How Do You Calculate the Momentum of an Alpha Particle?

[flyboy9]

Homework Statement

An alpha particle (a helium nucleus, containing 2 protons and 2 neutrons) starts out with kinetic energy of 10 MeV (10*10^6 eV), and heads in the +x direction straight toward a gold nucleus (containing 79 protons and 118 neutrons). The particles are initially far apart, and the gold nucleus is initially at rest. Answer the following questions about the collision.
What is the initial momentum of the alpha particle? (You may assume its speed is small compared to the speed of light).

Homework Equations

1 ev=1.69e-19 J= 1.69e-19 kg*m²/s₂
K=.5*m*v²
p=m*v
m_proton=m_neutron=1.7e-27 kg

The Attempt at a Solution

I solved for velocity of the particle then multiplied it by its mass to find momentum.
10*1.69e-19=.5*m*v²
3.38e-18=8.8e-27*v²
497058823.5=v²
22294.81607=v

P=mv
P=8.8e-27*22294.81607
P=1.96194e-22

 

[ManyNames]

um, yeh. It seems correct.

 

[flyboy9]

i thought so too... maybe the online program has the wrong answer.

 

[lightgrav]

10 Mega-eV ...

 

[flyboy9]

10 Mega-eV ...

what about it?

 

[lightgrav]

the solution you posted started with 10 eV , not 10 MeV ... Mega = 1 000 000 .

 

[flyboy9]

the solution you posted started with 10 eV , not 10 MeV ... Mega = 1 000 000 .

I would have never caught that. Thank you. The process seems correct to you?

 

[lightgrav]

the PROCESS looks okay, but where did you get m = 8.8e-27 kg ?

by the way, p^2 /2m is USUALLY more useful than ½mv^2 ... especially in collisions !

 

[flyboy9]

the PROCESS looks okay, but where did you get m = 8.8e-27 kg ?

by the way, p^2 /2m is USUALLY more useful than ½mv^2 ... especially in collisions !

I used numbers in the back of our book that our homework is based off of

 

[lightgrav]

proton mass is 1.673 e-27kg ... only keeping 2 digits leads to severe roundoff error.

But I was asking "what computation did you use" which yielded that as the answer.
... it is a "leading" question

 

[flyboy9]

proton mass is 1.673 e-27kg ... only keeping 2 digits leads to severe roundoff error.

But I was asking "what computation did you use" which yielded that as the answer.
... it is a "leading" question

This homework which is based off of our book only uses 2 significant digits for mass of proton and neutron.
total mass = 4*1.7e-27
= 6.8e-27

I apologize for the typo earlier as well.