1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Question about Newton's first law

  1. Oct 4, 2013 #1
    1. The problem statement, all variables and given/known data

    All external forces on a body cancel out.
    Which statement must be correct?

    A The body does not move.
    B The momentum of the body remains unchanged.
    C The speed of the body remains unchanged.
    D The total energy of the body remains unchanged.

    2. Relevant equations

    3. The attempt at a solution

    I know that B is the best answer and that A and D is wrong, but I don't know why C is incorrect.

    If you cancel out every forces on the body, surely the body will continue its motion with the same velocity in a straight line, and since the direction of motion is constant, then certainly the speed should be the same too.

    Now consider a circular object being applied a torque. My question is, since the lines of action do not meet, can you say that the forces "cancel out"?

    If you say that it does cancel out, then considering the changing of the velocity of the circular object, B would be wrong.

    If you say it doesn't cancel out, that means torques/moments are not considered in this question at all, so what is left is just bodies traveling in a straight line whose speed can't possibly change when there's no external force. So does this support my claim that C is correct?
    Last edited: Oct 4, 2013
  2. jcsd
  3. Oct 4, 2013 #2
    C is the most correct in my opinion.
  4. Oct 4, 2013 #3
    No - B is the best answer because

    Force is defined as the net rate of change of momentum.[STRIKE][/STRIKE]

    [tex] \sum_i \vec{F_i}=0 = \frac{d \vec{\ p}}{dt} [/tex]

    However, the momentum being constant does not mean that the speed has to be the same, nor the Energy to be the same.

    Mathematically you can see it like:

    [tex] \frac{ d\vec{p} }{dt } = 0 = m \frac{d \vec{v}} {dt} + \vec{v} \frac{d m }{dt } [/tex]
    [tex] \frac{d \vec v}{dt} = \frac{-1}{m} \left(\vec{v} \underbrace{\frac{d m}{dt}}_{\neq 0}\right) [/tex]

    This is a tricky point, however, physically you can think about it like this:

    If the moving object (say at a constant velocity) loses some its mass while travelling (or the "effective" mass of the object is changing for some reason) the velocity could increase while the momentum is constant (due to decreasing mass),

    Similarly, you can express Kinetic Energy as:

    [tex] E = \frac{|\vec{p}^2|}{2 m} [/tex] and even if the momentum is constant, a change in mass could change the total Energy ..., hence D isn't the correct answer either.

    Hope this helps,
    Last edited: Oct 4, 2013
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted