Physics 170 Week 11, Lecture 3
http://www.phas.ubc.ca/∼gordonws/170
Physics 170 203 Week 11, Lecture 3
1
Textbook Chapter 15:
Section 15.4
Physics 170 203 Week 11, Lecture 3
2
Learning Goals:
• We will analyze impact and collisions.
• We will solve some collision problems.
• After this lecture, students should be able to analyze a collision
process using the coefficient of restitution and conservation of
momentum.
Physics 170 203 Week 11, Lecture 3
3
Example:
The two disks A and B have a mass of 3 kg and 5 kg, respectively.
If they collide with the initial velocities shown, determine their
velocities just after impact. The coefficient of restitution is
e = 0.65.
Physics 170 203 Week 11, Lecture 3
4
Impact
Impact occurs when two bodies collide with each other during a
very short period of time, causing relatively large (impulsive) forces
to be exerted between the bodies.
Physics 170 203 Week 11, Lecture 3
5
Five stages of Impact
Approach
Deformation
Z
mA~v = mA~vA1 +
Physics 170 203 Week 11, Lecture 3
Z
P~ , mB ~v = mB ~vB1 −
P~
6
Maximum compression
Restitution
Z
Z
mA~vA2 = mA~v −
Physics 170 203 Week 11, Lecture 3
~ , mB ~vB2 = mB ~v +
R
~
R
7
Rebound
Physics 170 203 Week 11, Lecture 3
8
Kinematics of a collision
All forces are internal, so momentum is conserved,
mA~vA2 + mB ~vB2 = mA~vA1 + mB ~vB1
Consider components parallel to the line of collision.
Let v be the velocity at the instant of maximum deformation.
Z
Z
mA v = mA vA1 + P , mA vA2 = mA v − R
Coefficient of restitution
R
R
R
R
v − vA2
−v + vB2
R
R
e=
=
, e=
=
vA1 − v
−vB1 + v
P
P
eliminate v
e=
Physics 170 203 Week 11, Lecture 3
vB2 − vA2
vA1 − vB1
9
Coefficient of restitution:
e=
vB2 − vA2
vA1 − vB1
Elastic impact e = 1
Plastic impact e = 0
We will analyze collisions where we use the equation for e plus
conservation of momentum to solve, for example, for the final
velocities of particles in terms if initial velocities.
Physics 170 203 Week 11, Lecture 3
10
Example:
If we drop an object from height h, its velocity when it hits the
ground is (using conservation of energy)
p
vA1 = − 2gh
where h is the initial height. It then has a collision with the earth.
The velocity of the earth is
vB1 = 0
After the collision, the object has a velocity vA2 and the earth has
velocity vB2 = 0.
p
p
−vA2
e=
→ v2A = −ev1A = e 2gh = 2gh0
vA1
Height to which the object bounces is h0
q
0
e = hh
Physics 170 203 Week 11, Lecture 3
11
Example:
The two disks A and B have a mass of 3 kg and 5 kg, respectively.
If they collide with the initial velocities shown, determine their
velocities just after impact. The coefficient of restitution is
e = 0.65.
Physics 170 203 Week 11, Lecture 3
12
Strategy:
Choose coordinates where x-axis is horizontal, y-axis vertical.
Motion in y-direction is unaffected by collision
vA2y = 0 , vB2y = −7 sin 60 m/s
For motion in x-direction, use conservation of momentum and
restitution equation.
Physics 170 203 Week 11, Lecture 3
13
Example cont’d:
P
Conservation of momentum (mv)x
mA (6 m/s) − mB (7 m/s) cos 60 = mA vA2x + mB vB2x
Restitution equation:
e = 0.6 =
vB2x − vA2x
vB2x − vA2x
=
vA1x − vB1x
6 + 7 cos 60 m/s
The result is two linear equations
Physics 170 203 Week 11, Lecture 3
14
Example cont’d:
Two linear equations
mA vA2x + mB vB2x = mA (6 m/s) − mB (7 m/s) cos 60
−vA2x + vB2x = e(6 + 7 cos 60 m/s)
Multiply second by mA and add them together,
(mA +mB )vB2x = mA (6 m/s)−mB (7 m/s) cos 60+mA e(6+7 cos 60 m/s)
Multiply the second by mB and subtract it from the first,
(mA +mB )vA2x = mA (6 m/s)−mB (7 m/s) cos 60−mB e(6+7 cos 60 m/s)
Physics 170 203 Week 11, Lecture 3
15
Example cont’d:
We have derived the two equations
vB2x
mA (6 m/s) − mB (7 m/s) cos 60 + mA e(6 + 7 cos 60 m/s)
=
(mA + mB )
vA2x =
mA (6 m/s) − mB (7 m/s) cos 60 − mB e(6 + 7 cos 60 m/s)
(mA + mB )
Plug in the numbers: mA − 3 kg, mB = 5 kg, e = 0.65
vB2x = 2.38 m/s , vB2y = −6.06 m/s , vB = 6.51 m/s
v2Ax = −3.80 m/s , vA2y = 0 , vA = 3.80 m/s
Physics 170 203 Week 11, Lecture 3
16
Example:
Two smooth coins A and B, each having the same mass, slide on a
smooth surface with the motion shown. Determine the speed of
each coin after collision if they move off along the blue paths.
