Transcript 2D collision - Nassau BOCES

Conservation of Momentum
Conservation of momentum:
m1v1i  m2 v2i  m1v1 f  m2v2 f
Split into components:
m1v1ix  m2v2ix  m1v1 fx  m2v2 fx
m1v1iy  m2v2iy  m1v1 fy  m2v2 fy
If the collision is elastic, we can also use conservation of energy.
Two Methods
Method 1 - Graphical Method – Ruler and protractor.
1.
Calculate the momentum of each object.
2.
Choose an appropriate scale.

Draw each for the momenta before the interaction

Arrange them head to tail

Draw the resultant, the total momentum before
3.
Draw each for the momenta after the interaction

Arrange them head to tail

Draw the resultant, the total momentum after
4. Compare the magnitude and direction of the total momentum vectors before
and after. If they are identical momentum is conserved.
Method 2: Trigonometric Method
1.
2.
3.
4.
5.
Calculate the momentum of each from p = mv.
Use trig calculate the X and Y components of each object before.
Find the x and y components after the collision.

Total the X components before the collision
opp
sin  

Total the X components after the collision
hyp

Set them equal to each other

Total the Y components before the collision
adj
cos  

Total the Y components after the collision
hyp

Set them equal to each other in magnitude and direction
Solve the individual component equations for the unknowns.
opp
tan  
Make a chart and fill out as in the following example
ady
1. Fill out the chart
pxi
pyi
pxf
pyf
Mass 1
Mass 2
Total  p  p
yi
xi
 p xf  p yi
2. Solve the component equations
 pxi   pxf
 p yi   p yf
Two types of 2-D collisions
Type 1 – Glancing collision - One object hits another stationary
object they fly off in separate directions.
Ex: The masses of the green ball and red ball are 2 kg and 3 kg. If the initial velocity
of the green ball is 10m/s. Construct the component equations, don’t solve them.
pxi
pyi
pxf
pyf
Mass
1
Mass
2
Total
 pxi   pxf
 p yi   p yf
After the collision
Before the collision
for green mass
p xi  mvxi
 (2kg )(10 m / s )
m
 20kg
s
p yi  0
For red mass
p xi  0
p yi  0
for red mass
for green mass
pxf  mvrf cos 300
pxf  mvgf cos 600
 (3kg )(v f ) cos 600
 (2kg )(vgf ) cos 600
 1vrf
 1vgf
p yf  mv f sin 300
p yf  mvgf sin 600
 (3kg )vr f sin 600
 (2kg )(vgf ) sin 600
 1.73 vrf
 1.73vgf
Type 2 – Right angle collisions
Type 2 – Right angle collisions
1. Objects approach at 900, one with momentum in the x-direction
the other in the y-direction only.
2. They fly off together at an angle , with some final velocity
Right angle collisions
A 95 kg Physics student running south the hall with a speed of 3m/s
tackles a 90 kg Chemistry student moving west with a speed of 5 m/s.
If the collision is perfectly inelastic, calculate:
1.
The velocity of the students just after the collision
2.
The energy lost as a result of the collision.
Use Pythagorean
Theorem and trig
N
pti 
pxi 2  p yi 2
 (270kgm / s) 2  (285kgm / s) 2
m1=95kg
 72,900  81, 225
V1i=3m/s
 393 kgm / s
V2i=5m/s
W
m2=90kg
E

p yi 285
tan  

 46.50
pxi 270
S
Right angle collisions
An eastern kingbird and a bee approach each other at right angles.
The bird has a mass of .13 kg and a speed of .6 m/s, and the bee has a
mass of 5x10-3 kg and a speed of 15 m/s. If the bird catches the bee,
what’s the new speed of the bird?
Accident investigation. Two cars of
equal mass approach an intersection.
One is traveling east at 29 mi/h (13.0
m/s) the other north with unknown
speed. The vehicles collide and stick
together, leaving skid marks at an angle
of 55º north of east. The second driver
claims he was driving below the speed
limit of 35 mi/h (15.6 m/s).
1. Is he telling the truth?
2. What is the speed of the“combined
vehicles” after the collision?
3. How long are the skid marks (uk = 0.5)
Right angle collisions
13.0 m/s
?? m/s
Glancing Collision
A cue ball traveling at 4m/s makes a glancing, elastic collision with a
target ball of equal mass initially at rest. It is found that the cue ball
is deflected such that it makes an angle of 30 with respect to its
original direction of travel. Set up the conservation of momentum
equations in the x and y directions, don’t solve them.
Glancing Collision
A 2 kg ball is initially at rest on a horizontal frictionless surface. A
4 kg ball moving horizontally with a velocity of 10 m/s hits it. After the
collision the 4 kg ball has a velocity of 6 m/s at an angle of 37 to the
positive x-axis.
1.
Determine the speed and direction of the 2 kg ball after the
collision.
2.
How much K.E. was lost in the collision?
KE1 + KE 2 = KE1 + KE2
 + p2y
p1y + p 2y = p1y
1 2 1
1
mv1  mv12  mv22
2
2
2
 + p2x
p1x + p2x =p1x
0 = m1v1sinθ1 + m2 v2sinθ2 m1v1 = m1v1cosθ1 + m2 v2cosθ2
Two skaters collide and embrace, in a completely inelastic collision.
Thus, they stick together after impact, where the origin is placed at
the point of collision. Alfred, whose mass mA is 83 kg, is originally
moving east with speed vA = 6.2 km/h. Barbara, whose mass mB is 55
kg, is originally moving north with speed vB = 7.8 km/h.