Newton's experimental law -- Questions about the collision of two spheres

[hendrix7]

TL;DR Summary

Impact problems involving conservation of momentum and law of restitution help needed please.

The following is from the 2018 AQA AS Further Maths/Mechanics Paper 2 exam:

'Two smooth spheres A and B of equal radius are free to move on a smooth horizontal surface. The masses of A and B are m and 4m respectively. The coefficient of restitution between the spheres is e. The spheres are projected directly towards each other, each with speed u, and subsequently collide.
(a) Show that the speed of B immediately after the impact with A is u(3 - 2e)/5'

I have 2 questions regarding this problem:
1) Since momentum is a vector and we don't know the direction of motion of either sphere following collision, how can we form the equation with the correct signs?
2) Newton's experimental law does not involve vectors but I believe I still need to know whether to add or subtract the speeds on separation. Again, since we don't know the direction of either sphere on separation, how do we form this equation with the correct signs?
Thanks for any help.

 

[PeroK]

Summary:: Impact problems involving conservation of momentum and law of restitution help needed please.

The following is from the 2018 AQA AS Further Maths/Mechanics Paper 2 exam:

'Two smooth spheres A and B of equal radius are free to move on a smooth horizontal surface. The masses of A and B are m and 4m respectively. The coefficient of restitution between the spheres is e. The spheres are projected directly towards each other, each with speed u, and subsequently collide.
(a) Show that the speed of B immediately after the impact with A is u(3 - 2e)/5'

I have 2 questions regarding this problem:
1) Since momentum is a vector and we don't know the direction of motion of either sphere following collision, how can we form the equation with the correct signs?

The direction of motion after the collision is determined by conservation of momentum - and the constraints on energy conservation. In other words, you can work out the final velocities if you want.

 

[etotheipi]

1) Since momentum is a vector and we don't know the direction of motion of either sphere following collision, how can we form the equation with the correct signs?
2) Newton's experimental law does not involve vectors but I believe I still need to know whether to add or subtract the speeds on separation. Again, since we don't know the direction of either sphere on separation, how do we form this equation with the correct signs?

Momentum is a vector, yes, and if you work in a single dimension then you are working with a single component. That is to say for a closed system $\sum \vec{p}_i$ is a conserved quantity, but so is $\sum p_{i,x}$, where $\vec{p}_i = p_{i,x}\hat{x}$ and the $x$ axis is chosen to be oriented along the line of the motion.

In this question, call the final $x$ components of velocity $v_{A,x}$ and $v_{B,x}$. In Classical Physics, if the motion is in one dimension, then the speed of separation is equal to the magnitude of the relative velocity. Since $v_{B,x} > v_{A,x}$, the speed of separation will just be $v_{B,x} - v_{A,x}$.

 

[jbriggs444]

The direction of motion after the collision is determined by conservation of momentum - and the constraints on energy conservation. In other words, you can work out the final velocities if you want.

It is a three body collision. We do not know the coefficient of restitution with respect to the third body.

Edit: oops.

 

[PeroK]

It is a three body collision. We do not know the coefficient of restitution with respect to the third body.

What's the third body?

 

[jbriggs444]

What's the third body?

The surface.

Edit: scrapes egg off face.

 

[etotheipi]

The surface.

But the surface is smooth and the motion rectilinear?

 

[jbriggs444]

But the surface is smooth and the motion rectilinear?

The motion following the collision is not necessarily horizontal since the collision angle is not horizontal.
Aww, shoot. Unequal masses, not unequal radii.

 

[etotheipi]

Aww, shoot. Unequal masses, not unequal radii.

That is a good follow up question though! I dare say it's a little bit tricky for AS level

 

[hendrix7]

The direction of motion after the collision is determined by conservation of momentum - and the constraints on energy conservation. In other words, you can work out the final velocities if you want.

Thanks for reply.
You say that direction of motion is determined by conservation of momentum but how do I know if A continues moving in the same direction or motion is reversed? So:
initial momentum (i.e. just before collision) = mu
final momentum (after collision) could be: -mVa + 4mVb or mVa + 4mVb and the question doesn't tell me which it should be.

 

[PeroK]

Thanks for reply.
You say that direction of motion is determined by conservation of momentum but how do I know if A continues moving in the same direction or motion is reversed? So:
initial momentum (i.e. just before collision) = mu
final momentum (after collision) could be: -mVa + 4mVb or mVa + 4mVb and the question doesn't tell me which it should be.

You know the direction of the smaller mass $A$ must be reversed because the total momentum is in the direction of motion of the larger object. If $A$ keeps going in its original direction then the heavier object $B$ must change direction, which fails conservation of momentum.

Note, however, there is a mathematical solution were both objects pass through each other. That is a valid mathematical solution that you have to reject on the physical grounds that the objects are solid.

