A question on impulsive tensions

jiayingsim123

1. The problem statement, all variables and given/known data
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=...0to%22&f=false

Page 113 Exercise 5B Question 2


2. Relevant equations
Impulsive tension: mv-mu
g=10ms^-2
3. 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

I agree, but warn that you are courting trouble if you don't type this as √(2g)
Or you could have said the KE of the moving block becomes the KE of the pair.
Can you explain why you say v=0? (As I explained, I can't see the diagram at your link.)

azizlwl

NascentOxygen

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!
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

for the first metre, u = 0

for the second metre, u = (√2)g/2
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

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

TSny

Since I feel a little grumpy today , 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

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
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 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)