EGR 232 Engineering Dynamics
Lecture 18: Impact and Collisions
Fall 2013
Today:
Homework Due
Test on Monday
Impact and Collisions
Coefficient of Restitution
Read Chap 13, Sections 10 to 11
Homework: Chap. 13, Problems 146, 159, 165, 170
Lecture Outcomes:
Following today's lecture you should be able to
-- understand that momentum is conserved in a collision
-- explain the significance of the coefficient of restitution.
-- explain the difference between elastic and inelastic impact
-- explain the difference between direct and oblique central impact
-- predict the motion of bodies following different kinds of collisions.
Impact: The result of two bodies colliding with each other. Large forces act equally
but opposite directions on the two bodies over a short period of time, during which
momentum is exchanged between the two bodies.
Conservation of Momentum in Impact of Two Bodies:
mA v A  mB v B  mA v'A  mB v'B
(before impact)
(after impact)
Two Types of Impact:
vB
et
Line of
Impact
en
Line of
Impact
vB
vA
vA
Direct Central Impact
Oblique Central Impact
For Direct Central Impact
Velocity and Impact is only directed along the line of impact.
Line of
Impact
vB
'
A
mA v A  mB v B  mA v  mB v
'
B
ˆn  (mAv 'A  mBvB' )e
ˆn
(mAv A  mBvB )e
vA
For Oblique Central Impact:
Velocity and Impact are directed both along
the impact line and perpendicular to it.
mA v A  mB v B  mA v'A  mB v'B
Along line of impact:
ˆn
(mAv A cos  A  mBvB cos  B )e
et
Line of
Impact
θ'B
v'B
ˆn
 (mAv 'A cos  A'  mBv B' cos  B' )e
Perpendicular to line of impact:
Treat the balls separately:
Ball A:
ˆt  (mAv 'A sin  A' )e
ˆt
(mAv A sin  A )e
Ball B:
ˆt  (mBvB' sin B' )e
ˆt
(mBvB sin B )e
en
v'A
vB
θ'A
θA
θB
Momentum is conserved both along the line of impact and perpendicular to it.
Energy Considerations:
While momentum is always conserved in a collision, energy may be lost during the
compression and rebound of the material.
Impacts are often classified by the two extremes associated with energy lost in a
collision.
Elastic Collision: A collision in which no energy is lost. Both momentum and energy
are conserved.
1
1
2
mi v i  mf v f
 mv
mf vf2
i i  
2
2
Inelastic Collision: A collision in which the maximum amount of energy is lost.
In an inelastic collision, the bodies stick together when the move off at the same final
velocity.
1
1
2
mi v i  mf v f
 mv
(m1  m2 )vf2
i i 
2
2
Coefficient of Restitution:
The coefficient of restitution is a way of classifying impact that fall somewhere between
perfectly elastic and completely inelastic collisions.
e
vB'  v 'A
v A  vB
For perfectly elastic collisions:
e=1
For completely inelastic collisions:
e=0
For partially inelastic collisions
0 < e <1
One important point to understand is that the restitution equation is only valid for velocity
components along the line of the impact. For direct impact, this works for the entire
magnitude of the velocities. However for an oblique impact, the restitution equation only
applies to the components along the impact (along en)
v '  v 'An
e  Bn
v An  vBn
Example 1: Two small spheres (6 lb and 2
lb respectively) move toward each other
(velocities of 8 ft/s and 12 ft/s) and collide.
If the coefficient of restitution is 0.8, find
the velocities after impact and determine
the amount of energy lost in the collision.
6 lb
vA=8ft/s
2 lb
vB=12 ft/s
Example 1: Two small spheres (6 lb and 2
lb respectively) move toward each other
(velocities of 8 ft/s and 12 ft/s) and collide.
If the coefficient of restitution is 0.8, find
the velocities after impact and determine
the amount of energy lost in the collision.
6 lb
2 lb
vA=8ft/s
Solution:
Conservation. of Momentum:
Restitution Equation:
vB'  v 'A
v A  vB
e
mAv A  mBvB  mAv 'A  mBvB'
Let to the right be positive:
then
vA = 8 ft/s
vB = -12 ft
'
A
WAv A  WBvB WAv  WBv

g
g
WA = 6 lb
'
B
WB = 2 lb
e
'
A
6(8ft / s)  2(12m / s)  6v  2v
'
B
6v 'A  2v B'  24
'
B
e = 0.8
'
A
v v
v A  vB
vB'  v 'A
0.8 
8  (12)
'
vB  v 'A  16
Solving for the velocities:
6v 'A  2(v 'A  16)  24

v 'A  1ft / s
and
v B'  15ft / s
Energy before collision:
:
1
1
Tbefore  mAv 2A  mBvB2
2
2
1
6lb
1
2lb
Tbefore 
(8ft / s)2 
(12ft / s)2
2
2
2 (32.2ft / s )
2 (32.2ft / s )
Tbefore  5.963  4.472  10.43 ft  lb
Energy after collision
1
1
Tafter  mAv 'A2  mBvB' 2
2
2
1
6lb
1
2lb
Tafter 
(1ft / s)2 
(15ft / s)2
2
2 (32.2ft / s )
2 (32.2ft / s2 )
Tafter  0.093  6.988  7.081 ft  lb
T  Tbefore  Tafter  10.43  7.081  3.349 ft  lb
vB=12 ft/s
Example 2: The two disks of the same mass collide as shown in the figure below.
If the velocity of disk A is 10 in/s and the velocity of disk B is 16 in/s, determine the
velocities (magnitude and direction) of the disks following the collision.
Coefficient of restitution is 0.75.
Solution:
This is an oblique collision:
Assume positive direction is
up to the right along the
impact line.
Along the line of the impact:
apply conservation of momentum
m v A cos 20o  m(v B cos 20o )  m v 'A cos  A'  m v B' cos B'
'
10 cos 20o  16 cos 20o  v 'An  vBn
'
v 'An  v Bn
 5.64
Apply the restitution equation along the impact line.
'
vBn
 v 'An
'
e
e(v An  vBn )  vBn
 v 'An

