A question on impulsive tensions

[jiayingsim123]

Homework Statement

Two identical blocks, A and b, each of mass 2kg, are fastened to the ends of a rope 2m long. Block A rests on a smooth table, 1m from the edge. The rope passes over a pulley at the edge of the table and block B is held at the height of the tabletop, as shown. B is is then released.

*For visual purposes: the diagram can be found here:
http://books.google.com.my/books?id...B, each of mass 2kg, are fastened to"&f=false

Page 113 Exercise 5B Question 2

Homework Equations

Impulsive tension: mv-mu
g=10ms^-2

The Attempt at a Solution

mgh = 1/2mu^2 -> this is used to find the velocity of block B just before the top becomes taut.
u^2 = 4g
u = √2g
Take upward and rightward directions as negative.
Impulse is denoted as I
For block B: -I = 2v - 2√2g
For block A: -I=-2v

-2v-2v=-2√2g
4v=2√2g
v=√2g/2

Take the v as the initial velocity for both blocks A and B (initial velocity of Block A which is moving towards the edge of the table and initial velocity of Block B moving 1m downwards)

u=√2g/2
s=1m
v=0
t=?
s=0.5(u+v)t
t=0.894

However, the answer given in the book is 0.782s

Please include detailed explanations along with your solution, thanks! :D

 

[tiny-tim]

hi jiayingsim123!

(try using the X² button just above the Reply box )

i'm confused

the string goes taut immediately

the initial speed is zero

call the speed v, and the angle θ, and use conservation of energy and one other equation ​

 

[NascentOxygen]

That link shows nothing useful in my current old IE7, but from the wording and OP's approach, I'd infer that block B falls 1m before the rope goes taut.

I can't imagine what angle θ might be?

 

[tiny-tim]

Hi NascentOxygen!

i can't see the figure either ,

but I'm assuming it's the standard case of the string starting horizontal and stretched

 

[NascentOxygen]

u = √2g

I agree, but warn that you are courting trouble if you don't type this as √(2g)

Take upward and rightward directions as negative.
Impulse is denoted as I
For block B: -I = 2v - 2√2g
For block A: -I=-2v

-2v-2v=-2√2g
4v=2√2g
v=√2g/2

Or you could have said the KE of the moving block becomes the KE of the pair.

Take the v as the initial velocity for both blocks A and B (initial velocity of Block A which is moving towards the edge of the table and initial velocity of Block B moving 1m downwards)

u=√2g/2
s=1m
v=0
t=?
s=0.5(u+v)t
t=0.894

Can you explain why you say v=0? (As I explained, I can't see the diagram at your link.)

 

[azizlwl]

http://img266.imageshack.us/img266/650/dropb.jpg

 

[NascentOxygen]

Hi NascentOxygen!

i can't see the figure either ,

but I'm assuming it's the standard case of the string starting horizontal and stretched

Fair enough. Poster had better describe the setup then.

Ah, thanks azizlwl

 

[jiayingsim123]

Hi everyone, sorry for not replying sooner, I was a tad busy these few days.
The answer to that question is 0.782s, but I can never seem to get that answer.
After further thought, I think v is not 0 as the block doesn't stop after going down by 1m. I'm thinking along the lines of mgh=0.5mv^2 to get the final speed. But I'm really not too sure! Thanks azizlwl for posting the question! :)

 

[tiny-tim]

yes, thanks azizlwl!

mgh = 1/2mu^2 -> this is used to find the velocity of block B just before the top becomes taut.
u^2 = 4g
u = √2g
Take upward and rightward directions as negative.
Impulse is denoted as I
For block B: -I = 2v - 2√2g
For block A: -I=-2v

-2v-2v=-2√2g
4v=2√2g
v=√2g/2

looks ok so far …

now use s = ut + 1/2 at² twice, once for the first metre (a = g), and once for the second metre (a = g/2)

 

[jiayingsim123]

Hi tiny-tim, thanks. But what should be the value of u? And if you were to do it, would you get 0.782? Sorry for my persistent asking of questions but what is the rationale behind doing what you suggested? Thanks in advance! :)

 

[tiny-tim]

Hi tiny-tim, thanks. But what should be the value of u?

for the first metre, u = 0

for the second metre, u = (√2)g/2

And if you were to do it, would you get 0.782?

i've no idea

you tell us! ​

 

[jiayingsim123]

Hi tiny-tim, I tried doing what you suggested, but I got 0.774 instead of 0.782, which is actually really close.
This is what I did:
Take g=10m/s^2

For the first metre:
s=ut+0.5at^2
1=0.5gt^2
t=sqrt1/5
= 0.447

For the second metro
s=ut+0.5at^2
1= √(2g)/2t + 0.5(g/2)t^2
t=0.327

Total time elapsed=0.774

Sorry tiny-tim, but could you explain why you suggested this? I cannot really see the rationale, I'm sorry! :( Thanks in advance!

