Momentum and Energy in 2D
Name__________________________________________ Date ________
Explosions:
Nuclear decay
P3
M3
3
M1
P1
Large unstable nucleus
with no initial momentum
M2
2
P2
Conservation of Momentum:
If two particles are emitted at 900 to each
other then momentum vectors form a
P3
P2
3
2
P1
Note: If particles are not emitted at 900
Then must use Law of Sines:
P1 / sin(1800-(1 + 2)) = P2 / sin3 = P3 / sin2
Or Law of Cosines
P32 = P22 + P12 - 2P1P2cos2
P22 = P32 + P12 - 2P3P1cos3
P12 = P32 + P22 - 2P3P2cos(1800-(1 + 2))
_____________________________, so
Collisions:
V1f
1
M1
V1i
M2
2
V2f
Conservation of Momentum:
P1f
P2f
1
2
P1i
If the collision is elastic, the target particle is
initially stationary and the masses of the
colliding particles are the same, as in the game
of pool:
m1 = m2 = m
1/2m1v1i2 = 1/2m1v1f2 + 1/2m2v2f2
so,
1/2mv1i2 = 1/2mv1f2 + 1/2mv2f2
v1i2 = v1f2 + v2f2
Velocity vectors must form a _____________
_____________________________________
According to the ________________________
_____________________________________
Examples:
An unstable nucleus of mass 17 x 10-27 kg, initially at rest, disintegrates into three
particles. One of the particles, of mass 5.0 x 10-27 kg, moves along the positive yaxis with a speed of 6.0 x 106 m/s. Another particle of mass 8.4 x 10-27 kg, moves
along the positive x-axis with a speed of 4.0 x 106 m/s. Determine the third
particle’s speed and direction of motion. (Assume that mass is conserved)
Write down what you know
mnucl = 17 x 10-27 kg m1 = 5.0 x 10-27 kg m2 = 8.4 x 10-27 kg
m3 = mnucl – (m1 + m2) = 17 x 10-27 kg – (5.0 x 10-27 kg + 8.4 x 10-27 kg) = 3.6 x 10-27 kg
vnuci = 0 m/s v1 = 6.0 x 106 m/s @ y-axis v2 = 4.0 x 106 m/s @ x-axis v3 = ?
Draw a vector diagram of the momentum
P2
3
P1
P3
Solve for the momentum of the third particle and then find its velocity
Right angled triangle so use Pythagorean Theorem
P12 + P22 = P32 therefore:  (P12 + P22) = P3
 ((m1v1)2 + (m2v2)2) = P3
 ((5.0 x 10 –27 kg)(6.0x106m/s))2 + (8.4 x 10-27kg)(4.0 x 106 m/s))2 = P3
(2.0 x 10-39 kg2m2/s2) = 4.50 x 10-20 kgm/s = P3
v3 = P3 / m3 = 4.50 x 10-20 kgm/s / 3.6 x 10-27 kg = 12.5 x 106 m/s
Use trig to determine the angle relative to the positive x-direction (going counter
clockwise)
3 = tan-1 (P1 / P2 ) = tan-1 (m1v1 / m2v2)
= tan-1 ((5.0 x 10 –27 kg)(6.0x106m/s) / (8.4 x 10-27kg)(4.0 x 106 m/s))
= tan-1 (3 x 10-20 kgm/s / 3.36 x 10-20 kg/m/s)
= 41.760
Therefore:  = 1800 + 3 = 1800 + 41.760 = 221.80
A particle with a speed v1 = 2.64 x 106 m/s makes a glancing elastic collision with
another particle that is at rest. Both particles have the same mass. After the
collision the struck particle moves off at 100 to v1. What is the speed of each
particle after the collision and what angle is the incident particle scattered at?
Write down what you know
v1 = 2.64 x 106 m/s
v2 = 0 m/s
m1 = m2 = m
2 = 100
v1f = ? v2f = ?
Draw a vector diagram of the situation (note: masses are the same so can use velocity
vectors rather than momentum)
v1f
v2f
1
2
v1
Use trig to determine the speed of each particle after the collision
v1f = v1sin2 = (2.64 x 106 m/s) sin100 = 4.58 x 105 m/s
v2f = v1cos2 = (2.64 x 106 m/s) cos100 = 2.6 x 106 m/s
Determine the scattering angle of the incident particle using geometry
1 = 900 - 2 = 900 – 100 = 800 wrt the incident particle direction