v An  vBn
'
e(v A cos 20o  vB cos 20o )  vBn
 v 'An
'
0.75(10 cos 20o  (16 cos 20o ))  vBn
 v 'An
'
vBn
 v 'An  18.3
Combine to solve:
v 'An  (v 'An  18.3)  5.64
2v 'An  23.94
v 'An  11.97 in / s
'
v Bn
 6.33 in / s
Along the tangent direction, there is no change of velocity:
Let up and to the left be the positive direction:
v At  v 'At  v A sin(20o )  10 sin20o  3.4 in / s
'
vBt  vBt
 vB sin20o  16 sin20o  5.47 in / s
v'Bn
For Body A:
Magnitude:
v
'
A
 v
' 2
An
v'Bn
v
' 2
At
2
2
 11.97  3.4  12.44 in / s
Angle from line of impact after collision:
 3.4 
 15.85o

 11.97 
 A'  arc tan 
v'An
θ’A
therefore:
v 'A  12.44 in / s @ 35.85o
v'An
For Body B:
Magnitude:
'2
'2
vB'  vBn
 vBt
 6.332  5.472  8.37 in / s
Angle from line of impact after collision:
 5.47 
 40.83o

 6.33 
B'  arc tan 
vB'  8.37 in / s
@
60.83o
v'An
θ’B
v'Bn
Example 3:
A sphere rebounds as shown after striking an inclined plane with a vertical velocity of 5
m/s. Knowing that α = 30 degrees and e = 0.8 between the sphere and the plane,
determine the height, h, reached by the sphere.
Solution:
This problem is a collision problem and a projectile motion problem.
For the Collision: Momentum is conserved along a line normal to the plane of impact.
Momentum along the plane of the collision
v0y= v0 cos α
does not change during the collision:
α vO
'
m v ox  m v o cos   m v ox
'
v ox
 v o sin 
Momentum normal to plane is subject to
the restitution equation:
v '  v 'An
e  Bn
v An  vBn
where van = van' = 0
'
'
v Oy
0
v Oy
e 

0  v Oy
v O cos 
vO'
v0x= v0 sin α
v0y'
α
v0x'
v0y'
'
vOy
 e vO cos 
Therefore the velocity after the collision can be expressed as
'
'
'
'
'
'
v o'  v ox
 v ox
 (v ox
cos  iˆ  v ox
sin  jˆ)  (v oy
sin  iˆ  v oy
cos  jˆ)
'
'
'
'
 (v ox
cos   v oy
sin  )iˆ  (v ox
sin   v oy
cos  ) jˆ
v o'  (2.5 cos 30o  3.46 sin30o )iˆ  (2.5 sin30o  3.46 cos 30o ) jˆ
 3.895iˆ  1.746 jˆ
Applying the projectile motion equation in the y direction to the maximum point:
y  y0 
v y 2  v y 02
2a

α
v0x'
for vo = 5 and α = 30 degrees: then
'
'
and vOy
v ox
 5 sin30o  2.5
 e vO cos   0.8(5)cos30o  3.46
and
v y 2  v y 02  2a(y  y 0 ) 
α
02  1.7462
 0.155m
2(9.81)
Max Height
EGR232 Dynamics: Homework Set 18
Problem 13:146
Fall 2013
At an intersection car B was traveling south and car A was traveling 30 degrees north of
east when they slammed into each other. Upon investigation it was found that after the
crash the two cars got tuck and skidded off at an angle of 10 degrees north of east. Each
driver claimed that he was going at the speed limit of 50 km/h and that he tried to slow
down but couldn’t avoid the crash because the other driver was going a lot faster.
Knowing that the masses of cars A and B were 1500 kg and 1200 kg, respectively,
determine a) which car was going faster, b) the speed of the faster of the two cars if the
slower car was traveling at the speed limit.
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EGR232 Dynamics: Homework Set 18
Fall 2013
Problem 13.159
Two identical cars A and B are at rest on a loading dock with brakes released. Car C, of a
slightly different style but of the same weight, has been pushed by dockworkers and hits
car B with a velocity of 1.5 m/s. Knowing that the coefficient of restitution is 0.8 between
B and C and 0.5 between A and B, determine the velocity of each car after all collision
have taken place.
_______________________________________________________________________
Solution:
EGR232 Dynamics: Homework Set 18
Fall 2013
Problem 13.165
A 600 g ball is moving with a velocity of magnitude 6 m/s when it is hit as shown by a 1 kg ball B
which has a velocity of magnitude 4 m/s. Knowing that the coefficient of restitution is 0.8 and
assuming no friction, determine the velocity of each ball after impact.
-------------------------------------------------------------------------------------------------------------------
EGR232 Dynamics: Homework Set 18
Fall 2013
Problem 13:170
A girl throws a ball at an inclined wall from a height of 1.2 m, hitting the wall at A with a
horizontal velocity v0 of magnitude 15 m/s. Knowing that the coefficient of restitution between the
ball and the wall is 0.9 and neglecting friction, determine the distance d from the foot of the wall to
the point B where the ball will hit the ground after bouncing off the wall.
----------------------------------------------------------------------------------------------------------Solution:
1.2 m