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I’m sorry for so simple question, but I just need to be sure.

I understand, that the changing of the speed occurs only when the force is applied, I understand that if one punch a ball in the free space it will infinitely move with a constant velocity

Some point-like body with mass $m$ is situated in vacuum, and has initial velocity $v_1=0 \space m/s$.

Some force is now acting on a body for a infinitely short period of the time.

The acceleration that gives the application of this force to body equals $a=5 \space m/s^2$.

The velocity after will be $v_2=0+5 =5\space m/s$?


Also, if the force is acting for a non-infinitely short period of time how to calculate then?

I found this from https://physics.stackexchange.com/a/231120/255554

$$x=( x + \frac{|F| }{2m} t^{2} ) $$

Seems it can be applied for both of my cases, but I don’t know why there is 2 times mass


And, can you, please confirm, if 1 Newton is the force that during 1 second changes the 1 kg body velocity on 1 m/s, then 2 Newtons is the force that changes:

  • if mass is same: during 1 second velocity on 2 m/s
  • if mass is 2 kg: during 1 second velocity on 1 m/s

Am I understanding correctly?

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3 Answers 3

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Firstly, the body will only accelerate while the force is being applied, and it will move at a constant velocity the instant the force stops being applied.

Your final equation is just a variation on

$$x=\frac12at^2$$

Why that factor of ½ arises can be shown using elementary calculus, or by a geometrical argument.


Both statements about a 1 Newton force in your update are correct.

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  • $\begingroup$ Your equation doesn’t have initial x $\endgroup$
    – Artur
    Commented Mar 3, 2020 at 19:28
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    $\begingroup$ To be fair, they did say it was “a variation” of that equation. $\endgroup$ Commented Mar 3, 2020 at 19:32
  • $\begingroup$ @Artur The full form of that equation is $x=x_0 + ut + \frac12at^2$, where $x_0$ is the position at $t=0$ and $u$ is the speed at $t=0$. I should also mention that these equations are only valid when $a$ is constant. $\endgroup$
    – PM 2Ring
    Commented Mar 3, 2020 at 19:50
  • $\begingroup$ Thank You, but I actually didn’t understand from where appears 1/2. It somehow related to kinetic energy? If we a have a second Newton law $F=m\frac{dv}{dt}$ or $F=m\frac{dx^2}{dt}$. Just logically we have acceleration, say 5, then we have $5=\frac{v_1-v_0}{t_1-t_0}=\frac{v_1-0}{1-0}$ and $5=v_1$ $\endgroup$
    – Artur
    Commented Mar 3, 2020 at 20:08
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    $\begingroup$ @PM2Ring, now understood. This is the equal acceleration movement formula, you integrate it and get that formula. Thank You! $\endgroup$
    – Artur
    Commented Mar 3, 2020 at 21:53
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In order to determine the velocity as a result of the application of the force, you need to know the duration (time) of the application of the force on $m$, or the displacement $x$ of the mass $m$ during the application of the force, and the force as a function of time if not constant.

Your value of $v_2$ is based on a constant acceleration of 5 m/s$^2$ due to a constant force applied for a duration of 1 second, and comes from the equation:

$$v_{f}=v_{i}+at$$

where $v_{i}$ is the initial velocity and $v_f$ is the final velocity

You equation for displacement $x$ is based on the equation

$$x_{f}=x_{i}+\frac{at^2}{2}$$

Where $m_f$ and $m_i$ are the final and initial displacements

Substituting $$a=\frac{F}{m}$$

Gives

$$x_{f}=x_{i}+\frac{F}{2m}t^2$$

Hope this helps.

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  • $\begingroup$ Well, seems I understand correctly, can You also check my update? $\endgroup$
    – Artur
    Commented Mar 3, 2020 at 19:21
  • $\begingroup$ @Artur the update looks good $\endgroup$
    – Bob D
    Commented Mar 3, 2020 at 20:12
  • $\begingroup$ Thank You then. $\endgroup$
    – Artur
    Commented Mar 3, 2020 at 20:35
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You appear to be applying the impulse $5 m Ns$, where $m$ is the mass (in kg), which leads to the applied force $$F(t) = 5 m \delta (t) N,$$ that is, the impulse acts for an infinitesimally small time at the time $t = 0$.

Applying Newton's Law, $F = m a$ with $a = \frac{d v}{d t} = \frac{d^2 x}{d t^2}$ yields $$\frac{ d v}{d t} = 5 \delta (t).$$

This can be easily integrated to give $$ \int_{v = 0}^{v(t)} dv = \int_{t' = - \infty}^{t'= t} 5 \delta (t') dt,$$ since for times less than $t = 0$ the particle is at rest and for $t > 0$ the particle travels with speed $v$. This gives $v (t) = 5 m/s$ for times $t > 0$. So you are correct, the speed after the impulse is a constant $5 m/s$.

Note that the mass $m$ cancels out in the problem as specified.

For the more general case when the force has the form $F = G(t)$ then have the result $$ m \int_{v = 0}^{v(t)} dv = \int_{t' = - \infty}^{t' = t} G (t') dt,$$ again assuming that the particle is at rest in the far past.

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