Physics Proof?

Question

Prove that in the elastic collision of two objects of identical mass, with one being a target initially at rest, the angle between their final velocity vectors is always 90 degrees.

Answer 1

Sorry, can't be done. In fact, if the collision is head on, the angle between the resulting vectors will be indeterminable. This results because mv + M0 = m0 + Mv'; where m = M the masses of the two billiard balls. Thus, mv = Mv' and v = v', which proves the after impact velocity v' is precisely the same as the before impact velocity.

Although billiard balls are not perfectly elastic, they are very close to it. And we see this conservation of momentum all the time...the cue ball strikes the target ball head on. What happens? The cue ball stops dead in its tracks while the target ball takes off in the same direction the cue ball was going and at (almost) the same speed.

Thus, there is no angle between the two velocities because one of the two is zero (both before and after impact).

Answer 2

Your question isn't quite right: the angle between the final velocity vectors is always 90 degrees in a PERFECTLY elastic collision, not just any old elastic collision.

Let the direction of the moving ball define the x axis. Conservation of momentum tells us that:

mv = m v1 + m v2

where v, v1, and v2 are vectors. Cancel the m's to get

v = v1 + v2

In other words, the vector sum of v1 and v2 equals the original velocity. If v1 and v2 are nonzero, we can think of them as three sides of a triangle. Now, is it a RIGHT triangle?

Conservation of kinetic energy tells us that

1/2 m v^2 = 1/2 m ( v1 )^2 + 1/2 m ( v2 )^2

or

v^2 = v1^2 + v2^2

That's the Pythagorean Theorem, which only applies to right triangles. Therefore, the angle between v1 and v2 must be 90 degrees IF both final velocities are nonzero ( one possible outcome is that v1 = 0 and v2 = v ).

Answer 3

I don't think this is correct. If I am understanding this, you fire a projectile at a similar object at rest. Well, this is basically a 1 dimension problem and you will not have a 90 degree angle. How can that be?

Answer 4

This does happen and is a result of the complete conservation of kinetic energy and conservation of momentum. It won't happen unless the masses are exactly equal
mv=mv1+mv2 so v=v1+v2
.5mv*v=.5mv1*v1 + .5mv2*v2
so v^2=v1^2+v2^2 which means
(v1+v2)^2=v1^2+v2^2 which can only be true if
2v1v2=0 which is only true if the angle between
them is 90 degrees

Answer 5

You need to read your own question

a white snooker ball follows the struck ball into the pocket
(if they are in-line) or stops dead in it's tracks

where's the 90 degree vector?

Answer 6

Assume that the stationary mass A is at origin and the moving mass B is approaching origin from -ve x axis at

velocity v.

Now we have the following

a) The energy must be conserved
b) The momentum along x axis must be conserved
c) The momentum along y axis must be conserved

Let x1, x2 be the angles made by A and B with positive x axis after the collosion. Let v1 and v2 be their respective

velocities.

a) gives
m*v*v = m*v1*v1 + m*v2*v2 -------------------(1)
b) gives
m*v = m*v1*cos(x1) + m*v2*cos(x2) -----------------------(2)
c) gives
0 = m*v1*sin(x1) + m*v2*sin(x2) -------------------------------(3)

Squaring 3 gives:
m^2*v1^2*sin^2(x1) + m^2*v2^2*sin^2(x2) = -2*v1*v2*sin(x1)*sin(x2) ----------------------(4)

Squaring (2) and substituting LHS with RHS of (1) gives

m*v1*v1 + m*v2*v2 = m^2*v1^2*cos^2(x1) + m^2*v2^2*cos^2(x2) + 2*m1*m2*v1*v2*cos(x1)*cos(x2)

Using identity sin^2 + cos^2 = 1 gives
m^2*v1^2*sin^2(x1) + m^2*v2^2*sin^2(x2) = 2*m1*m2*v1*v2*cos(x1)*cos(x2)

But LHS = -2*v1*v2*sin(x1)*sin(x2) from (4)

Substituting gives:

-2*v1*v2*sin(x1)*sin(x2) = 2*m1*m2*v1*v2*cos(x1)*cos(x2)

OR

tan(x1) = -cot(x2)

OR

Sin(x1)/Cos(x1) + Cos(x2)/Sin(x2) = 0

Using common denominator and equating numerator to zero gives

sin(x1)sin(x2) + cos(x1)*cos(x2) = 0 (here we assuming that sin(x1)*cos(x2) is not zero for any solution)

But using the identity Cos(A-B) = cos(A)*Cos(B) + sin(A)*Sin(B) we get

Cos(x1-x2) = 0

Hence the solutions are x1-x2=90, x1-x2=270

That is the angle between the paths of the masses must be 90 degrees.

Strictly speaking, there is another solution where either A or B is stationary after the collision. So we need to include the clause "if both A and B have nonzero velocities after the collision"