# B Work/Impulse = Velocity?

1. Mar 23, 2016

### 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?

2. Mar 23, 2016

### Buzz Bloom

Hi lonely:

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
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) [Broken] 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

Last edited by a moderator: May 7, 2017
3. Mar 24, 2016

### 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.

4. Mar 24, 2016

### PeroK

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

$v \ne Work/Impulse$

5. Mar 24, 2016

### Staff: Mentor

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.

6. Mar 24, 2016

### 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!

7. Mar 24, 2016