Momentum and Collisions of a particle

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SUMMARY

The discussion focuses on the elastic collision between two particles, specifically analyzing the momentum and energy changes involved. Particle p, with a mass of 2 kg and an initial speed of 3 m/s, collides with particle N, initially at rest, and is deflected at a 90° angle with a final speed of 2.3 m/s. The key equations used include momentum conservation (mv=p) and the conservation of kinetic energy. The angle of the recoiling particle N relative to the incident direction of particle p is calculated using inverse tangent, emphasizing the importance of correct coordinate orientation in calculations.

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  • Understanding of elastic collisions in physics
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nastassja
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Homework Statement



A particle p traveling with a speed of vpi = 3 m/s hits and scatters elastically from another particle N, initially at rest. Particle p is deflected through 90°, leaving with a speed of vpf = 2.3 m/s, and a mass mp = 2 kg.
a) What angle (in degrees) does the recoiling N make to the incident-p direction?

b) What is the magnitude of the momentum of the recoiling N?

c) What is the change in the energy of the p?

d) What is the speed of the recoiling N?

e) What is the mass of the N?

Homework Equations


mv=p
mpv0p + mNv0N=mpvfp + mNvfN

(for x and y directions)


The Attempt at a Solution



a) Since N is at rest at first, and p is perpendicular to the x-axis after the collision, I assumed that the pfNx would equal the initial momentum of p, or 2kg*3m/s or 5kg*m/s.

For the y direction, the momentum of N has to be equal and opposite p's momentum, so it is = -4.6kg*m/s.

I used inverse tangent to get the angle = -47.4 which is wrong. Where did I go wrong?

b)
c)
d)
e)

I have c) and I am sure I know how to do the rest if I can get a) right. Thanks.
 
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Hi nastassja,

nastassja said:

Homework Statement



A particle p traveling with a speed of vpi = 3 m/s hits and scatters elastically from another particle N, initially at rest. Particle p is deflected through 90°, leaving with a speed of vpf = 2.3 m/s, and a mass mp = 2 kg.
a) What angle (in degrees) does the recoiling N make to the incident-p direction?

b) What is the magnitude of the momentum of the recoiling N?

c) What is the change in the energy of the p?

d) What is the speed of the recoiling N?

e) What is the mass of the N?

Homework Equations


mv=p
mpv0p + mNv0N=mpvfp + mNvfN

(for x and y directions)


The Attempt at a Solution



a) Since N is at rest at first, and p is perpendicular to the x-axis after the collision, I assumed that the pfNx would equal the initial momentum of p, or 2kg*3m/s or 5kg*m/s.

This should be 6 kg m/s.

For the y direction, the momentum of N has to be equal and opposite p's momentum, so it is = -4.6kg*m/s.

I used inverse tangent to get the angle = -47.4 which is wrong. Where did I go wrong?

Also, you performed the inverse tangent incorrectly. The initial particle is moving along the x axis; since they want the angle with the incident direction of the p particle, that means they are asking for the angle relative to the x-axis.

But you calculated -47.4 by using:

[tex] \tan^{-1} \left( \frac{5}{-4.6} \right)[/tex]

If you want the angle from the x-axis, you have to use y/x in the inverse tangent; but you used x/y.
 
I can't believe I made such a stupid mistake. Thanks.
 

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