[LiamSP]

Hi guys i'm having a bit of a problem with an impulsive tension question and i'm wondering whether you'd be able to help me out.

http://mathsathawthorn.pbworks.com/f/M3.pdf It's question 6 on that Paper.

I started by solving for the speed of block B just before the string became taught, I found its value to be 5.6 which agrees with the mark scheme.

Then, to calculate the velocity after the impulse I used conservation of momentum.

mass of Block A=7kg (stationary before impulse)
Mass of block B=3kg (velocity of 5.6 before impulse)

Using conservation of momentum: 7(0)+3(5.6)=7(-v)+3(v)
Where V is the common velocity post impulse and also taking downwards as +ve

thus v=-4.2 thus the speed is 4.2

however this is incorrect, the mark scheme has the value as 1.68 which is the result you get if you take momentum after impulse to be 7(v)+3(v)

Can anyone explain why ?

Thanks in advance

[Coursework.info]

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[njl94]

I think because it's on a string over the pulley, you can't take the directions to be opposite like that.
Consider only the direction relative to the string. Take "A to B" as positive and "B to A" as negative
If the string were horizontal, and B was accelerated horizontally at g, this problem is the same and you can see why the signs are the same.