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I konw that mass affects weight (force), so how does that relate to speed? F=ma. so how does all this affect speed?

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Speed of what? This question is imprecise--- mass affects acceleration, which indirectly affects speed. The mass times the speed is the magnitude of the momentum, and the different components of momentum along the x,y,z axis are conserved. Can you ask it more precisely? Like, if I apply the same force on two objects, one twice as heavy, what is the difference in the final speed. – Ron Maimon May 29 '12 at 20:54
Force affects change in speed. – ja72 May 30 '12 at 20:40

Mass doesn't affect speed directly. It determines how quickly an object can change speed (accelerate) under the action of a given force. Lighter objects need less time to change speed by a given amount under a given force.

Alternatively, mass determines how strong a force has to be to accelerate an object at a given rate. Lighter objects can do with weaker force to change speed by a given amount in a given amount of time.

Thus, mass is a measure of object's inertia which is resistance to changes in object's motion. This is what the equation

\begin{equation} F=ma \end{equation}


Note that speed is relative (i.e. depends on the choice of the frame of reference) and, in the framework of Newtonian mechanics, for each object there is a frame of reference in which it is moving at arbitrarily high velocity.

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Thank you so much! I feel smarter now because of your explanation! :) <3 – user15973 Nov 14 '12 at 14:44

Let us first assume that a constant force, $F$, is applied to a stationary body of mass $m$ from $t = 0$ to $t = T$.

Let us begin with the equation $F = ma$.

This can be re-written as $F = m\frac{dv}{dt}$.

After re-arranging the equation, we arrive at:

$\frac{dv}{dt} = F/m$. we can then integrate both sides with respect to time to give:

$v = \int_0^T \frac{F}{m}\mathrm dt$, where the limits of integration apply over the time the force is applied.

Since we have assumed that $F$ is constant between $t=0$ and $t=T$, we can calculate the integral to arrive at:

$v = \frac{F}{m}T$, assuming the mass is initially at rest before the force is applied.

As we can see, if the mass was doubled, the final velocity achieved at $t=T$ will be halved.

Secondly, one may consider how mass effects speed by considering kinetic energy. The formula for kinetic energy is given by:

$KE = \frac{1}{2}mv^2$.

Therefore, if a photon losses all its energy, $hf$, to a free particle of mass, $m$, the particle will travel at a speed of $\sqrt{2\frac{hf}{m}}$ after gaining the energy from the photon. If however a photon loses the same amount of energy to a particle of double the mass, $2m$, the particle will travel at a velocity of $\sqrt{\frac{hf}{m}}$, which is slower by a factor of $\sqrt{2}$.

In conclusion, we have seen that the final velocity of a heavy particle after the application of a constant force will be lower than that of a light particle. Similarly, a heavy particle will travel at a lower velocity than a light particle, given an equal amount of energy.

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protected by Qmechanic Jan 28 '13 at 23:05

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