Conservation of Energy and Momentum

AI Thread Summary
The discussion revolves around a physics problem involving a 1g object on a frictionless surface absorbing light with a frequency of 632nm and reaching a speed of 1mm/s. Two methods were employed to calculate the number of absorbed photons: one using conservation of energy and the other using conservation of momentum, resulting in different answers. The conservation of momentum approach is deemed more reliable due to the inelastic nature of the collision. Questions arise about the fate of the energy, suggesting it may have been converted to heat or electrical energy. Ultimately, the conversation emphasizes the complexities of energy conservation in such scenarios.
Glass
Messages
24
Reaction score
0

Homework Statement


I have a 1g object on a frictionless surface being hit with light of a frequency 632nm (directly parrallel to the surface and the object absorbs all the light). How many photons did the object absorb by the time it's moving at 1mm/s?


Homework Equations




The Attempt at a Solution


Well I worked it out two different ways and I got two different answers:

My first attempt is via conservation of energy:
n*h*c/lambda + m*c^2 = gamma*m*c^2
where
n is the number of photons
h is Planck's constant
c is the speed of light
lambda is the wavelength
m is the mass of the object
gamma is the 1/sqrt(1-v^2/c^2)
Then I solved for n.

My second attempt is via conservation of momentum:
n*h/lambda = gamma*m*v
Then I solved for n.

In both cases I got different answers. So I'm not sure what's wrong.
 
Physics news on Phys.org
Trust conservation of momentum before you trust a naive conservation of energy calc. The collision is inelastic.
 
Ok thanks. But what happened to the energy? Did it radiate as heat or something?
 
Some of it could have been converted into electrical energy if the surface is connected to a power grid. Conservation can't tell you where it went. But yes, most likely answer is heat.
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top