Momentum and Collisions of a particle

In summary, the conversation was about a particle p colliding elastically with another particle N, which was initially at rest. The p particle was deflected through 90°, leaving with a speed of 2.3 m/s and a mass of 2 kg. The conversation included questions about the angle of the recoiling N relative to the incident direction of p, the magnitude of the momentum of the recoiling N, the change in energy of p, the speed and mass of the recoiling N. The equations used were mv=p and mpv0p + mNv0N=mpvfp + mNvfN for the x and y directions. The person attempting the solution made a mistake in calculating the initial momentum
  • #1
nastassja
6
0

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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  • #2
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.
 
  • #3
I can't believe I made such a stupid mistake. Thanks.
 

Related to Momentum and Collisions of a particle

1. What is momentum and how is it calculated?

Momentum is a measure of an object's motion and is calculated by multiplying the object's mass by its velocity. The equation for momentum is p = m x v, where p is momentum, m is mass, and v is velocity.

2. How does momentum affect collisions between particles?

In a collision between two particles, the total momentum before the collision is equal to the total momentum after the collision. This means that the momentum of one particle may change, but the total momentum of both particles remains constant.

3. What is the difference between elastic and inelastic collisions?

In an elastic collision, both the total momentum and total kinetic energy of the particles are conserved. In an inelastic collision, only the total momentum is conserved, and some kinetic energy is lost as the particles stick together or deform upon impact.

4. How does the Law of Conservation of Momentum apply to collisions?

The Law of Conservation of Momentum states that in a closed system, the total momentum before a collision is equal to the total momentum after the collision. This means that momentum is always conserved in collisions between particles.

5. How can momentum be used to predict the outcome of a collision?

By calculating the momentum of each particle before a collision and analyzing the forces acting on them, we can predict the outcome of the collision and determine the final velocities of the particles. This is useful in understanding the behavior of particles in various situations, such as in car crashes or sports collisions.

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