How Do You Determine the Inertia of Cart A?

[emily081715]

Homework Statement

A 1-kg standard cart collides with a cart A of unknown inertia. Both carts appear to be rolling with significant wheel friction because their velocities change with time as shown graph below:

What is the inertia of cart A?

Homework Equations

i am unsure how to even solve for inertia but i know the equation is i=mr^2 except i have never used this equation

The Attempt at a Solution

i tried randomly rearranging formula i know how to use for momentum and got 1.6 kg but i have never solve a problem like this and I'm very confused on how to actually solve for inertia

 

[Simon Bridge]

i am unsure how to even solve for inertia but i know the equation is i=mr^2 except i have never used this equation

That is the equation for the moment of inertia I of a body, so you can tell something about it's rotational dynamics. When "inertia" is used by itself it usually means "mass"... look up "law of inertia".

I notice you got it right for your other problem.

 

[emily081715]

That is the equation for the moment of inertia I of a body, so you can tell something about it's rotational dynamics. When "inertia" is used by itself it usually means "mass"... look up "law of inertia".

I notice you got it right for your other problem.

Am I just using the equation F=ma? that is what i got when i looked it up. if that's the case i don't know acceleration or the mass of cart A or the F and have three unknown variables

 

[PeroK]

Am I just using the equation F=ma? that is what i got when i looked it up. if that's the case i don't know acceleration or the mass of cart A or the F and have three unknown variables

Does Newton's third law tell you anything about the forces involved?

 

[emily081715]

Does Newton's third law tell you anything about the forces involved?

Newtons third law says that for action there is an equal and opposite reaction. would this mean the the force acting on the standard cart is equal to the one on cart A?

 

[PeroK]

Newtons third law says that for action there is an equal and opposite reaction. would this mean the the force acting on the standard cart is equal to the one on cart A?

You need to be more precise. The force that the standard cart exerts on cart A is equal and opposite to the force that cart A exers on the standard cart.

But, do you think there are other forces involved? Hint: how long does the collision last?

 

[emily081715]

Does Newton's third law tell you anything about the forces involved?

i don't think there is any other forces acting on the object. the collision is very quick and doesn't even last a second

 

[PeroK]

i don't think there is any other forces acting on the object. the collision is very quick and doesn't even last a second

That's not right. Normally these problems involve an instantaneous collision. But not in this case. The collision clearly lasts a significant length of time. In fact, it's exactly one second.

 

[emily081715]

That's not right. Normally these problems involve an instantaneous collision. But not in this case. The collision clearly lasts a significant length of time. In fact, it's exactly one second.

so what does that mean?

 

[PeroK]

so what does that mean?

Do you think friction took a break while the carts got on with their collision?

 

[emily081715]

Do you think friction took a break while the carts got on with their collision?

No, so that means both carts have a force of friction acting on them. i am still unsure how to actually go about solving the question though

 

[PeroK]

No, so that means both carts have a force of friction acting on them. i am still unsure how to actually go about solving the question though

Well, this problem is not so easy. To solve this problem, I think you need to really understand what is going on. Then, you need to organise your thoughts. Hit the problem with exactly the right equations and, finally, solve those equations.

The crux of this problem is the relationship between forces and change in momentum. I'm not convinced you understand this well enough yet.

The problem would be much easier with an instantaneous collision. As it stands, I think this question might be a bit hard!

The other question I'm helping you with is really much easier than this one.

 

[emily081715]

Well, this problem is not so easy. To solve this problem, I think you need to really understand what is going on. Then, you need to organise your thoughts. Hit the problem with exactly the right equations and, finally, solve those equations.

The crux of this problem is the relationship between forces and change in momentum. I'm not convinced you understand this well enough yet.

The problem would be much easier with an instantaneous collision. As it stands, I think this question might be a bit hard!

The other question I'm helping you with is really much easier than this one.

both questions need to be answered though, can you keep working on this with me as well

 

[PeroK]

both questions need to be answered though, can you keep working on this with me as well

See if you do it without friction first. Ignore friction.

I'll give you one hint. You got $1.6kg$ for cart A. But, that means there is more momentum after the collison than before. So, that can't be right.

 

[emily081715]

See if you do it without friction first. Ignore friction.

