Solving Particles Colliding Homework: Mass of Struck Particle

  • Thread starter kateman
  • Start date
  • Tags
    Particles
In summary, the problem involves a neutron with a mass of 1.6 x 10^-27 and an unknown stationary particle. After an elastic collision, the neutron rebounds with a velocity of 0.7Vo in the opposite direction. To find the mass of the struck particle, equations for conservation of momentum and kinetic energy are used. The final equations should be m1v0 = m1(-0.7v0) + m2v2 for momentum and m1v0^2 = m1(0.7v0)^2 + m2v2^2 for kinetic energy. Care should be taken with the directions of velocities and squaring the entire velocity.
  • #1
kateman
114
0

Homework Statement


A neutron (m= 1.6 x10-27) moving with speed Vo strikes a stationary particle of unknown mass and rebounds elastically straight back along its path with a velocity of .7Vo. What is the mass of the struck particle?


Homework Equations


P initial = P final
Ke initial = Ke final

The Attempt at a Solution


alright i tried saying that the velocity imparted on the stationary particle was .3Vo but that gave me an answer of a mass of a neutron when the actual answer is 9.46 x 10-27.

i know: v1i = Vo
v2i = 0
v1f = .7Vo

but i am unsure as how to continue on from here. Please help.
 
Physics news on Phys.org
  • #2
Let mass of the stationary particle be m and its velocity after collision be V. Write down the equations for conservation of momentum and kinetic energy (keep in mind the +/- sign of the velocities). Solve these to obtain m. [If you are able to write down the equations but do not know how to solve them to get m, post again writing down upto the step you are able to reach]
 
  • #3
yeah i understand what your saying:

1/2m1Vo^s = 1/2m1v1f^s+ 1/2m2 .7Vo^s

m1Vo = m1v1f + m2 .7Vo

what i have done is rearrange it for one value and then substitute it into the other equation but this still gives me the wrong answer. any ideas?
 
  • #4
kateman said:
yeah i understand what your saying:

1/2m1Vo^s = 1/2m1v1f^s+ 1/2m2 .7Vo^s

m1Vo = m1v1f + m2 .7Vo
You should recall that momentum is a vector quantity and hence, direction is important.
 
  • #5
1/2m1Vo^s = 1/2m1v1f^s - 1/2m2 .7Vo^s

ahhh well a negative in place of a positive might help :P
thanks hootenanny!
 
  • #6
Careful though, I'm sure 1/2m2(0.7v02) should be 1/2m1(0.7v02). Assuming that m1 is your known mass
 
  • #7
argh its these small errors that really throw me off
thanks again :)
 
  • #8
kateman said:
argh its these small errors that really throw me off
thanks again :)
No problem :smile:
 
  • #9
alright so i used

m1Vo^2 = -m1 .7Vo^2 + m2 V2f

v2f = m1Vo^2 + m1 .7Vo^2/m2

then i subs the v2f into the kinetic formula equation and ended up with this:

m1Vo^2 + m1 .7Vo^2/m2 = (m1Vo^2 + m1 .7Vo^2/m2)^2

0 = m1Vo^2 + m1 .7Vo^2/m2

alright, have i gone wrong somewhere and if so, where?
 
  • #10
kateman said:
alright so i used

m1Vo^2 = -m1 .7Vo^2 + m2 V2f

v2f = m1Vo^2 + m1 .7Vo^2/m2

then i subs the v2f into the kinetic formula equation and ended up with this:

m1Vo^2 + m1 .7Vo^2/m2 = (m1Vo^2 + m1 .7Vo^2/m2)^2

0 = m1Vo^2 + m1 .7Vo^2/m2

alright, have i gone wrong somewhere and if so, where?
I'm not sure what you've done there, I think you may have got your equations mixed up. Kinetic energy is a scalar quantity; i.e. direction doesn't matter. And make sure you square the whole velocity. I'll start you off;

Momentum
[tex]m_1v_0 = m_1(-0.7\times v_0) + m_2v_2[/tex]

Kinetic Energy
[tex]m_1v_0^2 =m_1(0.7\times v_0)^2 + m_2v_2^2[/tex]
 
  • #11
hmm i see the directions with ke did stuff me up

if the .7Vo velocity is negative, could you not just say that m1.7Vo as a whole is negative, and then rearrange as such?

i try to rearrange it but keep stuffing it up.

what should i do, rearrange on for v2f and then substitute it in the other? if so i need some more guidance
 

1. What is the formula for calculating the mass of a struck particle in a collision?

The formula for calculating the mass of a struck particle in a collision is m1 = m2(v2-v1)/v1-v2, where m1 is the mass of the struck particle, m2 is the mass of the striking particle, v1 is the velocity of the struck particle before the collision, and v2 is the velocity of the struck particle after the collision.

2. How do I determine the velocity of the struck particle before and after the collision?

The velocity of the struck particle before and after the collision can be determined using the momentum conservation equation, which states that the total momentum of a system before a collision is equal to the total momentum after the collision. This can be written as m1v1 + m2v2 = m1v1' + m2v2', where v1 and v2 are the velocities before the collision, and v1' and v2' are the velocities after the collision.

3. Can I use this formula for any type of collision?

Yes, this formula can be used for any type of collision, as long as the collision is elastic and the total momentum is conserved.

4. What units should I use for the mass and velocity in this formula?

The mass should be in kilograms (kg) and the velocity should be in meters per second (m/s). This will give the mass of the struck particle in kilograms.

5. Are there any other factors I need to consider when using this formula?

Yes, it is important to make sure that the velocities are in the same direction and that the units are consistent. It is also important to take into account any external forces acting on the particles during the collision. Additionally, this formula assumes an elastic collision where no energy is lost during the collision. If the collision is inelastic, the formula will need to be modified to account for the loss of energy.

Similar threads

  • Introductory Physics Homework Help
Replies
2
Views
2K
  • Introductory Physics Homework Help
2
Replies
54
Views
8K
  • Introductory Physics Homework Help
Replies
3
Views
2K
  • Introductory Physics Homework Help
Replies
16
Views
4K
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
9
Views
5K
  • Introductory Physics Homework Help
Replies
4
Views
951
  • Introductory Physics Homework Help
Replies
9
Views
3K
  • Introductory Physics Homework Help
Replies
17
Views
3K
  • Introductory Physics Homework Help
Replies
2
Views
884
Back
Top