# How impulse momentum theorem is put to use?

Guys the below link is a question already asked in this website and I have referred the answers given by the users and some of them were good ,but I have few doubts and I wanted to ask it in a comment but my reputation wasn't 50,so i'm posting it as a question. Sorry for the duplication! the link is this: Why isn't momentum conserved in this pulley problem?. My doubt is this: In the solution give in the users question the impulse imparted to the partice should be mV-mv right? What I mean is that it should be final momentum -initial momentum right? And,impulse imparted for block should be -mV right? i.e it should negative because the block moves up. Please forgive my bad presentation of the problem.

• Pay attention to the sign convention. The question says the impulse is measured upwards, so final momentum is $-mV$ and initial momentum is $-mv$ therefore upwards impulse on particle change in momentum is $(-mV)-(-mv)=mv-mV$. See David White's answer for a clearer solution. – sammy gerbil Jun 19 '18 at 18:39
• Don't you think that david whites answer is wrong?...Because he took upwards positive on one side and downwards positive on the other side ,whereas everything should be taken either "up positive" or "down positive...i.e all calcualtions must be donw with one sign convention. – scisyhp Jun 20 '18 at 1:57
• @scisyhp Consider a table pulley problem, here a part of the system moves horizontally and the other vertically: one simply can't take "everything either 'up positive' or 'down positive'". If you want to ignore the force the pulley exerts, you have a constraint that essentially 'bends' the 1D space your problem now lives. What matters then is internal consistency. In the horiz. pulley example that means right and down are positive, in the vertical pulley problem you link that means one side down positive and the other side up positive. – stafusa Jun 20 '18 at 8:50

## 1 Answer

The outlined solution has up positive, but $v$ and $V$ are implicitly positive in a downward direction. With up, $v$, and $V$ positive, the ball's initial momentum is $-mv$ and it's final momentum is $-mV$. The change in momentum is thus $(-mV) - (-mv)$ or $mv-mV$.

Note that with up positive, it would perhaps have been better to write equation (ii) as $$\int(T-N)\,dt = -mV$$ Negating both sides results in equation (ii) as written in the outlined solution.

• In the equation (ii)(of yours) ,don't you think that it should be (+mV) because the impulsive force on pan will be more due to the ball(impulsive normal) than the tension(impulsive tension) in the string and hence the ball moves down .So it should be {integral(N-T)dt=-mV} ...i.e {integral(T-N)dt=mV}...considering the up direction as positive(+ve). – scisyhp Jun 20 '18 at 1:53