How Do You Determine the Inertia of Cart A?

In summary: 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...In summary, the conversation discusses the problem of calculating the inertia of cart A in a collision with a standard cart. The equation for moment of inertia, I=mr^2, is mentioned but the student is unsure how to use it. The concept of inertia is also discussed, with it being noted that it is usually used to refer to mass. The student attempts to use the equation F=ma, but realizes there are multiple unknown variables. The conversation
  • #1
emily081715
208
4

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:
Mazur1e.ch4.p28.jpg

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
 
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  • #2
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.
 
  • #3
Simon Bridge said:
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
 
  • #4
emily081715 said:
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?
 
  • #5
PeroK said:
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?
 
  • #6
emily081715 said:
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?
 
  • #7
PeroK said:
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
 
  • #8
emily081715 said:
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.
 
  • #9
PeroK said:
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?
 
  • #10
emily081715 said:
so what does that mean?

Do you think friction took a break while the carts got on with their collision?
 
  • #11
PeroK said:
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
 
  • #12
emily081715 said:
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.
 
  • #13
PeroK said:
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
 
  • #14
emily081715 said:
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.
 
  • #15
PeroK said:
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
 
  • #16
emily081715 said:
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.
 
  • #17
PeroK said:
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
 
  • #18
PeroK said:
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?
 
  • #19
Poster has been warned not to post multiple threads on the same question...

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
 
Last edited by a moderator:
  • #20
emily081715 said:

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.
 
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  • #21
berkeman said:
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
 
  • #22
emily081715 said:
It is asking for the inertia of cart A
momentum and kinetic energy is conserved in an elastic collision
 
  • #23
emily081715 said:
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?
 
  • #24
berkeman said:
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
 
  • #25
emily081715 said:
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)...
 
  • #26
berkeman said:
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
 
  • #27
emily081715 said:
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?
 
  • #28
berkeman said:
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
 
  • #29
emily081715 said:
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
 
  • #30
emily081715 said:
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! :smile:
 
  • #31
berkeman said:
Looks good so far! :smile:
by using what i did earlier i get the mass of cart A as 2.1 kg
 
  • #32
p=mv

pi=pf

m1v+m2v=m1v+ m2v

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

-4.2 =m2(2-4)
 
  • #33
emily081715 said:
p=mv

pi=pf

m1v+m2v=m1v+ m2v

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

-4.2 =m2(2-4)
Almost there...
 
  • #34
And BTW, there is a good visual way to check your final answer. Can you take a guess at what it is? :smile:
 
  • #35
what step have i missed? solving the rest of the equation i get an answer of 2.1 kg
 

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