Determining Final Velocity in 2-Dimensional Collision Problem

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Homework Help Overview

The problem involves a collision between a neutron and a helium nucleus, where the initial conditions and some final velocities are provided. The goal is to determine the final velocity of the neutron after the collision, utilizing principles of conservation of momentum in two dimensions.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the conversion of units from km/s to m/s and attempt to apply conservation of momentum equations in both x and y directions. There are questions about the correctness of the calculated velocities and whether the mass of the particles has been properly accounted for in the calculations.

Discussion Status

Some participants have provided calculations and expressed uncertainty about the results, particularly regarding the final velocity of the neutron. There is an acknowledgment of potential errors in the approach, and suggestions to reconsider the impact of mass in the calculations have been made.

Contextual Notes

There is a mention of a possible misunderstanding regarding the units of the final velocity, as well as the need to account for the masses of the particles involved in the collision.

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Homework Statement


A neutron of mass 1.7 X 10^-27 kg traveling at 2.2 km/s, hits a stationary helium nucleus of mass 6.6 X 10^-27 kg. After the collision, the velocity of the helium nucleus is 0.53 km/s at 52° to the original direction of motion of the neutron. Determine the final velocity of the neutron.


Homework Equations



pi1x+pi2x=pf1x+pf2x
pi1y+pi2y=pf1y+pf2y

The Attempt at a Solution



I changed all the km/s to m/s since the answer is in m/s

(1.7X10^-27(2200)-6.6X10^-27(530cos52))/1.7X10^-27=vficos∅
933.18m/s=vficos∅

Well, it does not matter what i get for the y component since the answer 1.9 m/s and 933 is just too high.
So, is the answer wrong or did i do something wrong.
 
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I think the answer you were given should have been 1.9 km/s rather than 1.9 m/s.
 
Well, I did solve it for km/s I got 1.05 km/s.
vf2y=0.53sin52
vf2y=0.418km/s
vf2x=0.933 km/s

If you use the Pythagorean theorem you get 1.05km/s
 
dnawiam said:
Well, I did solve it for km/s I got 1.05 km/s.
vf2y=0.53sin52
vf2y=0.418km/s

You did not account for the masses of the particles in this calculation.
 
TSny said:
You did not account for the masses of the particles in this calculation.

You are right. How silly of me. Well, I got 1.9 km /m for my answer. Thanks for your help.
 

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