Collisions in Two Dimensions

Collisions in Two Dimensions

The basic conservation law does not change when a collision becomes two-dimensional:

(1)

when the net external impulse during the collision is negligible.

What changes is that momentum is now a vector with independent components.

For two particles,

(2)

In Cartesian components,

m1u1x + m2u2x = m1v1x + m2v2x,		(3)
m1u1y + m2u2y = m1v1y + m2v2y.		(4)

A planar collision therefore provides two momentum equations rather than one.

1 Choose the coordinate axes deliberately

A good coordinate choice can simplify the algebra dramatically.

Common choices are:

• align the x axis with the initial velocity of one particle,
• align one axis with the line of impact,
• use Normal and tangential directions at the contact point,
• use the center of mass frame when symmetry is important.

The conservation law is independent of coordinates, but the component equations are not equally convenient in every coordinate system.

2 Momentum conservation by components

Suppose the collision takes place in the xy plane.

Then

(5)

and

(6)

Vector equality requires

Pxi	= Pxf,	(7)
Pyi	= Pyf.	(8)

Therefore

m1u1x + m2u2x	= m1v1x + m2v2x,	(9)
m1u1y + m2u2y	= m1v1y + m2v2y.	(10)

3 Velocity components from speed and direction

If a velocity of magnitude v makes angle 𝜃 with the positive x axis,

vx	= v cos 𝜃,	(11)
vy	= v sin 𝜃.	(12)

Thus

(13)

Signs should come from the geometry and the chosen axis directions.

A velocity below the positive x axis has

(14)

4 Example 1: perfectly inelastic collision in two dimensions

A 2.0 kg puck moves east at

(15)

A 3.0 kg puck moves north at

(16)

They collide and stick.

The initial momentum components are

Pxi	= (2.0)(4.0) = 8.0 kg m∕s,	(17)
Pyi	= (3.0)(2.0) = 6.0 kg m∕s.	(18)

The combined mass is

(19)

Therefore

vfx	= = 1.6 m∕s,	(20)
vfy	= = 1.2 m∕s.	(21)

The final speed is

(22)

The direction is

(23)

north of east.

Thus

(24)

5 Center of mass interpretation of sticking

For any perfectly inelastic collision,

(25)

Momentum conservation gives

(26)

Therefore

(27)

This result is dimension independent.

6 Elastic collisions in two dimensions

An elastic collision conserves both momentum and translational kinetic energy:

(28)

and

(29)

In two dimensions this gives three scalar equations:

• momentum in x,
• momentum in y,
• kinetic energy.

That may still be insufficient if four post-collision components are unknown. Additional geometry, contact constraints, or a measured scattering angle is often needed.

7 Equal masses with one initially at rest

A particularly important case is an elastic collision between equal masses,

(30)

with particle 2 initially at rest:

(31)

Momentum conservation gives

(32)

Cancel m:

(33)

Kinetic-energy conservation gives

(34)

Square the momentum relation:

u12	= |v 1 + v2|2	(35)
	= v12 + v 22 + 2v 1 ⋅ v2.	(36)

Compare with the energy relation:

(37)

Therefore

(38)

so

(39)

Hence the outgoing velocities are perpendicular:

(40)

when the particles scatter to opposite sides of the original direction.

8 Example 2: equal-mass elastic scattering

An incoming puck of speed

(41)

strikes an identical stationary puck elastically.

After collision, puck 1 moves at angle

(42)

above the original direction.

For equal masses with one initially at rest,

(43)

Thus puck 2 moves at

(44)

below the original direction.

From y-momentum conservation,

(45)

Hence

(46)

From the x equation,

(47)

Substitute the relation above:

(48)

Therefore

(49)

and

(50)

9 Collision geometry and the line of impact

For smooth rigid bodies such as frictionless spheres or disks, the collision impulse acts along the common normal at the contact point.

This direction is called the line of impact.

Define:

• e_n: unit vector normal to the contact surfaces along the line of impact,
• e_t: unit vector tangent to the contact surfaces.

Any velocity can be decomposed as

(51)

10 Smooth-contact assumption

If the contacting surfaces are idealized as smooth and frictionless, the collision impulse has no tangential component:

(52)

Therefore each particle’s tangential velocity component remains unchanged during the short collision:

v1t	= u1t,	(53)
v2t	= u2t.	(54)

Only the normal components change.

