39Yes it's (a), (c).
If the resultant of all the external forces acting on a system of particles is zero, then from an inertial frame one can surely say that
(A) linear momentum of the system does not change in time
(B) kinetic energy of the system does not change in time
(C) angular momentum of the system does not change in time
(D) potential energy of the system does not change in time
Sol. (A)
i think it must be a,c??
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9 Answers
39\sum{F} = 0 => \vec{P} = constant => \vec{r} \times \vec{P} = constant => \vec{L} = constant
Hence (A), (C).
Note : Some would berate me and quote the example of a force couple to disprove this. Let me remind you that this is a system of particles and not a force couple. You can't tell whether those external forces are forming a couple or not.
Open to debate though....let's see.
1Okay A is the only answer.
C is wrong becoz even though the sum of forces on a system might be zero, they could still produce a NET TORQUE, which would change both the angular P and the Rotational KE of the system.
And The formula, P times r, works for the case when ONE force alone acts on system, so that THERE IS ONLY ONE R. but when its forceS tat act on the system, u will have different r for each force, and as a net calculation, u Will get a change in angular momentum.
:)
1And to clarify, B is also wrong as
Though Linear KE wont change, if angular momentum changes, rotational KE will also change!
39Thanks a ton for the clarification. So final ans is (A) only.
1Answer is only A...
For a system of 'n' particles in which \sum_{1}^{n}{\vec{F_i}}=0:
\sum_{1}^{n}{\vec{P_i}}=constant
This certainly does NOT imply:
\sum_{1}^{n}{\vec{r_i}\times \vec{P_i}}=Constant
What Pritish had written is valid for a single particle system