A problem about momentum and the elastic collision of two particles

[Minming]

Homework Statement

For an elastic collision between a projectile particle of mass m1 and a target particle (at rest) of mass m2, show that the scattering angle, θ’, of the projectile (a) can take any value, 0 to 180°, for m1< m2, but (b) has a maximum angle φ given by cos²φ=1-(m2/m1)² for m1> m2.

Relevant Equations

total initial momentum = total final momentum
total initial kinetic energy = total final kinetic energy

 

[PeroK]

The equation you derived is probably not very useful. One idea is to use the law of cosines for the momentum vectors. You can draw a triangle with the initial momentum ($m_1v_1$) along the base, and the momentum of the two particles after the collision as the other two sides. You can do this two ways, giving you an expression involving either $\cos \alpha$ or $\cos \theta'_1$.

PS although there is nothing wrong with your notation, simply $\theta$ would have done just as well as $\theta'_1$.

 

[PeroK]

PPS I looked up my notes and I don't have a ready solution for this exact problem. But, I think you need to do more than manipulate equations. For part b) you might think about the problem in the centre-of-momentum (zero-momentum) frame, in order to establish an important point about the angle $\theta$.

In fact, a completely different approach would be to use the centre-of-momentum frame to establish the range of angles for $\theta$, then to transform back to the frame in which $m_2$ is at rest.

 

[haruspex]

Homework Statement: For an elastic collision between a projectile particle of mass m1 and a target particle (at rest) of mass m2, show that the scattering angle, θ’, of the projectile (a) can take any value, 0 to 180°, for m1< m2, but (b) has a maximum angle φ given by cos²φ=1-(m2/m1)² for m1> m2.
Relevant Equations: total initial momentum = total final momentum
total initial kinetic energy = total final kinetic energy

View attachment 356525

If you expand the LHS of your last equation some useful cancellations can be made.
I didn’t follow through to completion but it looks promising.

 

[kuruman]

. . . show that the scattering angle, θ’, of the projectile (a) can take any value, 0 to 180°, . . .

Any value? What about a scattering angle of 90°? How does that conserve linear momentum?

 

[PeroK]

Any value? What about a scattering angle of 90°? How does that conserve linear momentum?

If $m_1 < m_2$, then a scattering angle of 90° is possible.

 

[kuruman]

If $m_1 < m_2$, then a scattering angle of 90° is possible.

Yes, I see that now. It seems that I tried to do too much in my head.

 

[MatinSAR]

View attachment 356525

Your issue lies in not fully understanding the reasoning behind your approach. Why did you solve those three equations for the two angles? The goal is to analyze the angle $\theta '_1$, so you need to use those three equations to eliminate the variables $\alpha$ and $v'_{2}$. By doing this, you will derive an equation that can be directly used to answer both part (a) and part (b) of the problem easily.

 

[haruspex]

Your issue lies in not fully understanding the reasoning behind your approach. Why did you solve those three equations for the two angles? The goal is to analyze the angle $\theta '_1$, so you need to use those three equations to eliminate the variables $\alpha$ and $v'_{2}$. By doing this, you will derive an equation that can be directly used to answer both part (a) and part (b) of the problem easily.

That may work algebraically, but it does not finish the job in terms of logic.
In setting up the collision, the variable you can control is $\alpha$, not $v'_1$. If you find that a value of $\theta'_1$ can be achieved by a certain value of $v'_1/v$, how do you know that value of $v'_1/v$ can be achieved?

 

[MatinSAR]

That may work algebraically, but it does not finish the job in terms of logic.

But it still solves the problem, right? We have an expression for $v'_1/v_1$, which we can use to limit the possible angles. We know the ratio should not be imaginary. Do you agree?

In setting up the collision, the variable you can control is $\alpha$, not $v'_1$. If you find that a value of $\theta'_1$ can be achieved by a certain value of $v'_1/v$, how do you know that value of $v'_1/v$ can be achieved?

You suggest solving for angles, like what the OP did?

I replied to your post that day, but it was deleted by the mentors for breaking the rules (I said more than needed in that post, almost solving the problem in the way I suggested). They asked me to wait for the OP to solve the problem for himself. I hope this post doesn't violate any laws... @Nugatory

 

[PeroK]

But it still solves the problem, right? We have an expression for $v'_1/v_1$, which we can use to limit the possible angles. We know the ratio should not be imaginary. Do you agree?

You suggest solving for angles, like what the OP did?

I replied to your post that day, but it was deleted by the mentors for breaking the rules (I said more than needed in that post, almost solving the problem in the way I suggested). They asked me to wait for the OP to solve the problem for himself. I hope this post doesn't violate any laws... @Nugatory

It's a shame that the OP hasn't replied, as it would be good to compare the various approaches. I used the law of cosines and implicit differentiation.

 

[haruspex]

But it still solves the problem, right? We have an expression for $v'_1/v_1$, which we can use to limit the possible angles. We know the ratio should not be imaginary. Do you agree?

It may show a limit, but it won't show whether that limit can be reached. Maybe the value of $v_1’$ that achieves that limit for a $v$ that cannot itself be reached.

You suggest solving for angles, like what the OP did?

You know what the range of $\alpha$ can be, so if you have $v$ as a function of that you can be confident of your answer.
Indeed, suppose the stationary target were a cube, oriented square on to the velocity of the missile, while the missile is a cube oriented at 45°. Now $0<\alpha<\pi/4$, imposing a tighter limit on $v$.

 

[Minming]

PPS I looked up my notes and I don't have a ready solution for this exact problem. But, I think you need to do more than manipulate equations. For part b) you might think about the problem in the centre-of-momentum (zero-momentum) frame, in order to establish an important point about the angle $\theta$.

In fact, a completely different approach would be to use the centre-of-momentum frame to establish the range of angles for $\theta$, then to transform back to the frame in which $m_2$ is at rest.

Thanks. I'll study centre-of-momentum frame, and try to solve it.

 

[Minming]

It's a shame that the OP hasn't replied, as it would be good to compare the various approaches. I used the law of cosines and implicit differentiation.

Hi, the law of cosines and implicit differentiation is useful, I get the answer by dθ'/dα =0, and get sinθ'=m
₂/m₁, it's same to the answer(when m₁> m₂). thanks