Finally, can you see an argument (based on energy) to show that the larger mass $B$ cannot change direction?

 

[etotheipi]

initial momentum (i.e. just before collision) = mu

Is this right? Remember it is the momentum of the system of both particles that is conserved.

final momentum (after collision) could be: -mVa + 4mVb or mVa + 4mVb and the question doesn't tell me which it should be.

You don't need to know what directions they move in. You can just work with the $x$ component $v_{Ax}$ which can be determined to be $\pm v_A$ later on.

 

[hendrix7]

Is this right? Remember it is the momentum of the system of both particles that is conserved.
You don't need to know what directions they move in. You can just work with the $x$ component $v_{Ax}$ which can be determined to be $\pm v_A$ later on.

Thank you etotheipi for spotting my mistake. The initial momentum should be 5mu, I believe. I mistakenly thought B was initially at rest.

 

[PeroK]

Thank you etotheipi for spotting my mistake. The initial momentum should be 5mu, I believe. I mistakenly thought B was initially at rest.

The initial total momentum is not $5mu$ either. Remember you said momentum is a vector quantity.

 

[hendrix7]

The initial total momentum is not $5mu$ either. Remember you said momentum is a vector quantity.

 

[hendrix7]

My bad. Third time lucky: 3mu or -3mu depending on + direction chosen. Am I right now?

 

[hendrix7]

You know the direction of the smaller mass $A$ must be reversed because the total momentum is in the direction of motion of the larger object. If $A$ keeps going in its original direction then the heavier object $B$ must change direction, which fails conservation of momentum.

Note, however, there is a mathematical solution were both objects pass through each other. That is a valid mathematical solution that you have to reject on the physical grounds that the objects are solid.

Finally, can you see an argument (based on energy) to show that the larger mass $B$ cannot change direction?

No, I can't see it. Do you mean using conservation of energy?

 

[PeroK]

No, I can't see it. Do you mean using conservation of energy?

Using the fact that the total kinetic energy cannot increase as a result of the collision.

Hint: first look at what happens if the larger mass stops. Can you show that is impossible? Then, show that if the larger mass ends up changing direction then that is even more impossible!

 

[hendrix7]

Using the fact that the total kinetic energy cannot increase as a result of the collision.

Hint: first look at what happens if the larger mass stops. Can you show that is impossible? Then, show that if the larger mass ends up changing direction then that is even more impossible!

But kinetic energy doesn't have direction. We square the velocity, don't we?

 

[PeroK]

But kinetic energy doesn't have direction. We square the velocity, don't we?

Let me show you what I mean. We have two masses, $m$ and $4m$ moving towards each other with equal speed $u$ (equal and opposite velocity). Let's take the direction of $M$ as positive.

The initial momentum is $3mu$ and the initial KE is $\frac 5 2 mu^2$.

Now let's assume that mass $4m$ stops as a result of the collision. That means that the mass $m$ now has all the momentum. If we take its velocity to be $v$ we see that:
$$mv = 3mu, \ \ \text{hence} \ \ v = 3u$$
And the total KE after the collison is $\frac 9 2 mu^2$, which is more than it was before.

Finally, if the large mass rebounds, then it has some negative momentum (plus some KE), the small mass must have even more momentum, hence even more KE and again we have more energy after the collision than before.

Therefore, in this case, the large mass cannot stop or rebound but must be moving in the same direction after the collision.

That's one way to establish the direction of motion of the large mass after the collision.

 

[hendrix7]

Let me show you what I mean. We have two masses, $m$ and $4m$ moving towards each other with equal speed $u$ (equal and opposite velocity). Let's take the direction of $M$ as positive.

The initial momentum is $3mu$ and the initial KE is $\frac 5 2 mu^2$.

Now let's assume that mass $4m$ stops as a result of the collision. That means that the mass $m$ now has all the momentum. If we take its velocity to be $v$ we see that:
$$mv = 3mu, \ \ \text{hence} \ \ v = 3u$$
And the total KE after the collison is $\frac 9 2 mu^2$, which is more than it was before.

Finally, if the large mass rebounds, then it has some negative momentum (plus some KE), the small mass must have even more momentum, hence even more KE and again we have more energy after the collision than before.

Therefore, in this case, the large mass cannot stop or rebound but must be moving in the same direction after the collision.

That's one way to establish the direction of motion of the large mass after the collision.

Thank you so much for this explanation. I understand all that and it never occurred to me to look at these problems in these terms. Do you think the question I originally posted, and others similarly worded, expected candidates to look at the problem with these considerations?

 

[etotheipi]

Thank you so much for this explanation. I understand all that and it never occurred to me to look at these problems in these terms. Do you think the question I originally posted, and others similarly worded, expected candidates to look at the problem with these considerations?

