Elastic collision with particles, find the kinetic energy

[Mathias Girouard]

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.

 

[kuruman]

Hi Mathias Girouard and welcome to PF.

How about conserving momentum? You wrote the equation, but you did not use it.

 

[gneill]

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.

 

[neilparker62]

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 $(Δp)^2/(2m_2)$

 

[gneill]

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.

 

[neilparker62]

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.

 

[Nimesh Nayaju]

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