MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics

Physics 8.01

IC_W08D2-6 Table Problem: Elastic Two-Dimensional Collision Solution In the laboratory reference frame, an “incident” particle with mass m1 is initially moving with given initial speed

v1,i . The second “target” particle is of mass m2 = m1 and at rest.

After an elastic collision, the first particle moves off at an angle !1, f with respect to the

initial direction of motion of the incident particle with final speed v1, f . Particle two

moves off at an angle !2, f with final speed

v2, f . Find the equations that represent conservation of momentum and energy. Assume no external forces. You do not have to solve these equations for

!1, f , !2, f , and

v2, f .

Solution: Choose a set of positive unit vectors for the initial and final states as shown in figure above. We designate the respective speeds of each of the particles on the momentum flow diagrams. Initial State: The components of the total momentum total

0 1 1,0 2 2,0m m= +p v v! ! ! in the initial state are given by

px ,0

total = m1v1,0 (1)

py ,0

total = 0. (2) Final State: The components of the momentum total

1 1, 2 2,f f fm m= +p v v! ! ! in the final state are given by

px , f

total = m1 v1, f cos!1, f + m1 v2, f cos!2, f (3)

py , f

total = m1 v1, f sin!1, f " m1 v2, f sin!2, f . (4) There are no any external forces acting on the system, so each component of the total momentum remains constant during the collision,

px ,0

total = px , ftotal (5)

py ,0

total = py , ftotal (6)

or using our above results for the component of the momentum, we have that

m1 v1,0 = m1 v1, f cos!1, f + m1 v2, f cos!2, f (7)

0 = m1 v1, f sin!1, f " m1 v2, f sin!2, f . (8)

The collision is elastic; the kinetic energy is the same before and after the collision,

K0

total = K ftotal , (9)

or

12

m1v1,02 =

12

m1v1, f2 +

12

m1v2, f2 . (10)

We have three equations, two momentum equations for each component, Eqs. (7) and (8) and one energy equation Eq. (10), with three unknown quantities,

v1, f , v2, f and

!2, f

since we are already given v1,0 and

!1, f . Although the problem does not specifically ask us to find ,

v1, f , v2, f and

!2, f in terms of

v1,0 and

!1, f , we shall now do so. We first rewrite equations (7) , (8), and (10) canceling the factors of 1m and (1 / 2) , as

v2, f cos!2, f = v1,0 " v1, f cos!1, f (11)

v2, f sin!2, f = v1, f sin!1, f . (12)

v1,0

2 = v1, f2 + v2, f

2 (13) Add the squares of the expressions in Eqs. (11) and (12), yielding

v2, f

2 cos2!2, f + v2, f2 sin2!2, f = (v1,0 " v1, f cos!1, f )2 + v1, f

2 sin2!1, f . (14) We can use the identities

cos2!2, f + sin2!2, f = 1 and

cos2!1, f + sin2!1, f = 1 to simplify

Eq. (14), yielding 2 2 2

2, 1,0 1,0 1, 1, 1,2 cosf f f fv v v v v!= " + . (15) Substituting Eq. (15) into Eq. (13) yields

v1,0

2 = v1, f2 + (v1,0

2 ! 2v1,0 v1, f cos"1, f + v1, f2 ) . (16)

Eq. (16) simplifies to 2

1, 1,0 1, 1,0 2 2 cosf f fv v v != " , (17) which may be solved for the final speed of object 1,

v1, f = v1,0 cos!1, f . (18)

Divide Eq. (12) by Eq. (11), yielding

2, 2, 1, 1,

2, 2, 1,0 1, 1,

sin sincos cosf f f f

f f f f

v vv v v

! !! !

="

. (19)

Eq. (19) simplifies to

1, 1,2,

1,0 1, 1,

sintan

cosf f

ff f

vv v

!!

!=

". (20)

Thus object 2 moves at an angle

!2, f = tan"1

v1, f sin!1, f

v1,0 " v1, f cos!1, f

#

$%

&

'( (21)

We can now substitute Eq. (18) into Eq. (13) to find

v2, f = v1,0

2 ! v1, f2 = v1,0

2 ! (v1,0 cos"1, f ) = v1,0 1! cos2"1, f = v1,0 sin"1, f (22)

choosing the positive square root because

v2, f is the final speed of the target particle. Note that if

v1,0 = 3.0 m ! s"1 and

!1, f = 30! then

v1, f = v1,0 cos!1, f = (3.0 m " s#1)cos30! = 2.6 m " s#1 (23)

!2, f = tan"1v1, f sin!1, f

v1,0 " v1, f cos!1, f

#

$%

&

'(

!2, f = tan"1 (2.6 m ) s"1)sin30!

3.0 m ) s"1 " (2.6 m ) s"1)cos30!#

$%&

'(

= 60!.

(24)

v2, f = v1, 0 sin!1, f = (3.0 m " s#1)sin 30! = 1.5m " s#1 . (25)