Elastic collision with particles, find the kinetic energy

In summary, to find the fraction of the proton's kinetic energy transferred to the alpha particle in a completely elastic collision, one must apply the conservation of momentum and use the relationship between the final velocities of the two particles. Then, the fraction can be calculated by dividing the kinetic energy of the alpha particle by the initial kinetic energy of the proton.
  • #1
Mathias Girouard
1
0

Homework Statement



A proton strikes a stationary alpha particle (4He nucleus) head-on. Assuming the collision is completely elastic, what fraction of the proton’s kinetic energy is transferred to the alpha particle?

Homework Equations


Pi = Pf
Ki = Kf

The Attempt at a Solution


Tried finding the speed with:

mv^2 = m1v1^2 + 4 x m2v2^2

m is canceled out

So,
v^2 = v1^2 + 4 x v2^2

Kinda stuck from this point on! I have no idea what to do.
 
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  • #2
Hi Mathias Girouard and welcome to PF.

How about conserving momentum? You wrote the equation, but you did not use it.
 
  • #3
Working directly with KE in collision problems leads to tedious algebra (all those squares of velocity). Instead, use the entirely equivalent conservation rule that the speed of approach is equal to the speed of recession. That is, if the initial velocities of the objects are ##v_1## and ##v_2##, and their final velocities are ##v'_1## and ##v'_2##, then ##v_2 - v_1 = v'_1 - v'_2##. This is in fact a consequence of conservation of energy.

And as @kuruman suggests, you need to invoke conservation of momentum, too, to solve for the two velocities. Two equations in two unknowns.
 
  • #4
Alternate method:

Momentum transfer - elastic collision: $$Δp=2μΔv$$
where μ is the reduced mass of the colliding objects:$$μ=\frac{m_1m_2}{m_1+m_2}$$
and Δv their relative velocity along the collision line. Then energy transfer to the stationary object ##m_2## is just ##(Δp)^2/(2m_2)##
 
Last edited:
  • #5
neilparker62 said:
Alternate method:

Momentum transfer - elastic collision: $$Δp=2μΔv$$
where μ is the reduced mass of the colliding objects:$$μ=\frac{m_1m_2}{m_1+m_2}$$
and Δv their relative velocity along the collision line. Then energy transfer to the stationary object ##m_2## is just ##(Δp)^2/(2m_2)##
Is this something you'd expect a student to memorize? It may be an elegant result, but it's one that's pretty specific in its application.
 
  • #6
gneill said:
Is this something you'd expect a student to memorize? It may be an elegant result, but it's one that's pretty specific in its application.
Thanks for the kind compliment.

The way I look at it is that it's something like the quadratic formula. A student should be able to derive the formula (and hence understand the underlying principles of conservation of momentum as well as the applicable energy equations) but thereafter use it to solve collision problems just like he/she would use the quadratic formula to solve quadratic equations rather than plough through completion of square all the time.

I wouldn't say the formula is all that specific since elastic collisions crop up quite frequently in physics problems. Furthermore it has a more general form namely: $$ Δp = (1+e)μΔv $$ where e is the coefficient of restitution. Thus for perfectly elastic collisions we have ## Δp = 2μΔv ## and for perfectly inelastic collisions ## Δp = μΔv ##. Between the latter 2 , I would say one can deal with a wide range of collision problems in a fairly straightforward fashion.

Use of the formula does relate to the underlying principle of collisions as "Newton 3 events" if I may put it that way. Equal and opposite forces result in equal and opposite impulses. When you use the formula all you are doing is calculating exactly what that "equal an opposite" impulse is.

I am experimenting with it's application in 2D collison problems - bit more tricky but that's generally the case anyway.
 
  • #7
Mathias Girouard said:

Homework Statement



A proton strikes a stationary alpha particle (4He nucleus) head-on. Assuming the collision is completely elastic, what fraction of the proton’s kinetic energy is transferred to the alpha particle?

Homework Equations


Pi = Pf
Ki = Kf

The Attempt at a Solution


Tried finding the speed with:

mv^2 = m1v1^2 + 4 x m2v2^2

m is canceled out

So,
v^2 = v1^2 + 4 x v2^2

Kinda stuck from this point on! I have no idea what to do.

You would have to similarly apply the conservation of momentum (since, it's an elastic collision), which would give you two different equations involving the initial and final velocities. Therefore, you'd be able to represent the final velocity of proton to the final velocity of alpha particle. Then, solve for the fraction of proton's kinetic energy transferred to the alpha particle.

Pi = Pf
mv = m(-v1) + 4mv2
v = 4v2 - v1
v1^2 = 16v2^2 - 8v1v2 + v1^2
v1^2 + 4 x v2^2 = 16v2^2 - 8v1v2 + v1^2
12v2^2 = 8v1v2
3/2v2 = v1

Fraction of proton's kinetic energy transferred to the alpha particle = (K.E alpha) / (K.E initial proton)Good Luck
 

What is an elastic collision?

An elastic collision is a type of collision between particles where the total kinetic energy of the system is conserved. This means that the total kinetic energy before and after the collision remains the same.

How do you calculate the kinetic energy in an elastic collision?

The formula for calculating kinetic energy in an elastic collision is 1/2mv^2, where m is the mass of the particle and v is the velocity. In an elastic collision, the kinetic energy of the particles before the collision is equal to the kinetic energy after the collision.

What factors affect the kinetic energy in an elastic collision?

The kinetic energy in an elastic collision is affected by the mass and velocity of the particles involved. The higher the mass and velocity, the higher the kinetic energy will be. Additionally, the angle and direction of the collision can also affect the kinetic energy.

Can the kinetic energy in an elastic collision be negative?

No, the kinetic energy in an elastic collision cannot be negative. Kinetic energy is a measure of the energy an object possesses due to its motion, and it is always a positive value. In an elastic collision, the kinetic energy is conserved, so it cannot be negative.

What is the difference between elastic and inelastic collisions?

In an elastic collision, the total kinetic energy of the system is conserved, while in an inelastic collision, some kinetic energy is lost. In an elastic collision, the objects bounce off each other, while in an inelastic collision, they stick together or deform upon impact.

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