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 .
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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 p0total  m1v1,0  m2 v2,0 in the initial
state are given by
px,total
 m1v1, 0
0
(1)
p y,total
 0.
0
(2)
Final State: The components of the momentum ptotal
 m1v1, f  m2 v 2, f in the final state
f
are given by
px,totalf  m1 v1, f cos1, f  m1 v2, f cos 2, f
(3)
p y,totalf  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,
pxtotal
 pxtotal
,0
,f
(5)
p ytotal
 p ytotal
,0
,f
(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 f total ,
(9)
1
1
1
m1v1,2 0  m1v1,2 f  m1v2,2 f .
2
2
2
(10)
or
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 m1 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)
v1,2 0  v1,2 f  v2,2 f
(13)
Add the squares of the expressions in Eqs. (11) and (12), yielding
v2,2 f cos 2  2, f  v2,2 f sin 2  2, f  (v1, 0  v1, f cos1, f )2  v1,2 f sin 2 1, f .
(14)
We can use the identities cos 2  2, f  sin 2  2, f  1 and cos 2 1, f  sin 2 1, f  1 to simplify
Eq. (14), yielding
2
v2,2 f  v1,0
 2v1,0v1, f cos1, f  v1,2 f .
(15)
Substituting Eq. (15) into Eq. (13) yields
v1,2 0  v1,2 f  (v1,2 0  2v1, 0 v1, f cos1, f  v1,2 f ) .
(16)
0  2v1,2 f  2v1,0v1, f cos1, f ,
(17)
Eq. (16) simplifies to
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
v2, f sin  2, f
v2, f cos2, f

v1, f sin 1, f
.
v1,0  v1, f cos1, f
(19)
Eq. (19) simplifies to
tan  2, f 
v1, f sin 1, f
v1,0  v1, f cos1, f
.
(20)
Thus object 2 moves at an angle

 2, f  tan 1 
v1, f sin 1, f
 v1, 0  v1, f cos1, f
We can now substitute Eq. (18) into Eq. (13) to find



(21)
2
2
v2, f  v1,0
 v1,2 f  v1,0
 (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  30o then
v1, f  v1, 0 cos1, f  (3.0 m  s 1 )cos30o  2.6 m  s 1

 2, f  tan 1 
v1, f sin 1, f
 v1, 0  v1, f cos1, f
(23)





(2.6 m  s 1 )sin30o
1
1
o
 3.0 m  s  (2.6 m  s )cos30 
 2, f  tan 1 
(24)
 60o.
v2, f  v1, 0 sin1, f  (3.0 m  s1 )sin 30o  1.5m  s1 .
(25)