Physics 170 203 Week 11, Lecture 3
17
Strategy:
The line of impact has not been defined.
We will apply the conservation of momentum along the x and y
axes, respectively.
This gives two equations. Since the final angles are specified, we
need to find two quantities, the final speeds. We will solve the two
equations for the conservation of momentum.
Physics 170 203 Week 11, Lecture 3
18
Example cont’d:
Initial momentum is mA~vA + mB ~vB = m(~vA + ~vB ). We have found
that
µ
¶
4
3
~vA1 = (0.5 f t/s) − î − ĵ
5
5
³
´
~vB1 = (0.8 f t/s) − sin 30 î + cos 30 ĵ
Initial momentum=
£
¤
£
¤
4
3
m −(0.5) 5 − (0.8) sin 30 î + m −(0.5) 5 + (0.8) cos 30 ĵ f t/s
Physics 170 203 Week 11, Lecture 3
19
Example cont’d:
Final momentum is mA~vA2 + mB ~vB2 = m(~vA2 + ~vB2 ). We have
found that
³
´
~vA2 = vA2 − sin 45 î + cos 45 ĵ
³
´
~vB2 = vB2 − cos 30 î − sin 30 ĵ
Final momentum
= m [−vA2 sin 45 − vB2 cos 30] î + m [vA2 cos 45 − vB2 sin 30] ĵ
Physics 170 203 Week 11, Lecture 3
20
Example cont’d:
Initial momentum=final momentum
¸
·
¸
·
3
4
m −(0.5) − (0.8) sin 30 î + m −(0.5) + (0.8) cos 30 ĵ f t/s
5
5
= m [−vA2 sin 45 − vB2 cos 30] î + m [vA2 cos 45 − vB2 sin 30] ĵ
m cancels
This results in two linear equations, one equation for each
component.
Physics 170 203 Week 11, Lecture 3
21
Example cont’d:
Initial momentum=final momentum
4
−vA2 sin 45 − vB2 cos 30 = −(0.5) − (0.8) sin 30 f t/s
5
3
vA2 cos 45 − vB2 sin 30 = −(0.5) + (0.8) cos 30 f t/s
5
We can write this as a matrix equation
·
¸·
¸
¸ ·
−vA2 sin 45 −vB2 cos 30
vA2
−(0.5) 45 − (0.8) sin 30
f t/s
=
3
vA2 cos 45 −vB2 sin 30
−(0.5) 5 + (0.8) cos 30
vB2
Physics 170 203 Week 11, Lecture 3
22
Example cont’d:
We need to solve the matrix equation
¸ ·
¸ ·
·
¸
vA2
−(0.5) 54 − (0.8) sin 30
− sin 45 − cos 30
=
f t/s
3
cos 45 − sin 30
vB2
−(0.5) 5 + (0.8) cos 30
·
vA2
vB2
¸
·
− sin 45
=
cos 45
Physics 170 203 Week 11, Lecture 3
− cos 30
− sin 30
¸−1 ·
−(0.5) 45
−(0.5) 35
− (0.8) sin 30
+ (0.8) cos 30
¸
f t/s
23
Example cont’d:
Matrix inversion
·
¸ ·
vA2
− sin 45
=
vB2
cos 45
·
·
vA2
vB2
¸
− cos 30
− sin 30
¸−1 ·
−(0.5) 45 − (0.8) sin 30
−(0.5) 35 + (0.8) cos 30
¸
f t/s
¸
− sin 30 cos 30
¸
·
4
−(0.5) 5 − (0.8) sin 30
− cos 45 − sin 45
=
f t/s
cos 45 cos 30 + sin 45 sin 30 −(0.5) 53 + (0.8) cos 30
Physics 170 203 Week 11, Lecture 3
24
Finding the answer from the Matrix inverse
·
¸
− sin 30 cos 30
·
¸
·
¸
4
vA2
−(0.5)
−
(0.8)
sin
30
− cos 45 − sin 45
5
=
f t/s
3
cos 45 cos 30 + sin 45 sin 30 −(0.5) 5 + (0.8) cos 30
vB2
µ
·
¶
4
1
sin 30 (0.5) + (0.8) sin 30
vA2 =
cos 15
5
µ
¶¸
3
+ cos 30 −(0.5) + (0.8) cos 30
f t/s
5
vA2 = 0.766 f t/s
·
µ
¶
1
4
vB2 =
cos 45 (0.5) + (0.8) sin 30
cos 15
5
¶¸
µ
3
f t/s
− sin 45 −(0.5) + (0.8) cos 30
5
vB2 = 0.298 f t/s
Physics 170 203 Week 11, Lecture 3
25
For the next lecture, please read
Textbook Chapter 15:
Section 15.5-6
Physics 170 203 Week 11, Lecture 3
26