it is released from a still position, giving it a velocity:$J\quad =\quad \int { Fdt } \quad =\quad 1\int { \delta (t)dt } \quad =\quad 1\\ J\quad =\quad m\Delta v\quad =\quad m({ v }_{ b }-\quad 0)\\ 1\quad =\quad m{ v }_{ b }\\ { v }_{ b }=\frac { 1 }{ m }$
Similarly, for the cart’s velocity due to the normal force:${ v }_{ c }=\frac { 1 }{ M }$When the ball hits the cart at t = 2, it experiences a change in velocity and thus an impulse:$\Delta v\quad =\quad ({ v }_{ c }\quad -\quad { v }_{ b })\quad =\quad (\frac { 1 }{ M } \quad -\quad \frac { 1 }{ m } )\\ J\quad =\quad F\int { \delta (t-2)dt } \quad =\quad m\Delta v\quad =\quad m(\frac { 1 }{ M } \quad -\quad \frac { 1 }{ m } )$

The impulsive force has componets:${ F }_{ x }=\quad m(\frac { 1 }{ M } \quad -\quad \frac { 1 }{ m } )\cos { (60) } \\ \\ { F }_{ y }=\quad m(\frac { 1 }{ M } \quad -\quad \frac { 1 }{ m } )\sin { (60) }$c) After the collision, we have:$m{ v }_{ f }\cos { (60) } \quad +\quad M{ v }_{ c }\quad =\quad (M\quad +\quad m)v\\ \\ v\quad =\quad \frac { \cos { (60) } \quad +\quad 1 }{ (M\quad +\quad m) } \quad =\quad 0.15m/s$

 

[gneill]

a) The cart will move to the right after the ball is released. When the tension in the cord is relieved, the normal force from the wall will push the cart.

Ever watch a slinky drop?
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[haruspex]

When the tension in the cord is relieved, the normal force from the wall will push the cart.

This is inadequate. If the barrier is rigid the normal force will do no work. Reread what I wrote before.

For b and c, if the impulse is purely normal to the surface of the block, what happens to the component of the ball's velocity that was parallel to that surface?

 

[octu]

Ever watch a slinky drop?

Ok, so looking at the slinky I can assume that when the ball is released, the tension in the cord will not be relieved near the cart.

If the barrier is rigid the normal force will do no work

And if the barrier is rigid, then the normal force will do no work.

So, another attempted solution:

a) The cart will not move when the ball is released, assuming the barrier is rigid.

b) The ball has a velocity before it hits the cart given by:

$J\quad =\quad \int { Fdt } \quad =\quad 1\int { \delta (t)dt } \quad =\quad 1\\ J\quad =\quad m\Delta v\quad =\quad m({ v }_{ b }-0)\\ 1\quad =\quad m{ v }_{ b }\\ { v }_{ b }=\frac { 1 }{ m }$

When the ball hits the stationary cart we have an impulsive force that can be broken down into x and y components:

$J\quad =\quad F\int { \delta (t-2)dt } \quad =\quad m{ v }_{ b }\\ F\quad =\quad m{ v }_{ b }\quad =\quad 1N\\ { F }_{ x }=\quad m{ v }_{ b }\cos { (60) } =\quad 0.5N\\ { F }_{ y }=\quad m{ v }_{ b }\sin { (60) } =\quad 0.87N$

For b and c, if the impulse is purely normal to the surface of the block, what happens to the component of the ball's velocity that was parallel to that surface?

I'm taking x to be horizontal and y to be vertical, and I'm thinking the y component of the force just goes away. It would increase the normal force of the ball on the track and thus the friction, but if the objects are totally rigid this wouldn't happen, right? All I can think of is that the component of the ball's velocity that is parallel to the surface would make it squish in a certain way.

c) To find the final velocity of the cart we can use the x component of the impulsive force:

$\int { { F }_{ x }\delta (t-2)dt } \quad =\quad (M+m)v\\ v\quad =\quad \frac { m{ v }_{ b }\cos { (60) } }{ (M+m) } \quad =\quad 0.05m/s$

 

[gneill]

Ok, so looking at the slinky I can assume that when the ball is released, the tension in the cord will not be relieved near the cart.

Not instantaneously, no.

 

[octu]

what do you guys think of my solution? Is it valid to use the Dirac delta function to find the speed of the ball?

 

[haruspex]

a) The cart will not move when the ball is released, assuming the barrier is rigid.

Yes.

b) The ball has a velocity before it hits the cart given by:

If you plug in actual values, you must specify the units. Much better still, do not plug in values until the end. Create variables to represent them instead and work entirely algebraically. It makes it much easier to follow the working and to spot errors.

Is it valid to use the Dirac delta function to find the speed of the ball?

It doesn't make any sense to me. You provide no reasoning behind your initial equation.
It is an elastic cord. What kind of motion will the ball have up until it hits the cart?

Edit: it just occurred to me that there is not enough information. You will have to assume that the elastic has zero relaxed length.

