speed increases after elastic collision in 2d?

Question

I always heard momentum remains the same after elastic collisions, I never hear people talk about the speed increase (the length of the velocities before and after).

When I look at this example https://en.wikipedia.org/wiki/Elastic_collision#Two-dimensional

The initial red velocity becomes the 2 blue velocities after collision. I can see ONLY component-wise the velocity remains the same. However: the length of the two blue vectors will always be greater than the red, so the total speed increases on each collision (except head on)? Why is no one talking about this?

Answer 1

[ Part 1 of 2 ]

The text :

the length of the two blue vectors will always be greater than the red

should be :

the total scalar length of the two blue vectors will always be greater than the red scalar length .

Reason :

scalar is magnitude without direction. vector is magnitude with direction.

Ref : speed vs velocity.

So, when we want to add up :

lengthBlueTop + lengthBlueBottom = lengthRed

We have to add the magnitute AND the direction. If we do scalar addition, we only add up the magnitude (distance traveled), angle/direction is not considered. like this (scalar) :

LHS = lengthBlueTop + lengthBlueBottom
RHS = lengthRed

LHS = 0.0239 m + 0.0122 m 

= 0.0361 m

RHS = 0.02845 m

0.0361 m > 0.02845 m 
or 
lengthBlueTop(scalar) + lengthBlueBottom(scalar) > lengthRed(scalar)

with difference of 0.0361-0.02845 = 0.00765 m (26.8892 %, roughly)

Which describes the original statement. Great.

Now we do the proper vector (magnitude + direction) addition :

LHS = 0.0239∠-62° m + 0.0122∠28° m 

= ( 0.0239cos(-62°) + 0.0239sin(-62°) i ) + ( 0.0122cos(28°) + 0.0122sin(28°) i ) = (0.02199 - 0.01537 i)

= 0.026829∠-0.610024°

RHS = 0.02845∠0° m

0.026829∠-0.610024° ≈ 0.02845∠0° m 
or     
lengthBlueTop + lengthBlueBottom ≈ lengthRed

with difference of = 0.001621∠0.610024° m (5.697715 %, roughly)

Considering the human (manual) pixel based measurement error + limited time/pixel data precision from the ref gif animation frames, I think this is sufface.

[ Part 2 of 2 ]

As for the :

speed increases after elastic collision in 2d?

I think just table out all velocity and see..

		LHS (before)	RHS (after)
m1 displacement	r [m]	0.02845	0.0122
	deg from +x	0°	-62°
m2 displacement	r [m]	0	0.239
	deg from +x	0°	28°
m1 velocity	v1 [m/s]	0.01778125∠0°	0.007625∠-62°
m2 velocity	v2 [m/s]	0∠0°	0.149375∠28°

If we can observe, final m2 velocity IS higher than initial m1 velocity. But as we can see in the momentum conservation calculation (below), it was offset by the very slow m1 final velocity.

To key towards understanding vector maniputation.. is to break it up into its component (x,y axis). Then (it took me few years..) we can actually see through it. A 70km/hr car (1600 kg approx) knocking a 50kg person/thing will throw it away at a much faster speed.. See?

Hope it is understandable/acceptable. Please share your thoughts. (^_^)\

--

_{**everything below this line is extra info for the number used above. I just leave it as a reference/extra reading.}

---

Extra

OP mentioned :

momentum remains the same after elastic collisions

Yup. u r correct : TotalMomentumBefore = TotalMomentumAfter (assuming perfect elastic collision).

Since momentum = mass*velocity , (OP already know this, I just recite for clarity..) the above system momentum relation is defined by :

m1*v1i + m2*v2i = m1*v1f + m2*v2f    ('f' denote 'final/after')

Observing :

m1  = m2 = some kg

v1i = some m/s, at +x direction.       <--- The red arrow
v2i = 0 m/s

v1f = some m/s, at +x & +y direction.  <--- The top blue arrow
v2f = some m/s, at +x & -y direction.  <--- The bottom blue arrow

details :

Arrow	length	angle	period
Red	1.12"(0.02845 m)	+90 deg from +y axis & 0 deg from +x axis (parrallel)	20 x 8 x (1/100) s = 1.6 s
Blue,Top	0.48"(0.0122 m)	+28 deg from +y axis & -62 deg from +x axis	1.6 s
Blue,Bottom	0.94"(0.0239 m)	+152 deg from +y axis & +28 deg from +x axis	1.6 s

m1 = m2 = 13.20 g = 0.0132 kg

reconfirm (re-calculate, to verify precision ) :

m1*v1i + m2*v2i = m1*v1f + m2*v2f

LHSx = (0.0132 kg)*(0.02845∠0° m / 1.6 s) + (0.0132 kg)*(0 m/ 1.6 s) 
    = (0.0132 kg)*(0.02845*cos(0°) m / 1.6 s) + (0.0132 kg)*(0 m/ 1.6 s) 

= 2.35×10^-4 kg m/s

At LHS the y axis velocity is 0 m . bcoz 0.02845*sin(0°) is 0 . So I write : LHSy = (0.0132 kg)(0.02845sin(0°) m / 1.6 s) + (0.0132 kg)*(0 m/ 1.6 s) = 0 kg m/s

Meanwhile,

RHS = (0.0132 kg)*(0.0122∠-62° m / 1.6 s) + (0.0132 kg)*(0.0239∠28° m/ 1.6 s) 

Here, 0.0122∠-62° & 0.0239∠28° have both x & y component. We do for x axis 1st. :

RHSx = (0.0132 kg)*(0.0122*cos(-62°) m / 1.6 s) + (0.0132 kg)*(0.0239*cos(28°) m/ 1.6 s) 

= 2.213×10^-4 kg m/s

As for the y axis :

RHSy = (0.0132 kg)*(0.0122*sin(-62°) m/ 1.6 s) + (0.0132 kg)*(0.0239*sin(28°) m / 1.6 s)

= 3.699×10^-6 kg m/s

		LHS (before)	RHS (after)
m1 displacement	r [m]	0.02845	0.0122
	deg from +x	0°	-62°
	x [m]	0.02845	0.005727553
	y [m]	0	-0.010771961
m2 displacement	r [m]	0	0.239
	deg from +x	0°	28°
	x [m]	0	0.112203704
	y [m]	0	0.112203704
momentum	x [kg m/s]	0.00023500	0.00022100
	y	0.00000000	0.00000369
	Total	0.00023500	0.00022469

There is 4.38 % (0.00001031) difference between momentum before & after. Within the tolerance we accepted before. Conservation of momentum is valid for this case. ( :