I'll give you one hint. You got $1.6kg$ for cart A. But, that means there is more momentum after the collison than before. So, that can't be right.

i got 0.65kg

 

[PeroK]

i got 0.65kg

From the graph you should be able to see that the mass of cart A is significantly greater than that of the standard cart. This shows the gap between your knowledge of the subject and the knowledge required to solve a problem like this.

I'm really not sure I can help you through this one.

 

[emily081715]

From the graph you should be able to see that the mass of cart A is significantly greater than that of the standard cart. This shows the gap between your knowledge of the subject and the knowledge required to solve a problem like this.

I'm really not sure I can help you through this one.

can you break down the steps on what i should do to find the answer

 

[emily081715]

From the graph you should be able to see that the mass of cart A is significantly greater than that of the standard cart. This shows the gap between your knowledge of the subject and the knowledge required to solve a problem like this.

I'm really not sure I can help you through this one.

should there be less momentum after the collision?

 

[emily081715]

Homework Statement

A 1-kg standard cart collides with a cart A of unknown inertia. Both carts appear to be rolling with significant wheel friction because their velocities change with time
https://s.yimg.com/hd/answers/i/74904644501142d99a5374e203fbdddf_A.jpeg?a=answers&mr=0&x=1474920445&s=c19c3ca196cd197a4ef3be7b13349b07

Homework Equations

the law of inertia
F=ma
P=mv

The Attempt at a Solution

i know that inertia in this case means mass, but i am unsure how to solve for it. i tried and got 1.6kg but that can't be right because the momentums before is less then the momentum after the collision. can someone break down the steps that should be taken to solve this question

 

[berkeman]

Homework Statement

A 1-kg standard cart collides with a cart A of unknown inertia. Both carts appear to be rolling with significant wheel friction because their velocities change with time
https://s.yimg.com/hd/answers/i/74904644501142d99a5374e203fbdddf_A.jpeg?a=answers&mr=0&x=1474920445&s=c19c3ca196cd197a4ef3be7b13349b07

Homework Equations

the law of inertia
F=ma
P=mv

The Attempt at a Solution

i know that inertia in this case means mass, but i am unsure how to solve for it. i tried and got 1.6kg but that can't be right because the momentums before is less then the momentum after the collision. can someone break down the steps that should be taken to solve this question

What does the question ask for? That seems to be missing from your problem statement.

And what quantity is conserved in elastic collisions like this? How can you use this to solve the problem.

 

[emily081715]

What does the question ask for? That seems to be missing from your problem statement.

And what quantity is conserved in elastic collisions like this? How can you use this to solve the problem.

It is asking for the inertia of cart A

 

[emily081715]

It is asking for the inertia of cart A

momentum and kinetic energy is conserved in an elastic collision

 

[berkeman]

It is asking for the inertia of cart A

Then go for it. What quantity is conserved in elastic collisions? And how can you use that to solve for the problem?

 

[emily081715]

Then go for it. What quantity is conserved in elastic collisions? And how can you use that to solve for the problem?

the problem is i don't know how to go for it

 

[berkeman]

the problem is i don't know how to go for it

Please try harder. At least one of the equations you mentioned can be used to solve the problem. You need to write the equations out and use the graph to get some values for known variables. Then solve for the unknown variable (mass of the 2nd cart)...

 

[emily081715]

Please try harder. At least one of the equations you mentioned can be used to solve the problem. You need to write the equations out and use the graph to get some values for known variables. Then solve for the unknown variable (mass of the 2nd cart)...

i tried using the equation pi=pf.
m1v1+m2v1=m1v1+ m2v2

1(1)+m2(5)=m1(3.5)+m2(1.2)

1-3.5=m2(1.2)-m2(5)

-2.5=m2(1.2-5)
m2= 0.66kg
i know that this can't work though and don't see the part I'm missing

 

[berkeman]

m1v1+m2v1=m1v1+ m2v2

1(1)+m2(5)=m1(3.5)+m2(1.2)

That's the equation I would have chosen to use also.

Your first equation has a small typo that I've bolded, and your numbers in the 2nd equation do not seem to match the graph. Also, it is a good idea to keep quantities unique. In your first equation you should use some subscripts for the velocities so you can keep straight which mass' velocity it is, and whether it was just before or just after the collision. Something like V1i, V1f or similar notation.