 

[octu]

There is a certain elastic energy stored in the cord that gives the ball its velocity. We assume that when the ball hits the cart, all of the elastic energy has been converted into kinetic energy. X naught is the fully stretched length of the cord. Assume also that the cord’s unstretched length is negligible in comparison to its stretched length.
$(1)\quad \frac { 1 }{ 2 } k{ { x }_{ 0 } }^{ 2 }=\frac { 1 }{ 2 } m{ { v }_{ b } }^{ 2 }\quad \Rightarrow \quad k{ { x }_{ 0 } }^{ 2 }=m{ { v }_{ b } }^{ 2 }\\ (2)\quad { F }_{ s }=k{ x }_{ 0 }$We can find the velocity of the ball by treating the cord-ball system as we would a simple harmonic oscillator with a period T that has completed ¼ of a period after 2 seconds.$\omega =\frac { 2\pi }{ T } =\sqrt { \frac { k }{ m } } \\ T=2\pi \sqrt { \frac { m }{ k } } \\ k={ \left( \frac { 2\pi }{ T } \right) }^{ 2 }m$And using this value with equations (1) and (2):${ v }_{ b }=\frac { { F }_{ s } }{ m } \frac { T }{ 2\pi }$With this velocity, we can find the impulse and the impulsive force the instant the ball hits the cart:$J={ F }_{ i }\int { \delta (t-2)dt } =m{ v }_{ b }\\ { F }_{ ix }={ F }_{ i }\cos { (60) } ={ F }_{ s }\frac { T }{ 2\pi } \cos { (60) } \\ { F }_{ iy }={ F }_{ i }\sin { (60) } ={ F }_{ s }\frac { T }{ 2\pi } \sin { (60) }$Now, assume the impulsive force in the y-direction has no effect on the future motion of the cart (because the cart is constrained to move only in the x-direction). We find the following value for the velocity of the cart, after the ball sticks to it:$\int { { F }_{ ix }\delta (t-2)dt } = (M+m)v\\ v=\frac { { F }_{ s }\frac { T }{ 2\pi } \cos { (60) } }{ (M+m) }$

Plugging in values (including T = 8) yields v = 0.58m/s.

Comments? What do you guys think? It does seem like some assumptions are necessary when it comes to the cord. Maybe it would be ok to use the Dirac delta function to find the speed of the ball if we assume that the cord's fully stretched length is barely longer than its relaxed length. Does that make sense to you?

 

[haruspex]

Looks much better, but I'm not following the way you resolve the impulse.
The momentum of the approaching ball is horizontal, yet you multiply by cos(60) to get an x component.

There are two reasonable ways to proceed.
If we ignore the later information about the ball sticking, find the components normal to and parallel to the surface. The second of those does not result in an impulsive reaction between the objects. Then consider the consequences of the normal component for the cart.

Alternatively, we use the fact that the ball sticks and find the impulse the cart must exert on it to achieve that. The equal and opposite impulse applies to the cart.

 

[octu]

Let’s proceed in the 2ndway you mentioned. I’m not sure how I would go about proceeding in the 1st way without knowing more about the ball. I think my issue was that I was considering the impulsive force of the ball hitting the cart to be normal to the cart surface. Now I see that the impulse force should be completely horizontal, because the ball approaches in the horizontal direction.We can resolve the impulse force into components parallel and perpendicular to the cart surface.$J={ F }_{ i }\int { \delta (t-2)dt } =m{ v }_{ b }\\ { F }_{ i\bot }={ F }_{ i }\cos { (60) } \\ { F }_{ i\parallel }={ F }_{ i }\sin { (60) }$The parallel component is not important to us. We have an impulsive reaction from the perpendicular component. The perpendicular force can be resolved into an x-component and a y-component.${ F }_{ iy }={ F }_{ i\bot }\sin { (60) } \\ { F }_{ ix }={ F }_{ i\bot }\cos { (60) } ={ F }_{ i }\cos ^{ 2 }{ (60) }$The y-component does not contribute to the forward motion of the cart. Using the x-component of the perpendicular impulse force, we can find the resulting velocity of the cart and ball after the ball is stuck to the cart.$\int { { F }_{ ix }\delta (t-2)dt } =(M+m)v\\ \\ v=\frac { { F }_{ s }\frac { T }{ 2\pi } \cos ^{ 2 }{ (60) } }{ (M+m) }$Plugging in the numbers, we get v = 0.29m/s.

What do you think?

 

[haruspex]

I should have made clear that the two approaches I described will not lead to the same answer. This is why I really do not like the question. It is not clear what is to be assumed for part b.

I was considering the impulsive force of the ball hitting the cart to be normal to the cart surface. Now I see that the impulse force should be completely horizontal, because the ball approaches in the horizontal direction.

No, it doesn't follow from that.
If the ball bounces off, the impulsive force is normal to the surface; if the ball sticks, the ball acquires no vertical velocity, so the impulse must be horizontal.

The parallel component is not important to us.

Why not? An impulse on the block that is parallel to the surface would have a horizontal component and therefore result in horizontal motion.

