Conversion of Mass to kinetic energy.

AI Thread Summary
A 2.0 kg lump of uranium has half its mass converted into kinetic energy, raising the question of how fast the remaining half moves. The relevant equations include the Lorentz factor (gamma) and kinetic energy formulas. The initial attempt yielded a velocity of 0.745c, but the expected answer is 0.87c. The discussion highlights confusion over algebraic manipulation and the correct application of the equations. Ultimately, the correct answer is confirmed to be option c, 0.87c.
theshanthan
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Homework Statement



A lump of uranium has a mass of 2.0 kg, and begins at rest. Half of the lump’s mass is going to be totally converted into kinetic energy of the other half. After this is done, how fast is the remaining half going?

a. 0.60 c
b. 0.80 c
c. 0.87 c
d. 1.0c

Homework Equations



gamma = 1/ sqrt(1-(v^2/c^2)
KE = (gamma - 1)mc^2
(KE + mc^2)initial = (KE + mc^2)final

The Attempt at a Solution



0 + 2c^2 = C^2 + KE
KE = C^2

(gamma - 1)mc^2 = c^2
gamma = 3/2

V = 0.745

But according to the answer I am wrong. I have a final and I am trying to study for it, but this is not making any sense. as I posted above the answer is C. Any help would be appreciated!
 
Physics news on Phys.org
Solving (γ-1)m1c2 = m2c2 with m1 = m2 does not give γ = 3/2.
 
Thank you! Thanks a lot! I feel smart and dumb at the same time. Smart because I figured out the way to solve it all by myself and dumb because I just can't do super simple algebra!
 
Thread 'Variable mass system : water sprayed into a moving container'

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}...
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