Is Velocity Equal to Work Divided by Impulse in Physics?

[lonelypancreas]

Since Work = F*delta(d) and Impulse = F*delta(t) then dividing work over impulse, through simple cancellation of F we can say that it is now equal to delta d / delta t which is equal to velocity right? My question is, does this make sense "physics-wise" since I onlu arrived at my answer through cancellations?

 

[Buzz Bloom]

Hi lonely:

Your equations should be:

delta(Work) = F*delta(d)
delta(Impulse) = F*delta(t)​

delta(Work) is an infinitesimal of Work = an infinitesimal of energy = ΔE
delta(Impulse) is an infinitesimal of Impulse = an infinitesimal of momentum = Δp​

From

https://en.wikipedia.org/wiki/Photon#Physical_properties​

In empty space, the photon moves at c (the speed of light) and its energy and momentum are related by E = pc, where p is the https://www.physicsforums.com/javascript:void(0) of the momentum vector p.

Consider a photon which has momentum p and energy pc.
The result of dividing the photon's energy by its momentum gives its velocity c.

Hope this helps.

Regards,
Buzz

 

[lonelypancreas]

E = pc is the same as E = (mc)c --> E = mc^2 right? I still haven't encountered relativity so I'm not quite familiar with that famous equation but I think through your answer, it made sense. Thanks.

 

[PeroK]

Since Work = F*delta(d) and Impulse = F*delta(t) then dividing work over impulse, through simple cancellation of F we can say that it is now equal to delta d / delta t which is equal to velocity right? My question is, does this make sense "physics-wise" since I onlu arrived at my answer through cancellations?

If you have $KE = \frac{1}{2}mv^2$ and $p = mv$, then

$\frac{KE}{p} = \frac{1}{2}v$

And:

$KE = \frac{p^2}{2m}$

You might like to think about why

$v \ne Work/Impulse$

 

[jtbell]

E = pc is the same as E = (mc)c --> E = mc^2 right?

No, because p ≠ mc.

The general relativistic relationship between energy, momentum and (rest) mass is E² = (pc)² + (mc²)². Set m = 0 and you get E = pc.

 

[PeroK]

Given this was a post on classical mechanics in the Classical Physics section, I'm not sure how we ended up talking about relativity!

 

[Vanadium 50]

Given this was a post on classical mechanics in the Classical Physics section, I'm not sure how we ended up talking about relativity!

Because this is PF. You can't write down an inclined plane problem without someone chiming in about GR.