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

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The discussion revolves around the speed of an alpha particle emitted from an atomic nucleus, as well as a separate scenario involving a collision between two cars. The subject areas include nuclear physics and classical mechanics.

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  • Mixed

Approaches and Questions Raised

  • The original poster attempts to apply conservation of momentum to find the speed of the alpha particle but questions their calculation after receiving feedback about the mass change of the nucleus. Another participant raises a collision problem involving two cars and seeks clarification on how to derive the speed of the Toyota at impact, expressing confusion about the necessary equations.

Discussion Status

Participants are actively engaging with each other's posts, providing corrections and suggestions. Some guidance has been offered regarding the application of equations of motion and the need to consider changes in mass during the alpha particle emission scenario. The conversation is ongoing, with participants exploring different interpretations and approaches.

Contextual Notes

In the first problem, there is a critical assumption regarding the mass of the nucleus after the emission of the alpha particle. In the second problem, the original poster expresses uncertainty about how to derive a second equation needed to solve for the unknown speed.

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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 [tex]222u * 420 m/s + 0 = 222u * 350m/s + 4 u * X m/s[/tex] 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
[tex]1.0 * 10^3 v_1 = (1.0x10^3 + 2.2 * 10^3) v'[/tex]

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

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

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