How Do You Solve a Glancing Elastic Collision in Physics?

[air4ce]

Homework Statement

A cue ball traveling @ 4m/s makes a glancing elastic collision with a target ball of equal bass that is initially at rest. The cue ball is deflected @an angle of 30°. Find the angle between the two velocity vectors; and the speeds of the two balls post collision.

Homework Equations

$$KE= 1/2mv^2 \\<br /> P=mv$$

The Attempt at a Solution

$m_1=m_2 \\<br /> m_1*v_1=m_1*v_{1f} + m_1*v_{2f} \\<br /> v_{1f}=4-v_{2f}$
Stuck here. I know I have to use Kinetic Energy, but how?

 

[SHISHKABOB]

What's the definition of an elastic collision?

 

[air4ce]

What's the definition of an elastic collision?

One where both momentum and kinetic energy are conserved.

 

[SHISHKABOB]

So use conservation of kinetic energy to apply another constraint to the problem.

 

[air4ce]

$.5m_1v_{1o}^2 =.5m_1v_{1f}^2+.5m_1v_{2f}^2 \\<br /> v_{1o}^2=v_{1f}^2+v_{2f}^2 \\<br /> v_{1f}=\sqrt{v_1^2-v_2f^2} \\$
Am I on the right track?

 

[SHISHKABOB]

Yes and also don't forget to separate conservation of momentum into both x and y. Write a conservation equation for both directions and use the results of that as well.

 

[air4ce]

So the horizontal momentum cancels and the vertical momentum totals the momentum of the cue ball?
Since the angle of cue ball is 30, I use the two equations to find $v_{1fy}$ and then solve for the horizontal? That's going to be equal and opposite to the struck ball's x velocity, so I have both velocities, use pythagorean theorem and I'm done? Check the logic please!

 

[SHISHKABOB]

well when we write out the momentum in x and y we get

$x: mv_{1,x} = mv^{'}_{1,x} + mv^{'}_{2,x}$

and

$y: 0 = mv^{'}_{1,y} + mv^{'}_{2,y}$

where the prime denotes "after the collision"

so

$mv_{1} = mv^{'}_{1} + mv^{'}_{2}$

is really only half of it. But can you see how you can use pythagorean's theorem to relate the x and y components to the magnitude of the vector?

 

[air4ce]

I think your X and Y's are flipped, but yeah I get the point. But I only know two of the 4 quantities, the ones before the collision, and one KE equation. How does that work?

 

[gneill]

Since all the masses are equal we can do away with them in the various expressions. Similarly, we can do away with the (1/2)m's in the kinetic energy terms. Call the initial velocities v and final velocities u, with appropriate subscripts for the x and y components, of course.

The given departure angle for the cue ball gives you a relationship between its y and x velocities as it departs. Namely, the ratio is tan(30°), or $1/\sqrt{3}$, so that:
$$\frac{u1_y}{u1_x} = \frac{1}{\sqrt{3}}$$
Now, what does conservation of momentum have to say about the relationship between the y-velocities u1_y and u2_y ?

And what does it say about about the sum of the x-velocities?

That's three equations already, and conservation of KE hasn't been considered yet...