Oblique Collision Trigonometry problem

[Darth Frodo]

Homework Statement

A smooth sphere collides with an equal sphere which is at rest. Before the impact the line of motion of the first sphere makes an angle A with their line of centers, after the impact this is angle B. The coefficient of restitution is 1/4. Show that 8TanA = 3TanB

Homework Equations

-e = (V1 - V2) / (U1 - U2)

M1U1 + M2U2 = M1V1 + M2V2

The Attempt at a Solution

Initial Velocity 1: (XcosA i + XsinA j)
Initial Velocity 2: (0i + 0j)

Final Velocity 1: (YcosB i + YsinB j)
Final Velocity 2: (P i + 0 j) XsinA = YsinB. J components are equal.

Coefficient of restitution
-1/4 = [YcosB - p] / [XsinA]
-XsinA = 4YcosB - 4pConservation of momentum in i direction

masses are equal.
U1 + U2 = V1 + V2

XcosA = YcosB + p
XcosA - YcosB = p-XsinA = 4YcosB - 4( XcosA - YcosB )
-XsinA = 8YcosB - 4XcosA

-XsinA / -XcosA = 8YcosB /-XcosA -4XcosA /-XcosA
Tan A = (8YcosB) / (-XcosA) - 4

 

[ehild]

Homework Equations

-e = (V1 - V2) / (U1 - U2)

This equation holds for one-dimensional motion. In general, the coefficient of restitution is (speed of separation)/(speed of aproach). You need to calculate with the magnitude of relative velocities before and after the collision.

The Attempt at a Solution

Initial Velocity 1: (XcosA i + XsinA j)
Initial Velocity 2: (0i + 0j)

Final Velocity 1: (YcosB i + YsinB j)
Final Velocity 2: (P i + 0 j)

Was it given that the second ball moves in the direction of the line connecting their centres?

ehild

 

[Darth Frodo]

I thought that was the equation for relative velocity? According to my book anyway. It shows the derivation.

As for moving only on the i axis, according to my book "the bounce / impact takes place along the i axis therefore the j component of the velocity remains unchanged. I'm not having a problem with the physics, just the trig. I can't whittle it down to 8Tan A = 3TanB

 

[ehild]

You took the line connecting the centres of the spheres as x-axis and the y-axis is perpendicular to this line. Nothing ensures that the second ball will move along the x axis.
In case of bounce from the ground or from a wall, the force during impact is normal to the wall, so only the normal component of the velocity changes. It is not the case here.

The coefficient of restitution gives the relation between the magnitudes of the relative velocities, not only of the x components. See:https://www.physicsforums.com/library.php?do=view_item&itemid=270

ehild

 

[Darth Frodo]

"You took the line connecting the centres of the spheres as x-axis and the y-axis is perpendicular to this line. Nothing ensures that the second ball will move along the x axis."

But the j component of the velocity never changes. If it's originally 0 doesn't it stay 0?

This is what my book says. See picture attached.

 

[ehild]

OK, it is an approximation, which is not always true in reality. Can you kick a ball so as it starts to spin? It is because the interaction between your foot and the ball has tangential component to, not only radial. When friction is negligible between the colliding balls, however, the approximation is valid. And the balls in this problem are perfectly smooth. So you are right, the y components do not change.
As for the coefficient of restitution, I see, there are different definitions in different books.

You made a mistake when writing up the coefficient of restitution:

Coefficient of restitution
-1/4 = [YcosB - p] / [XsinA]

It has to be X cosA.

And use the equation for the y components of velocity, too: XsinA=YsinB

ehild

 

[Darth Frodo]

Yeah, all the problems I'm doing at the minute are High School and are under "Ideal conditions"

Yes, that mistake messed me up a bit. With that error overcome, I have retried the problem.

Attempt 2

Ball 1 (XcosA i + XsinA j) ----> (YcosB i + YsinB j)
Ball 2 (0i + 0j) ----> (Pi + 0j)


Momentum is conserved in the i direction. Masses are equal.

u1 + u2 = v1 + v2
XcosA = YcosB + P
P = XcosA - YcosB



Coefficient of restitution.

-e = $\frac{v1 - v2}{u1 - u2}$

$\frac{-1}{4}$ = $\frac{YcosB - P}{XcosA}$

-XcosA = 4YcosB - 4P

P = $\frac{-XcosA -4YcosB}{-4}$



So now I have 3 equations.

1. XsinA = YsinB

2. P = XcosA - YcosB

3. P = $\frac{-XcosA - 4YcosB}{-4}$


From that I get

XcosA - YcosB = $\frac{-XcosA - 4YcosB}{-4}$

-4XcosA + 4YcosB = -XcosA - 4YcosB

3XcosA = 8YcosB.



And that is as far as I can go.

 

[ehild]

3XcosA = 8YcosB.*

And that is as far as I can go.

And you have the equation for the y components:

XsinA = Y sinB**

Just divide eq. ** with eq * what you get ? You know, do not you, that sinA/cosA=tanA?

ehild

 

[Darth Frodo]

Yeah I understand that Tan A = Sin A / Cos A

But I have never come across dividing one equation with another.

Would you mind explaining please.

 

[ehild]

It is possible to multiply, divide add or subtract two equations.
If A=B and C=D then A+C=B+D, A-C=B-D, A*C=B*D, and A/C=B/D (here you must be sure that neither C nor D are zero)

Just a primitive example: 2=2, 3=3, dividing the equations you get 2/3=2/3.

Your equations are:
xsinA=ysinB and
3xcosA = 8ycosB

Divide:

$$\frac{xsinA}{3xcosA}=\frac{ysinB}{8ycosB}$$

Simplify:

$$\frac{sinA}{3cosA}=\frac{sinB}{8cosB}$$

$$\frac{tanA}{3}=\frac{tanB}{8}$$
Multiply the whole equation by 24, and you get the solution.

If you do not like to divide equations, you can express sinA from the first equation, cosA from the second one, and find tanA=sinA/cosA.

ehild

 

[Darth Frodo]

Excellent, thank you!

I never realized you could do that.