 

[hutchphd]

Now I see that the impulse force should be completely horizontal, because the ball approaches in the horizontal direction.

1) The details of the impulse do not matter. There are no external vector forces on the system in the x direction therefore x momentum (the integral of the impulse) is conserved. So that is easy (yes?).
2) Suppose the collision is elastic instead...

 

[haruspex]

The details of the impulse do not matter. There are no external vector forces on the system in the x direction therefore x momentum (the integral of the impulse) is conserved. So that is easy (yes?).

Part b asks specifically about the impulsive forces. Problem is, this is before we are told to assume the ball sticks to the cart. So do we assume that in part b or that it bounces off? It changes the answer.
The question says "the instant the ball hits the cart", but in reality it takes time for impulsive forces to build, so that is not helpful either.

 

[hutchphd]

Part b asks specifically about the impulsive forces. Problem is, this is before we are told to assume the ball sticks to the cart. So do we assume that in part b or that it bounces off? It changes the answer.
The question says "the instant the ball hits the cart", but in reality it takes time for impulsive forces to build, so that is not helpful either.

I confess the term "impulsive force" has no meaning for me. Sorry. Maybe the question (B) just asks for a recitation of all forces involved and then part C wants numbers? Not a clearly stated problem...but the elastic (and fast impulse) case should be included for completeness methinks.

 

[haruspex]

I confess the term "impulsive force" has no meaning for me. Sorry.I

They are the forces responsible for delivering the impulse. The term accentuates the fact that the forces are likely to be large but short lived, and not easily known - whereas the net impulse they deliver can often be known. It distinguishes them from other, more moderate, forces that may be present, such as gravity or spring force.
For the purpose of the question, you could simply interpret it as "impulse".

 

[hutchphd]

It distinguishes them from other, more moderate, forces that may be present, such as gravity or spring force.

Pedagogically I dislike the term...it immediately implies to my poor brain that there may exist non=impulsive forces! Impulse is a defined quantity...Perhaps short-duration forces.

 

[octu]

Ok. So we have a value for the velocity of the ball immediately before it hits the cart.${ v }_{ b }=\frac { { F }_{ s } }{ m } \frac { T }{ 2\pi }$Part b can be answered in two different ways based on the two possible assumptions regarding the impact: either the ball is assumed to stick, or it is assumed to bounce. I will treat each assumption individually.1) Ball sticks to cartb) Here, the impulsive force is completely horizontal. This is evident because the ball acquires no vertical velocity.$J={ F }_{ i }\int { \delta (t-2)dt } =m{ v }_{ b }$Equal and opposite impulsive forces will be applied to the cart and ball.c) x-momentum is conserved, so$m{ v }_{ b }=(m+M)v\\ v=\frac { m{ v }_{ b } }{ (m+M) }$2) Ball bounces off cartb) Here, the ball approaches horizontally and hits the angled wall of the cart, resulting in a force parallel to the cart surface and a force perpendicular to the cart surface:${ F }_{ i\bot }={ F }_{ i }\cos { (60) } \\ { F }_{ i\parallel }={ F }_{ i }\sin { (60) } \\ \\$
c) the question asks to assume that the ball sticks, so even if we answer b using assumption 2 we should still get the answer for c:$m{ v }_{ b }=(m+M)v\\ v=\frac { m{ v }_{ b } }{ (m+M) }$
What do you think of this?What would happen if the ball didn’t stick? I think it would bounce off the surface with an angle equal to the angle of incidence. The initial x-momentum of the ball would be split between the ball and the cart, giving each some x-velocity. The ball would gain momentum in the y-direction, and the cart would gain the opposite momentum in the y-direction. However, the cart is constrained to move only in the x-direction.

 

[haruspex]

What would happen if the ball didn’t stick? I think it would bounce off the surface with an angle equal to the angle of incidence.

Ah, no. Whether you think of it as instantaneous (the cart instantly acquiring velocity) or spread over a short period, the 'give' in the cart robs the ball of some of its bounce.
The right way to analyse this is through the different impulses:
- there is an impulse between ball and cart normal to the cart's surface
- there is a vertical impulse between cart and ground
You can write equations relating these to the resulting velocities and use conservation laws to solve.

Indeed, I do not understand what you wrote under 2b above. What is F_i there? Why would there be an impulse parallel to the surface?

 

[octu]

Whether you think of it as instantaneous (the cart instantly acquiring velocity) or spread over a short period, the 'give' in the cart robs the ball of some of its bounce.

Ok ,that makes sense.

Indeed, I do not understand what you wrote under 2b above. What is Fi there? Why would there be an impulse parallel to the surface?

I think I am missing something here. I'm using F_i to represent the total horizontal impulse force (is it right to call it that?), which I then break into two components, one parallel to the surface, and one perpendicular.

So is the perpendicular force here the impulse force (equal and opposite forces on ball and cart), and the parallel component is not important? Is impulse force always perpendicular to the surface?

 

[haruspex]

Is impulse force always perpendicular to the surface?

In the absence stickiness or friction, yes.