What is the speed of an alpha particle emitted from an atomic nucleus?

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An atomic nucleus moving at 420 m/s emits an alpha particle, resulting in the new nucleus slowing to 350 m/s. The mass of the original nucleus is 222 u, and after emission, it becomes 218 u. The calculation for the alpha particle's speed was initially incorrect due to not adjusting the nucleus's mass after emission. In a separate scenario, a Toyota collides with a Cadillac, and the user struggles to find the post-collision speed due to a lack of equations. The discussion emphasizes the importance of correctly applying conservation of momentum and accounting for changes in mass during particle emissions.
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"An atomic nucleus initially moving at 420 m/s emits an alpha particle in the direction of its velocity, and the new nucleus slows to 350 m/s. If the alpha particle has a mass of 4.0u and the original nucleus has a mass of 222 u, what speed does the alpha particle have when it is emitted?"

I did 222u * 420 m/s + 0 = 222u * 350m/s + 4 u * X m/s and I get 3885 m/s, but the answer says it's 4200. What did I do wrong?
 
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The mistake is in the RHS of your equation. The new nucleus does not have mass 222 u anymore, its new mass is 218 u.
 
Ah, I see. Thanks!
 
Here's another one...

A 1.0x10^3 kg Toyota collides into the rear end of a 2.2x10^3 kg Cadillac stopped at a red light. THe bumpers lock, the brakes are locked, and the cars skid forward 2.8m before stopping. The coefficient of kinetic friction between the tires are .40. What's the speed of the Toyota at impact?

I know that the equation is
1.0 * 10^3 v_1 = (1.0x10^3 + 2.2 * 10^3) v'

and that the force lost to friction is
3.2*10^3 kg * g * .40 = 35,123N

but I don't know how to find the v` of the equation...as I don't know where to get a second equation.
v_1 - v_2 = v_2' - v_1'
doesn't work since it just ends up equaling 0...how do I do this?
 
\vec{F}_{friction}=m_{total}\vec{a}
Any equations of motion you can use that include displacement, acceleration, and velocities?
 
Ah...I get it! Tricky :-p Thank you!
 
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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