Momentum in elastic collision

In summary: v1 = (v - v1)/2v1² = (v² - 2*v*v + v*v²)/4v1 = (v - v1)/2*(mv² - mv)v1² = (v² - 2*v*v + v*v²)/4*(mv² - mv)
  • #1
Magma828
28
0

Homework Statement



Uranium fission within a nuclear reactor produces high-speed neutrons. These must be slowed down for them to trigger a chain reaction within the uranium. In a heavy-water reactor the neutrons of mass m make multiple elastic collisions with deuterium nuclei of 2m, losing speed at each collision.

(a) Suppose a fast-moving neutron of speed v makes a head-on elastic collision with a stationary deuterium nucleus. Show that its speed is reduced to v/3.

Homework Equations



p=mv - We probably have to use this
Ek=0.5mv2 - Not sure if we have to use this, but the question IS in a section about kinetic energy. It just seems logical to use p=mv because using kinetic energy makes no sense.

The Attempt at a Solution



Ok it took me absolutely AGES to do this, but I don't think it's right:

PD1 = mv = 2m x 0 = 0 N s
PN1 = mv = 1m x v = mv

PD1 + PN1 = PD2 + PN2 if no external forces are applied

0 + mv = PD2 + PN2
mv = PD2 + PN2
PD2 = mv - PN2

PD2 = PN2 - not sure

Substitute above two equations together to show PN2:
PN2 = mv - PN2
2PN2 = mv
PN2 = mv/2
vN2 = PN2/2m
vN2 = (mv/2)/2m
vN2 = mv/4m - my rearranging may be wrong
vN2 = v/3m

But the answer is just v/3 not v/3m. Is this still right? Or is it just coincidence I got an answer so similar?
 
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  • #2
v/3m cannot be right. it has units of m.kg/s
 
  • #3
The answer i got was -v/3. So the neutron bounces backwards with speed v/3 and the deuterium goes forward with 2v/3. I didn't read through your answer very thoroughly, but anywhere with mass on the bottom will be wrong. All the masses cancel out in my calculations
 
  • #4
dacruick said:
The answer i got was -v/3. So the neutron bounces backwards with speed v/3 and the deuterium goes forward with 2v/3. I didn't read through your answer very thoroughly, but anywhere with mass on the bottom will be wrong. All the masses cancel out in my calculations

Could you post your calculation please? You don't even have to explain it, but I've been trying to do this all night and I don't have a clue. Got an exam on thursday l0L!
 
  • #5
bah fine. I'm a lazy individual.
v is initial velocity
v1 is final velocity of the neutron
v2 is the final velocity of the deuterium

mv = m . v1 + 2m . v2 ------------- (momentum)
mv² = m . v1² + 2m . v2² ------------- (energy)

All the masses cancel:

v = v1 + 2 . v2 ------------- (momentum)
v² = v1² + 2 . v2² ------------- (energy)

isolate the momentum equation for v2
v2 = (v - v1)/2
v2² = (v² - 2*v1*v + v1²)/4

sub that into your energy equation and isolate for v1.
 

1. What is momentum in elastic collision?

Momentum in elastic collision is a physical quantity that describes the motion of an object. It is defined as the product of an object's mass and velocity.

2. What is the law of conservation of momentum?

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 the total momentum of the objects involved in an elastic collision remains constant.

3. How is elastic collision different from inelastic collision?

In an elastic collision, both kinetic energy and momentum are conserved, meaning that the objects involved bounce off each other without any loss of energy. In an inelastic collision, some kinetic energy is lost and converted into other forms of energy, such as heat or sound.

4. How do you calculate the final velocities in an elastic collision?

To calculate the final velocities in an elastic collision, you can use the equations of conservation of momentum and conservation of kinetic energy. These equations involve the masses and initial velocities of the objects involved in the collision.

5. Can momentum in an elastic collision be negative?

Yes, momentum in an elastic collision can be negative. This can occur when the initial velocities of the objects are in opposite directions, causing one object to have a negative velocity and therefore a negative momentum. However, the total momentum of the system will still be conserved.

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