This reduces an oblique two-dimensional impact to a one-dimensional collision problem along the normal direction.

11 Momentum along the line of impact

Along the normal direction,

(55)

Along the tangent direction, for smooth bodies,

v1t = u1t,		(56)
v2t = u2t.		(57)

Thus the impact calculation can be performed entirely in the normal direction and then recombined with the unchanged tangential components.

12 Coefficient of restitution in oblique impact

For an oblique collision, the coefficient of restitution applies to the normal relative velocity, not to the entire velocity vector.

The correct relation is

(58)

Equivalently,

(59)

For ordinary passive impacts,

(60)

13 General smooth oblique-collision solution

For the normal components,

m1u1n + m2u2n	= m1v1n + m2v2n,	(61)
v2n − v1n	= e(u1n − u2n).	(62)

This is exactly the same algebra as a one-dimensional collision.

Thus

(63)

and

(64)

Then restore the tangential components:

v1	= v1nen + u1tet,	(65)
v2	= v2nen + u2tet.	(66)

14 Example 3: smooth oblique impact

Two identical pucks collide.

Let the normal direction be the x axis and the tangential direction be the y axis.

Initially,

u1	= (4ex + 3ey) m∕s,	(67)
u2	= (0ex + 1ey) m∕s.	(68)

Let

(69)

For equal masses, the normal restitution formulas give

v1x	= u1x + u2x,	(70)
v2x	= u1x + u2x.	(71)

Thus

v1x	= (4) = 1.0 m∕s,	(72)
v2x	= (4) = 3.0 m∕s.	(73)

Smooth contact leaves the tangential components unchanged:

v1y	= 3.0 m∕s,	(74)
v2y	= 1.0 m∕s.	(75)

Therefore

v1 = (1ex + 3ey) m∕s,		(76)
v2 = (3ex + 1ey) m∕s.		(77)

15 Impulse geometry

For a smooth collision, particle 1 receives an impulse

(78)

and particle 2 receives

(79)

Thus

m1(v1 − u1)	= Jen,	(80)
m2(v2 − u2)	= −Jen.	(81)

Because the impulse has no tangential component,

(82)

for each smooth body.

16 Finding the impulse magnitude

For particle 1,

(83)

Using the restitution solution,

(84)

With the reduced mass

(85)

this becomes

(86)

The sign depends on the chosen normal direction.

17 Energy loss in smooth oblique impact

Only the normal relative motion is affected by a smooth collision.

The normal relative kinetic energy before impact is

(87)

After impact,

(88)

Since the normal relative speed is multiplied by e,

(89)

Therefore

(90)

for the smooth, nonrotating point-mass model.

Tangential kinetic energy is unchanged under the smooth-contact assumption.

18 Center of mass frame in two dimensions

The center of mass velocity is

(91)

Define center of mass-frame velocities:

u1′	= u1 − VCM,	(92)
u2′	= u2 − VCM.	(93)

Then

(94)

Therefore the two incoming momenta are equal and opposite in the center of mass frame.

The same is true after collision.

For an elastic collision,

(95)

and the magnitudes remain unchanged while the directions can rotate.

19 Scattering angle in the center of mass frame

In the center of mass frame, an elastic collision can often be described by a single scattering angle χ.

If the initial momentum is

(96)

and the final momentum is

(97)

then

(98)

For an elastic collision,

(99)

so only the direction changes.

This viewpoint becomes fundamental in scattering theory.

20 When momentum and energy are not enough

In a general two-dimensional elastic collision there may be more unknowns than conservation equations.

For two outgoing velocity vectors, there can be four scalar unknown components.

Momentum conservation gives two equations.

Kinetic-energy conservation gives one more.

A fourth relation may come from:

• known contact geometry,
• a known scattering angle,
• the line-of-impact condition,
• smooth-contact constraints,
• measured direction of one outgoing body.

Conservation laws constrain the solution, but geometry completes it.

21 Rough contact and spin

The smooth-contact model assumes

(100)

If friction acts during impact, there can be a tangential impulse:

(101)

Then:

• tangential translational velocities can change,
• angular velocity can change,
• rotational kinetic energy must be included,
• angular momentum becomes important.