 

[pcm]

but I got 0.774 instead of 0.782, which is actually really close.

Take g=10m/s^2

i get 0.782 if i take g=9.8 m/s^2.

 

[TSny]

Since I feel a little grumpy today :grumpy: , I'm going to complain about how the problem is stated. Although it may seem natural to assume that both masses come to the same speed immediately after each receives the impulse from the rope, that's an assumption that should have been stated in the problem. It's similar to having a 2-body collision problem and just assuming that it's completely inelastic.

It's interesting to rework the problem assuming an elastic (but still very stiff) rope where no mechanical energy is lost in the "collision".

Of course you could have a case anywhere in between the completely inelastic and elastic cases.

(Now I feel a little better )

 

[tiny-tim]

Sorry tiny-tim, but could you explain why you suggested this? I cannot really see the rationale, I'm sorry! :( Thanks in advance!

hi jiayingsim123!

because for the 1st metre, the mass m is falling on its own, with acceleration g

and for the 2nd metre, the total mass 2m is falling, with external force only mg, so the acceleration is g/2

… It's similar to having a 2-body collision problem and just assuming that it's completely inelastic.

no, the energy lost in the collision isn't relevant, conservation of momentum determines the speeds

(though i agree that if the rope is stretchy, that will add slightly to the time taken)

 

[TSny]

hi jiayingsim123!

no, the energy lost in the collision isn't relevant, conservation of momentum determines the speeds

Hi tiny-tim. If no mechanical energy is lost in the collision, what would the speed of each mass be right after the collision?

 

[tiny-tim]

it is lost

i'm only saying it isn't relevant …

the motion is determined by conservation of momentum and by the constraint v₁ = v₂ (ie the collision is completely inelastic) ​

 

[TSny]

it is lost

i'm only saying it isn't relevant …

the motion is determined by conservation of momentum and by the constraint v₁ = v₂ (ie the collision is completely inelastic) ​

But aren't you adding the constraint v₁ = v₂ as an assumption that is not stated in the problem?

Maybe this constraint is implied in the wording of the question, but I don't see where. (Hope I'm not being too cantankerous in my grouchy mood today.)

 

[tiny-tim]

But aren't you adding the constraint v₁ = v₂ as an assumption that is not stated in the problem?

i don't think so …

the rope has a given length, so we assume it's a fixed length​

 

[TSny]

the rope has a given length, so we assume it's a fixed length​

Well, I don't see how assuming a fixed length makes much difference. When two steel balls collide, they don't deform noticeably during the collision, yet the collision might be close to elastic. Likewise, I think I can imagine a rope that doesn't stretch noticeably but still doesn't absorb much mechanical energy (sort of like an extremely stiff bungee cord).

Anyway, I'll let it go.

 

[jiayingsim123]

Hi tiny-tim and TSny, thanks for helping out! You guys really made my day! Just two more questions though directed to tiny-tim:

Okay, just to verify my understanding of the question.
Question 1: So when block B is released, block A doesn't move, and time is already at that point, taken into account? So time has started ticking the moment when block B is released (though block A doesn't move yet) until block A reaches the edge of the table?

Question 2:
I thought the speed √(2g)/2 is the speed at which block A jerks into motion, but does that mean that it is also its initial speed. I thought if a body jerks, it stops instantaneously after?

Sorry I'm still a tad confused on this topic, can someone shed some light on the jerking of motion? Thanks in advance again! :)

 

[tiny-tim]

hi jiayingsim123!

(just got up :zzz:)

Question 1: So when block B is released, block A doesn't move, and time is already at that point, taken into account? So time has started ticking the moment when block B is released (though block A doesn't move yet) until block A reaches the edge of the table?

the question says "B is then released. How long does it take for A …", so the question is telling you to start the time from the word "then"

Question 2:
I thought the speed √(2g)/2 is the speed at which block A jerks into motion, but does that mean that it is also its initial speed. I thought if a body jerks, it stops instantaneously after?

i've no idea where you got that idea from

once a body jerks into motion, it will stay at that speed unless there are other forces acting on it (Newton's first law)

in any collision (of which this is one), there is a sudden change of velocity …

we talk about the velocity "immediately before" the collision, and the velocity "immediately after" the collision …​

but before and after the collision, the velocity changes smoothly

(and so we can only use conservation of momentum during a collision, but we can usually use conservation of energy (also) both before and after the collision)

 

[jiayingsim123]

Thanks so much tiny-tim, you've been so kind! :) I posted another question on impulsive tension, just can't seem to grasp the topic well! :(