Can you say what the initial and final velocities are for each cart just before and just after the collision?

 

[emily081715]

That's the equation I would have chosen to use also.

Your first equation has a small typo that I've bolded, and your numbers in the 2nd equation do not seem to match the graph. Also, it is a good idea to keep quantities unique. In your first equation you should use some subscripts for the velocities so you can keep straight which mass' velocity it is, and whether it was just before or just after the collision. Something like V1i, V1f or similar notation.

Can you say what the initial and final velocities are for each cart just before and just after the collision?

for the standard cart: at 5 seconds (before)=0m/s, and at 6 seconds (after)= 4.2m/s
for cart A: at 5 seconds=4m/s and at 6 seconds=2m/s

 

[emily081715]

for the standard cart: at 5 seconds (before)=0m/s, and at 6 seconds (after)= 4.2m/s
for cart A: at 5 seconds=4m/s and at 6 seconds=2m/s

should i use these as the velocities

 

[berkeman]

for the standard cart: at 5 seconds (before)=0m/s, and at 6 seconds (after)= 4.2m/s
for cart A: at 5 seconds=4m/s and at 6 seconds=2m/s

Looks good so far!

 

[emily081715]

Looks good so far!

by using what i did earlier i get the mass of cart A as 2.1 kg

 

[emily081715]

p=mv

pi=pf

m1v+m2v=m1v+ m2v

1(0)+m2(4)=m1(4.2)+m2(2)

-4.2 =m2(2-4)

 

[berkeman]

p=mv

pi=pf

m1v+m2v=m1v+ m2v

1(0)+m2(4)=m1(4.2)+m2(2)

-4.2 =m2(2-4)

Almost there...

 

[berkeman]

And BTW, there is a good visual way to check your final answer. Can you take a guess at what it is?

 

[emily081715]

what step have i missed? solving the rest of the equation i get an answer of 2.1 kg

 

[berkeman]

what step have i missed? solving the rest of the equation i get an answer of 2.1 kg

I don't think you missed anything, and based on a visual inspection of the graph, that looks to be the correct answer. Why can I tell that with a visual inspection?

 

[emily081715]

I don't think you missed anything, and based on a visual inspection of the graph, that looks to be the correct answer. Why can I tell that with a visual inspection?

the answer should've been 2.50

 

[berkeman]

the answer should've been 2.50

Interesting. I could see maybe 2.2 if the estimate of v=4.2 should be v=4.4 instead, but to get all the way up to 2.5m/s would take something else.

Maybe we have to take the decreasing velocity due to friction into account. What is the full problem statement exactly?

 

[SammyS]

I estimated something like 2.55 to 2.7 kg .

 

[I like Serena]

Hi emily081715!

Yes. We can see that the momentum of both carts decrease at a constant rate throughout.
So what's the total momentum before the collision, what's the momentum after the collision, and how much momentum did we lose during the collision?

 

[jbriggs444]

As a result of the collision, cart A lost momentum and the standard cart gained momentum. Can you look at the graph and determine how much less momentum cart A had as a result of the collision than it would have had if the collision had not taken place?

 

[Simon Bridge]

Emily has other, more basic, problems that should be tackled before this one... get the basics down first.
Recommend anyone trying to help takes a look at past threads by this member first.

 

[I like Serena]

Hi Simon Bridge,

I don't get what you mean.
Just from Emily's previous posts in this thread she seems the have a firm grasp on the total momentum, and on setting up equations and solving them.
The one thing missing is the loss due to friction - for which she did not get a clear response.
And her request to break it down went unanswered.

 

[berkeman]

Hi Simon Bridge,

I don't get what you mean.
Just from Emily's previous posts in this thread she seems the have a firm grasp on the total momentum, and on setting up equations and solving them.
The one thing missing is the loss due to friction - for which she did not get a clear response.
And her request to break it down went unanswered.

The problem apparently is that she re-posted this question after she did not get enough hand-holding in her first thread attempt, and I took the bait. The problem turned out to be more complicated than I expected when I first replied, which is fine, but it would have been good if I hadn't wasted so much time assuming it was a simple question when the OP knew already that it was not.

Thread closed.