Such rough-body collisions require rigid-body impact mechanics beyond the point-particle model developed here.

22 Example 4: checking a proposed elastic result

Two equal masses undergo an isolated elastic collision.

Initially,

(102)

A proposed result is

v1	= (3ex + 1ey) m∕s,	(103)
v2	= (2ex − 1ey) m∕s.	(104)

Momentum is conserved:

(105)

But the final squared speeds are

v12	= 10,	(106)
v22	= 5.	(107)

Thus

(108)

while initially

(109)

Therefore kinetic energy is not conserved.

The proposed velocities could describe an inelastic collision, but not an elastic one.

23 A reliable workflow for planar collision problems

1. Define the two-body system.
2. Check whether external impulse is negligible.
3. Choose useful axes.
4. Resolve every known velocity into components.
5. Write momentum conservation in each relevant direction.
6. Identify the collision model: sticking, elastic, restitution, smooth oblique impact, or another constraint.
7. Add the required second relation or geometric constraint.
8. Solve the scalar equations.
9. Reconstruct velocity vectors.
10. Check momentum, kinetic energy, restitution, and geometry.

24 Common mistakes

1. Treating momentum conservation as a scalar equation in two dimensions.
2. Using speeds instead of signed components.
3. Conserving kinetic energy for an inelastic collision.
4. Assuming momentum conservation and kinetic-energy conservation uniquely determine every two-dimensional elastic collision.
5. Applying the coefficient of restitution to the full velocity magnitude in an oblique collision.
6. Forgetting that restitution acts along the line of impact.
7. Changing tangential velocity components in an ideal smooth collision.
8. Forgetting that frictional or rough contact can invalidate the smooth-contact assumption.
9. Mixing laboratory-frame and center of mass-frame velocities.
10. Assuming equal masses always leave at right angles without checking the required conditions: elastic collision and one mass initially at rest.
11. Forgetting that external impulse can invalidate the two-body momentum balance.
12. Ignoring rotational motion when the impact is off-center and contact is rough.

25 Practice exercises

1. A 2.0 kg puck moves east at 5.0 m∕s and a 3.0 kg puck moves north at 4.0 m∕s. They stick. Find the final velocity vector, speed, and direction.
2. Derive the vector formula for the final velocity of a perfectly inelastic two-body collision.
3. An equal-mass elastic collision has one puck initially at rest. Prove that the final velocity vectors are perpendicular.
4. In the previous case, one outgoing puck moves at 25^∘ above the initial direction. Find the direction of the other outgoing puck.
5. A 1.0 kg puck moving at 8.0 m∕s hits an identical stationary puck elastically. The first leaves at 45^∘. Find both final speeds.
6. For a smooth collision, show that zero tangential impulse implies unchanged tangential velocity components.
7. Two masses m₁ and m₂ collide obliquely with restitution e. Derive the normal-component formulas.
8. A smooth impact has m₁ = m₂, e = 0.80, u_1n = 5 m∕s, u_2n = 0, u_1t = 2 m∕s, and u_2t = −1 m∕s. Find both final velocity components.
9. Derive the impulse formula

   (110)

10. Derive the kinetic-energy loss for a smooth oblique collision.
11. Show that total momentum is zero in the center of mass frame.
12. Explain why an elastic collision in the center of mass frame changes directions but not momentum magnitudes.
13. Give an example where momentum and kinetic-energy conservation are insufficient to determine a unique two-dimensional collision.
14. Explain why rough contact can create spin.
15. Check whether the proposed final velocities (4, 0) and (1, 0) from an initial equal-mass state (5, 0) and (0, 0) can represent an elastic collision.

26 Summary

For a two-dimensional collision with negligible external impulse,

(111)

Momentum conservation applies componentwise:

Pxi = Pxf,		(112)
Pyi = Pyf.		(113)

For a perfectly inelastic collision,

(114)

For equal masses in an elastic collision with one initially at rest,

(115)

For smooth oblique impact,

v1t = u1t,		(116)
v2t = u2t,		(117)

and restitution applies along the line of impact:

(118)

The center of mass frame makes elastic scattering especially simple because the two particle momenta remain equal and opposite.

The next article develops center of mass motion and system dynamics more systematically.

References

References