## A question on impulsive tensions

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:

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

 Blog Entries: 27 Recognitions: Gold Member Homework Help Science Advisor 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
 Recognitions: Homework Help 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?

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## A question on impulsive tensions

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

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 Quote by jiayingsim123 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.)

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 Quote by 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
Fair enough. Poster had better describe the setup then.

Ah, thanks azizlwl
 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! :)

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yes, thanks azizlwl!
 Quote by jiayingsim123 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)
 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! :)

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 Quote by jiayingsim123 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!
 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!

 Quote by jiayingsim123 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.
 Recognitions: Gold Member Homework Help 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 )

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 Quote by jiayingsim123 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
 Quote by TSny … 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)

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 Quote by tiny-tim 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?
 Blog Entries: 27 Recognitions: Gold Member Homework Help Science Advisor 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)

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