@PeroK has given you some very good insight, but if you do flick through the mark schemes you will see they just expect you to work with the components. You will notice that these sorts of questions are quite formulaic: write one equation for momentum, write one equation for coefficient of restitution, solve the system. You don't need to determine what the directions will be, they just fall out of the equations.

 

[PeroK]

Thank you so much for this explanation. I understand all that and it never occurred to me to look at these problems in these terms. Do you think the question I originally posted, and others similarly worded, expected candidates to look at the problem with these considerations?

I don't know. You could do what I did (I just like to figure out what's happening first, if I can, before writing down lots of equations). You could do it using vector equations. You could use the centre of momentum frame. You could use the rest frame of the large mass. Or the rest frame of the small mass. There are a lot of options.

It never hurts to have more than one way to solve a problem!

 

[PeroK]

@PeroK has given you some very good insight, but if you do flick through the mark schemes you will see they just expect you to work with the components. You will notice that these sorts of questions are quite formulaic: write one equation for momentum, write one equation for coefficient of restitution, solve the system. You don't need to determine what the directions will be, they just fall out of the equations.

That's a good point. I was responding to the original doubt about how to tackle the problem without knowing the final directions of motion.

You're right that you don't need to know this in advance. You can just trust the vector equations.

 

[hendrix7]

@PeroK has given you some very good insight, but if you do flick through the mark schemes you will see they just expect you to work with the components. You will notice that these sorts of questions are quite formulaic: write one equation for momentum, write one equation for coefficient of restitution, solve the system. You don't need to determine what the directions will be, they just fall out of the equations.

That's what I thought but now, after looking at this question and a couple of others that are similar, I'm not so sure. Like I say in my original post, how do I know whether to add or subtract the speeds of separation without some knowledge of directions? I'll get different answers depending on what I choose, won't I?

 

[PeroK]

That's what I thought but now, after looking at this question and a couple of others that are similar, I'm not so sure. Like I say in my original post, how do I know whether to add or subtract the speeds of separation without some knowledge of directions? I'll get different answers depending on what I choose, won't I?

I just had a look at this. I suggest you do have to confirm first that the small mass rebounds (changes direction). You don't have to worry about the direction of the large mass.

 

[etotheipi]

That's what I thought but now, after looking at this question and a couple of others that are similar, I'm not so sure. Like I say in my original post, how do I know whether to add or subtract the speeds of separation without some knowledge of directions? I'll get different answers depending on what I choose, won't I?

Well a speed of approach/separation is strictly greater or equal to zero. If ball B is on the right of ball A and $\hat{x}$ points to the right, the speed of approach will be the velocity of ball A moving to the right in the frame of ball B ($u_{Ax} - u_{Bx}$), and after the collision the speed of separation will be the velocity of ball B moving to the right in the frame of ball A ($v_{Bx} - v_{Ax}$).

 

[PeroK]

That's what I thought but now, after looking at this question and a couple of others that are similar, I'm not so sure. Like I say in my original post, how do I know whether to add or subtract the speeds of separation without some knowledge of directions? I'll get different answers depending on what I choose, won't I?

PS the good thing about using the centre of momentum frame is that both masses definitely rebound. Then you really know where you are!

 

[PeroK]

PS the good thing about using the centre of momentum frame is that both masses definitely rebound. Then you really know where you are!

@hendrix7 if you haven't seen this before, here's how to use the CoM frame:

The velocity of the CoM frame is given by total momentum divided by total mass, so $V_{CoM} = \frac{3u}{5}$.

In that frame the velocities of the masses are: $u - \frac{3u}{5} = \frac{2u}{5}$ for the large mass, and $-u - \frac{3u}{5} = -\frac{8u}{5}$ for the small mass.

The next bit is easy. The velocities after the collision must be respectively: $-\frac{2ue}{5}$ and $+\frac{8ue}{5}$. This is because the total momentum is zero in this frame.

Then we add back on $\frac{3u}{5}$ to get velocities in the original frame. This gives the velocity of the large mass as $- \frac{2ue}{5} + \frac{3u}{5} = \frac{u(3-2e)}{5}$.

 

[haruspex]

how can we form the equation with the correct signs?

Just choose which directions are positive for which variables and be consistent.
No need to figure out the directions separately in advance, nor involve energy, nor go to the CoM frame.

Taking 4m as being initially in the positive direction for all velocities, and writing the final velocities as v₁ and v₄:
NEL: (u-(-u))e = v₁ - v₄
Momentum: 4u-u = v₁ + 4v₄
Subtracting first from second:
3u - 2ue = 5v₄

 

[A.T.]

I just like to figure out what's happening first, if I can, before writing down lots of equations

"Never make a calculation until you know the answer"
John Archibald Wheeler