0
$\begingroup$

I found the statement below on a physics textbook:

Whenever two objects of equal mass collide elastically in a glancing collision and one of them is initially at rest, their final velocities are perpendicular to each other.

I have been trying to find the proof by looking up Introduction to Mechanics & University Physics for Scientists and Engineers, but I ended up with nothing helpful.

Could anyone mathematically prove or disprove it?

$\endgroup$
5
  • 2
    $\begingroup$ Proofs can be found in any junior-level mechanics textbook. Nor is this a trivial fact, you can use it as a way of showing that alpha particles are helium nuclei: physics.stackexchange.com/questions/266333/… $\endgroup$ Commented Nov 19, 2019 at 0:52
  • $\begingroup$ I don't think I've even seen this topic covered in an introductory text like the one you named. Not that the math is much harder than that you find in such tomes, but it's a intricate and lengthy development. $\endgroup$ Commented Nov 19, 2019 at 3:02
  • 1
    $\begingroup$ This was a problem given to me as a bonus question in my 11th grade physics class. $\endgroup$ Commented Nov 20, 2019 at 18:38
  • $\begingroup$ @AaronStevens I am wondering if anyone in your class got perfect on that question. :) $\endgroup$
    – Leo Liu
    Commented Nov 20, 2019 at 20:18
  • $\begingroup$ I did. Although I did it much slower and less efficiently than I would today. $\endgroup$ Commented Nov 20, 2019 at 21:26

3 Answers 3

2
$\begingroup$

Let me label the initially moving ball $A$ and the initially at rest ball $B$

Momentum and energy are of course conserved:

$$m_a\vec v_a = m_a\vec v_a' + m_b\vec v_b'$$

$$m_av_a^2 = m_av_a'^2 + m_bv_b'^2 \rightarrow v_a^2 = v_a'^2 + v_b'^2$$

Where primed velocities mean velocities after the elastic collision.

One can cancel the masses out in the above equations.

As said, algebra gets messy in this problem so let me use a geometrical approach:

Note that momentum is a vector, which means that we can describe conservation of momentum by a (velocity) vector diagram (note masses get canceled out and thus play no role):

enter image description here

Note that, as momentum is conserved, the momentum lost by the incoming ball after the collision is the momentum the rest ball has gained. This means that the (vector) sum of the final velocities must be equal to the total initial velocity (i.e. it is equal to the initial velocity of the incoming particle). This means that the vector diagram can be redrawn as follows:

enter image description here

However the key of the problem lies on the energy conservation equation.

$$v_a^2 = v_a'^2 + v_b'^2$$

Think about this equation while having a look at the above triangle*. Doesn't ring the bell? Yeah, Pythagoras' theorem!

Based on Pythagoras' theorem, our triangle must have a right angle:

enter image description here

Then $\alpha + \beta = 90°$. Thus the angle between the two final velocities is $90°$.

*Don't get stuck thinking that energy is not a vector and thus we shouldn't think of the diagram; notice the energy equation represents the magnitude of the vectors. Think like this if you wish: 1) let's draw the velocity vectors based on conservation of momentum 2) Now let's think of Pythagoras' theorem for such a triangle.

$\endgroup$
1
  • 1
    $\begingroup$ Wow, I love the intuition within your answer! Thank you. $\endgroup$
    – Leo Liu
    Commented Nov 20, 2019 at 20:16
1
$\begingroup$

Let's call the initial velocity vector of the moving particle v. After hitting the rest particle it turns to v1 and the rest particle v2.

Because the masses are the same we can fixed them as 1.

By moment conservation: v = v$_1$ + v$_2$

By energy conservation: v$^2$ = v$_1$$^2$ + v$_2$$^2$

Squaring the first equation:

v$^2$ = v$_1$$^2$ + 2v$_1$.v$_2$ + v$_2$$^2$

Subtracting the equations we get:

v$_1$.v$_2$ = 0

For the scalar product of two non zero vectors be zero, they must be perpendicular.

Of course it is also possible that v$_1$ = 0. The moving particle stops, "transfering" all its velocity to the rest particle.

$\endgroup$
-2
$\begingroup$

Partial answer because I'm skipping a lot of steps. Use conservation of energy and momentum.

Momentum before collision in x-direction: $mv$
Momentum before collision in y-direction: $0$
Momentum after collision in x-direction: $mv_1 \mathrm{cos}(\alpha) + mv_2 \mathrm{cos}(\beta)$
Momentum after collision in y-direction: $mv_1 \mathrm{sin}(\alpha) + mv_2 \mathrm{sin}(\beta)$

Energy before collision: $\frac{1}{2} mv^2$
Energy after collision: $\frac{1}{2} mv_1^2 + \frac{1}{2} mv_2^2$

Solving these, you should get these three equations:

$v^2 = v_1^2 + v_2^2$
$v_1^2 \mathrm{cos}^2\alpha + 2v_1v_2\mathrm{cos}(\alpha)\mathrm{cos}\beta+v_2^2\mathrm{cos}^2\beta=v^2$
$v_1^2\mathrm{sin}^2\alpha + v_2^2\mathrm{sin}^2\beta + 2v_1v_2\mathrm{sin}\alpha\mathrm{sin}\beta = 0$

Then add the last two equations, apply some trigonometric identities, and you get this:

$\mathrm{cos}\alpha\mathrm{cos}\beta+\mathrm{sin}\alpha\mathrm{sin}\beta=0$

which from the trigonometric identity for the sum of an angle gives you $\alpha-\beta=\pi$, i.e. they are perpendicular.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.